# 3.3: Surface Integrals

Proving the volume and the surface area of a sphere by using integrals
Proving the volume and the surface area of a sphere by using integrals

3.3: Surface Integrals

Page ID
91906

We are now going to define two types of integrals over surfaces.

• Integrals that look like $$\iint_{S} \rho\,\text{d}S$$ are used to compute the area and, when $$\rho$$ is, for example, a mass density, the mass of the surface $$S\text{.}$$
• Integrals that look like $$\iint_{S} \vecs{F} \cdot\hat{\textbf{n}}\,\text{d}S\text{,}$$ with $$\hat{\textbf{n}}(x,y,z)$$ being a unit vector that is perpendicular to $$S$$ at $$(x,y,z)\text{,}$$ are called flux integrals. We shall see in §3.4, that when $$\vecs{v}$$ is the velocity field of a moving fluid and $$\rho$$ is the density of the fluid, then $$\iint_{S} \rho\vecs{v} \cdot\hat{\textbf{n}}\,\text{d}S$$ is the rate at which fluid is crossing the surface $$S\text{.}$$

## Parametrized Surfaces

Suppose that we wish to integrate over part, $$S\text{,}$$ of a surface that is parametrized by $$\vecs{r} (u,v)\text{.}$$ We start by cutting $$S$$ up into small pieces by drawing a bunch of curves of constant $$u$$ (the blue curves in the figure below) and a bunch of curves of constant $$v$$ (the red curves in the figure below).

Concentrate on any one the small pieces. Here is a greatly magnified sketch.

For example, the lower red curve was constructed by holding $$v$$ fixed at the value $$v_0\text{,}$$ varying $$u$$ and sketching $$\vecs{r} (u,v_0)\text{,}$$ and the upper red curve was constructed by holding $$v$$ fixed at the slightly larger value $$v_0+\text{d}v\text{,}$$ varying $$u$$ and sketching $$\vecs{r} (u,v_0+\text{d}v)\text{.}$$ So the four intersection points in the figure are

\begin{alignat*}{2} P_2&=\vecs{r} (u_0, v_0+\text{d}v) &\qquad P_3&=\vecs{r} (u_0+\text{d}u, v_0+\text{d}v)\\ P_0&=\vecs{r} (u_0, v_0) & P_1&=\vecs{r} (u_0+\text{d}u, v_0) \end{alignat*}

Now if

$\textbf{R}(t) = \vecs{r} (u_0+t\text{d}U\,,\,v_0+t\text{d}V) \nonumber$

(where $$\text{d}U$$ and $$\text{d}V$$ are any small constants) then, by Taylor expansion,

\begin{align*} \vecs{r} \big(u_0+\text{d}U\,,\,v_0+\text{d}V\big) &=\textbf{R}(1)\\ &\approx \big[\textbf{R}(0) +\textbf{R}'(0)\,\big(t-0\big)\big]_{t=1}\\ &=\vecs{r} (u_0\,,\,v_0) +\frac{\partial\vecs{r} }{\partial u}(u_0\,,\,v_0)\,\text{d}U +\frac{\partial\vecs{r} }{\partial v}(u_0\,,\,v_0)\,\text{d}V \end{align*}

Applying this three times, once with $$\text{d}U=\text{d}u\text{,}$$ $$\text{d}V=0\text{,}$$ once with $$\text{d}U=0$$ $$\text{d}V=\text{d}v\text{,}$$ and once with $$\text{d}U=\text{d}u\text{,}$$ $$\text{d}V=\text{d}v\text{,}$$

\begin{alignat*}{2} P_0&=\vecs{r} (u_0\,,\,v_0)\\ P_1&=\vecs{r} (u_0+\text{d}u, v_0) &&\approx \vecs{r} (u_0\,,\,v_0) +\frac{\partial\vecs{r} }{\partial u}(u_0\,,\,v_0)\,\text{d}u\\ P_2&=\vecs{r} (u_0, v_0+\text{d}v) &&\approx \vecs{r} (u_0\,,\,v_0) +\frac{\partial\vecs{r} }{\partial v}(u_0\,,\,v_0)\,\text{d}v\\ P_3&=\vecs{r} (u_0+\text{d}u, v_0+\text{d}v) &&\approx\vecs{r} (u_0\,,\,v_0) +\frac{\partial\vecs{r} }{\partial u}(u_0\,,\,v_0)\,\text{d}u +\frac{\partial\vecs{r} }{\partial v}(u_0\,,\,v_0)\,\text{d}v \end{alignat*}

We have dropped all Taylor expansion terms that are of degree two or higher in $$\text{d}u\text{,}$$ $$\text{d}v\text{.}$$ The reason is that, in defining the integral, we take the limit $$\text{d}u,\text{d}v\rightarrow 0\text{.}$$ Because of that limit, all of the dropped terms contribute exactly $$0$$ to the integral. We shall not prove this. But we shall show, in the optional §3.3.5, why this is the case.

The small piece of our surface with corners $$P_0\text{,}$$ $$P_1\text{,}$$ $$P_2\text{,}$$ $$P_3$$ is approximately a parallelogram with sides

\begin{align*} \overrightarrow{P_0P_1} \approx \overrightarrow{P_2P_3} &\approx \frac{\partial\vecs{r} }{\partial u}(u_0\,,\,v_0)\,\text{d}u\\ \overrightarrow{P_0P_2} \approx \overrightarrow{P_1P_3} &\approx \frac{\partial\vecs{r} }{\partial v}(u_0\,,\,v_0)\,\text{d}v \end{align*}

Denote by $$\theta$$ the angle between the vectors $$\overrightarrow{P_0P_1}$$ and $$\overrightarrow{P_0P_2}\text{.}$$ The base of the parallelogram, $$\overrightarrow{P_0P_1}\text{,}$$ has length $$\big|\overrightarrow{P_0P_1}\big|\text{,}$$ and the height of the parallelogram is $$\big|\overrightarrow{P_0P_2}\big|\,\sin\theta\text{.}$$ So the area of the parallelogram is 1

\begin{align*} |\overrightarrow{P_0P_1}|\ |\overrightarrow{P_0P_2}| \ \sin\theta &= \big|\overrightarrow{P_0P_1}\times\overrightarrow{P_0P_2}\big|\\ &\approx \bigg|\frac{\partial\vecs{r} }{\partial u}(u_0\,,\,v_0)\times \frac{\partial\vecs{r} }{\partial v}(u_0\,,\,v_0)\bigg| \text{d}u\text{d}v \end{align*}

Furthermore, $$\frac{\partial\vecs{r} }{\partial u}(u_0\,,\,v_0)$$ and $$\frac{\partial\vecs{r} }{\partial v}(u_0\,,\,v_0)$$ are tangent vectors to the curves $$\vecs{r} (t\,,\,v_0)$$ and $$\vecs{r} (u_0\,,\,t)$$ respectively. Both of these curves lie in $$S\text{.}$$ So $$\frac{\partial\vecs{r} }{\partial u}(u_0\,,\,v_0)$$ and $$\frac{\partial\vecs{r} }{\partial v}(u_0\,,\,v_0)$$ are tangent vectors to $$S$$ at $$(u_0,v_0)$$ and the cross product $$\frac{\partial\vecs{r} }{\partial u}(u_0\,,\,v_0)\times \frac{\partial\vecs{r} }{\partial v}(u_0\,,\,v_0)$$ is perpendicular to $$S$$ at $$(u_0,v_0)\text{.}$$ We have found both $$\text{d}S$$ and $$\hat{\textbf{n}}\,\text{d}S\text{,}$$ where $$\hat{\textbf{n}}$$ is a unit normal vector to the surface.

For the parametrized surface $$\vecs{r} (u,v)\text{,}$$

\begin{align*} \hat{\textbf{n}}\, \text{d}S & = \pm\frac{\partial\vecs{r} }{\partial u}(u\,,\,v)\times \frac{\partial\vecs{r} }{\partial v}(u\,,\,v)\ \text{d}u\text{d}v\\ \text{d}S&= \bigg|\frac{\partial\vecs{r} }{\partial u}(u\,,\,v)\times \frac{\partial\vecs{r} }{\partial v}(u\,,\,v)\bigg|\ \text{d}u\text{d}v \end{align*}

The $$\pm$$ sign in 3.3.1 is there because there are two unit normal vectors at each point of a surface, one on each side of the surface. Typically, the application itself tells you which of the two normal vectors should be used. We shall see many examples shortly.

## Graphs

The surface which is the graph $$z = f(x,y)$$ can be parametrized by

$\vecs{r} (x,y) = x\,\hat{\pmb{\imath}} + y\,\hat{\pmb{\jmath}} + f(x,y)\,\hat{\mathbf{k}} \nonumber$

As

$\begin{gather*} \frac{\partial\vecs{r} }{\partial x} = \hat{\pmb{\imath}} + \frac{\partial f}{\partial x}\,\hat{\mathbf{k}} \qquad\text{and}\qquad \frac{\partial\vecs{r} }{\partial y} = \hat{\pmb{\jmath}} + \frac{\partial f}{\partial y}\,\hat{\mathbf{k}} \end{gather*}$

we have

$\frac{\partial\vecs{r} }{\partial x}\times \frac{\partial\vecs{r} }{\partial y} =\det\left[\begin{matrix} \hat{\pmb{\imath}} & \hat{\pmb{\jmath}} & \hat{\mathbf{k}} \\ 1 & 0 & \frac{\partial f}{\partial x} \\ 0 & 1 & \frac{\partial f}{\partial y} \end{matrix}\right] = -f_x(x,y)\,\hat{\pmb{\imath}} – f_y(x,y)\,\hat{\pmb{\jmath}} + \hat{\mathbf{k}} \nonumber$

So, 3.3.1 gives the following.

For the surface $$z=f(x,y)\text{,}$$

\begin{align*} \hat{\textbf{n}}\, \text{d}S & = \pm\big[-f_x(x,y)\,\hat{\pmb{\imath}} – f_y(x,y)\,\hat{\pmb{\jmath}} + \hat{\mathbf{k}}\big]\ \text{d}x\text{d}y\\ \text{d}S&= \sqrt{1 + f_x(x,y)^2 + f_y(x,y)^2}\ \text{d}x\text{d}y \end{align*}

Similarly, for the surface $$x=g(y,z)\text{,}$$

\begin{align*} \hat{\textbf{n}}\, \text{d}S & = \pm\big[\hat{\pmb{\imath}} -g_y(y,z)\,\hat{\pmb{\jmath}} – g_z(y,z)\,\hat{\mathbf{k}}\big]\ \text{d}y\text{d}z\\ \text{d}S&= \sqrt{1 + g_y(y,z)^2 + g_z(y,z)^2}\ \text{d}y\text{d}z \end{align*}

and for the surface $$y=h(x,z)\text{,}$$

\begin{align*} \hat{\textbf{n}}\, \text{d}S & = \pm\big[-h_x(x,z)\,\hat{\pmb{\imath}} +\hat{\pmb{\jmath}} – h_z(x,z)\,\hat{\mathbf{k}}\big]\ \text{d}x\text{d}z\\ \text{d}S&= \sqrt{1 + h_x(x,z)^2 + h_z(x,z)^2}\ \text{d}x\text{d}z \end{align*}

Again, in any given application, some care must be taken in choosing the sign in 3.3.2, so as to get the appropriate normal vector.

The formulae like $$\text{d}S= \sqrt{1 + f_x(x,y)^2 + f_y(x,y)^2}\ \text{d}x\text{d}y$$ in 3.3.2 have geometric interpretations. The red parallelogram in the sketch

represents a little piece of our surface. It has area $$\text{d}S= \sqrt{1 + f_x(x,y)^2 + f_y(x,y)^2}\ \text{d}x\text{d}y\text{.}$$ The blue parallelogram in the same sketch represents the projection of the red parallelogram onto the $$xy$$-plane. It has area $$\text{d}x\text{d}y\text{.}$$ The vector $$\hat{\textbf{n}}$$ in the sketch is a unit normal for the red parallelogram. We have seen that it is parallel to

$\frac{\partial\vecs{r} }{\partial x}\times \frac{\partial\vecs{r} }{\partial y} = -f_x(x,y)\,\hat{\pmb{\imath}} – f_y(x,y)\,\hat{\pmb{\jmath}} + \hat{\mathbf{k}} \nonumber$

so that the angle $$\theta$$ between $$\hat{\textbf{n}}$$ and $$\hat{\mathbf{k}}$$ obeys

\begin{align*} \cos\theta &= \frac{(-f_x(x,y)\,\hat{\pmb{\imath}} – f_y(x,y)\,\hat{\pmb{\jmath}} + \hat{\mathbf{k}})\cdot\hat{\mathbf{k}}} {\big|-f_x(x,y)\,\hat{\pmb{\imath}} – f_y(x,y)\,\hat{\pmb{\jmath}} + \hat{\mathbf{k}}\big|\,|\hat{\mathbf{k}}|}\\ &=\frac{1}{\sqrt{1 + f_x(x,y)^2 + f_y(x,y)^2}} \end{align*}

The geometric interpretation of $$\text{d}S= \sqrt{1 + f_x(x,y)^2 + f_y(x,y)^2}\ \text{d}x\text{d}y$$ is that the area $$\text{d}S$$ of a little piece of surface is the area of its projection on the $$xy$$-plane times the factor $$\frac{1}{\cos\theta}$$ where $$\theta$$ is the angle between $$\hat{\textbf{n}}$$ (which is perpendicular to the surface) and $$\hat{\mathbf{k}}$$ (which is perpendicular to the $$xy$$-plane). Notice that

• when $$\theta$$ is close to zero, which corresponds the $$f$$ being almost constant and our surface being almost parallel to the $$xy$$-plane, $$\text{d}S$$ reduces to almost $$\text{d}x\text{d}y\text{.}$$
• On the other hand, in the limit $$\theta\rightarrow\frac{\pi}{2}\text{,}$$ which corresponds to $$f_x$$ and/or $$f_y$$ becoming infinite and our surface becoming perpendicular to the $$xy$$-plane, $$\text{d}S$$ becomes “infinity times” $$\text{d}x\text{d}y\text{.}$$ In this case, we should represent our surface either in the form $$x=g(y,z)$$ or in the form $$y=h(x,z)\text{,}$$ rather than in the form $$z=f(x,y)\text{.}$$

## Surfaces Given by Implicit Equations

Finally suppose that the surface is given by the equation $$G(x,y,z)=K\text{,}$$ with $$K$$ a constant. Suppose further that at some point on the surface $$\frac{\partial G}{\partial z} \ne 0\text{.}$$ Then near that point we may solve 2 the equation $$G(x,y,z)=K$$ for $$z$$ as a function of $$x$$ and $$y\text{.}$$ That is, the surface also obeys $$z=f(x,z)$$ for a function $$f(x,y)$$ that satisfies

$G\big(x,y,f(x,y)\big) = K \nonumber$

near the point. Differentiating this with respect to $$x$$ and $$y$$ gives, by the chain rule,

\begin{alignat*}{2} 0&=\frac{\partial}{\partial x}\Big[G\big(x,y,f(x,y)\big)\Big] &\ =\ G_x\big(x,y,f(x,y)\big) + G_z\big(x,y,f(x,y)\big)\,f_x(x,y)\\ 0&=\frac{\partial}{\partial y}\Big[G\big(x,y,f(x,y)\big)\Big] &\ =\ G_y\big(x,y,f(x,y)\big) + G_z\big(x,y,f(x,y)\big)\,f_y(x,y) \end{alignat*}

which implies

$\begin{gather*} f_x(x,y) = -\frac{G_x\big(x,y,f(x,y)\big)}{G_z\big(x,y,f(x,y)\big)} \qquad f_y(x,y) = -\frac{G_y\big(x,y,f(x,y)\big)}{G_z\big(x,y,f(x,y)\big)} \end{gather*}$

and

\begin{align*} -f_x(x,y)\,\hat{\pmb{\imath}} – f_y(x,y)\,\hat{\pmb{\jmath}} + \hat{\mathbf{k}} &=\frac{G_x\big(x,y,f(x,y)\big)}{G_z\big(x,y,f(x,y)\big)}\,\hat{\pmb{\imath}}+ \frac{G_y\big(x,y,f(x,y)\big)}{G_z\big(x,y,f(x,y)\big)}\,\hat{\pmb{\jmath}}+\hat{\mathbf{k}}\\ &=\frac{\vecs{ \nabla} G\big(x,y,f(x,y)\big)}{G_z\big(x,y,f(x,y)\big)} \end{align*}

So, by (3.3.1),

For the surface $$G(x,y,z)=K\text{,}$$ when $$G_z(x,y,z)\ne 0\text{,}$$

\begin{align*} \hat{\textbf{n}}\, \text{d}S & = \pm\frac{\vecs{ \nabla} G\big(x,y,z\big)}{\vecs{ \nabla} G\big(x,y,z\big)\cdot\hat{\mathbf{k}}}\ \text{d}x\text{d}y\\ \text{d}S&= \bigg|\frac{\vecs{ \nabla} G\big(x,y,z\big)}{\vecs{ \nabla} G\big(x,y,z\big)\cdot\hat{\mathbf{k}}}\bigg|\ \text{d}x\text{d}y \end{align*}

Similarly, for the surface $$G(x,y,z)=K\text{,}$$ when $$G_x(x,y,z)\ne 0\text{,}$$

\begin{align*} \hat{\textbf{n}}\, \text{d}S & = \pm\frac{\vecs{ \nabla} G\big(x,y,z\big)}{\vecs{ \nabla} G\big(x,y,z\big)\cdot\hat{\pmb{\imath}}}\ \text{d}y\text{d}z\\ \text{d}S&= \bigg|\frac{\vecs{ \nabla} G\big(x,y,z\big)}{\vecs{ \nabla} G\big(x,y,z\big)\cdot\hat{\pmb{\imath}}}\bigg|\ \text{d}y\text{d}z \end{align*}

and for the surface $$G(x,y,z)=K\text{,}$$ when $$G_y(x,y,z)\ne 0\text{,}$$

\begin{align*} \hat{\textbf{n}}\, \text{d}S & = \pm\frac{\vecs{ \nabla} G\big(x,y,z\big)}{\vecs{ \nabla} G\big(x,y,z\big)\cdot\hat{\pmb{\jmath}}}\ \text{d}x\text{d}z\\ \text{d}S&= \bigg|\frac{\vecs{ \nabla} G\big(x,y,z\big)}{\vecs{ \nabla} G\big(x,y,z\big)\cdot\hat{\pmb{\jmath}}}\bigg|\ \text{d}x\text{d}z \end{align*}

If, for some point $$(x_0,y_0,z_0)$$ we have $$G_x(x_0,y_0,z_0)= G_y(x_0,y_0,z_0)= G_z(x_0,y_0,z_0)= 0\text{,}$$ we also have problem! Often this is a sign that our surface is not smooth at $$(x_0,y_0,z_0)$$ and in fact does not have a normal vector there. For an example of this, see Example 3.2.2.

## Examples of $$\iint_{S} \rho\,\text{d}S$$

We’ll start by computing, in several different ways, the surface area of the hemisphere

$x^2+y^2+z^2=a^2\qquad z\ge 0 \nonumber$

(with $$a \gt 0$$). You probably know, from high school, that the answer is $$\frac{1}{2}\times 4\pi a^2=2\pi a^2\text{.}$$ But you have probably not seen a derivation of this answer. Note that, since $$x^2+y^2 = a^2-z^2$$ on the hemisphere, the set of $$(x,y)$$’s for which there is a $$z$$ with $$(x,y,z)$$ on the hemisphere is exactly $$\left \{(x,y)\in\mathbb{R}^2|x^2+y^2\le a^2 \right \} \text{.}$$

Let’s parametrize the hemisphere $$x^2+y^2+z^2=a^2\text{,}$$ $$z\ge 0\text{,}$$ using as parameters the polar coordinates $$r\text{,}$$ $$\theta$$ of the cylindrical coordinates 3

\begin{align*} x &= r\cos\theta\\ y &= r\sin\theta\\ z &= z \end{align*}

and then apply 3.3.1. In cylindrical coordinates the equation $$x^2+y^2+z^2=a^2$$ becomes $$r^2+z^2=a^2\text{,}$$ and the condition $$x^2+y^2\le a^2$$ is $$0\le r\le a\text{,}$$ $$0\le\theta \lt 2\pi\text{.}$$

So the hemisphere can be parametrized by

\begin{align*} &\big(x(r,\theta)\,,\, y(r,\theta)\,,\, z(r,\theta)\big) =\big(r\cos\theta\,,\,r\sin\theta\,,\,\sqrt{a^2-r^2}\,\big)\\ &\hskip1in\text{with }\ 0\le r\le a,\ 0\le\theta \lt 2\pi \end{align*}

Note that we selected the positive solution $$z=\sqrt{a^2-r^2}$$ of $$r^2+z^2=a^2$$ in order to satisfy the condition that $$z\ge 0\text{.}$$ Since

\begin{align*} \Big(\frac{\partial x}{\partial r}\,,\,\frac{\partial y}{\partial r}\,,\, \frac{\partial z}{\partial r}\Big) &=\Big(\cos\theta\,,\,\sin\theta\,,\,-\frac{r}{\sqrt{a^2-r^2}}\Big)\\ \Big(\frac{\partial x}{\partial\theta}\,,\,\frac{\partial y}{\partial\theta} \,,\, \frac{\partial z}{\partial\theta}\Big) &=(-r\sin\theta\,,\,r\cos\theta\,,\,0) \end{align*}

3.3.1 yields

\begin{align*} \hat{\textbf{n}}\,\text{d}S &=\pm\Big(\frac{\partial x}{\partial r}\,,\,\frac{\partial y}{\partial r} \,,\,\frac{\partial z}{\partial r}\Big) \times \Big(\frac{\partial x}{\partial\theta}\,,\,\frac{\partial y}{\partial\theta} \,,\, \frac{\partial z}{\partial\theta}\Big) \ \text{d}r\text{d}\theta\\ &=\pm \det\left[\begin{matrix} \hat{\pmb{\imath}} & \hat{\pmb{\jmath}} & \hat{\mathbf{k}} \\ \cos\theta & \sin\theta & -\frac{r}{\sqrt{a^2-r^2}} \\ -r\sin\theta & r\cos\theta & 0 \end{matrix}\right]\ \text{d}r\text{d}\theta\\ &=\pm\Big(\frac{r^2\cos\theta}{\sqrt{a^2-r^2}}\,,\, \frac{r^2\sin\theta}{\sqrt{a^2-r^2}}\,,\, r\Big)\ \text{d}r\text{d}\theta\\ \text{d}S&=\sqrt{ \frac{r^4}{a^2-r^2} +r^2}\ \text{d}r\text{d}\theta =\sqrt{ \frac{a^2r^2}{a^2-r^2}}\ \text{d}r\text{d}\theta =\frac{ar}{\sqrt{a^2-r^2}}\ \text{d}r\text{d}\theta \end{align*}

So the area of the hemisphere is

\begin{align*} \int_0^a \text{d}r\int_0^{2\pi}\text{d}\theta\ \frac{ar}{\sqrt{a^2-r^2}} &=2\pi a\int_0^a \text{d}r\ \frac{r}{\sqrt{a^2-r^2}}\\ &=2\pi a\int_{a^2}^0 \frac{-\text{d}u/2}{\sqrt{u}}\\ &\hskip1in\text{with } u=a^2-r^2,\ \text{d}u = -2r\,\text{d}r\\ &=2\pi a\Big[-\sqrt{u}\Big]_{a^2}^0\\ &=2\pi a^2 \end{align*}

as it should be.

This time we’ll compute the area of the hemisphere by using that, if $$(x,y,z)$$ is on the hemisphere, then $$G(x,y,z) = a^2$$ with $$G(x,y,z) = x^2 + y^2 + z^2\text{.}$$ Since

$\vecs{ \nabla} G(x,y,z) = \big(2x\,,\,2y\,,\,2z\big) \nonumber$

3.3.3 yields

\begin{align*} \text{d}S &= \bigg|\frac{\vecs{ \nabla} G\big(x,y,z\big)}{\vecs{ \nabla} G\big(x,y,z\big)\cdot\hat{\mathbf{k}}}\bigg| \ \text{d}x\text{d}y\\ &= \bigg|\frac{\big(2x\,,\,2y\,,\,2z\big)}{2z}\bigg| \ \text{d}x\text{d}y\\ &= \frac{\sqrt{x^2+y^2+z^2}}{|z|} \ \text{d}x\text{d}y\\ &= \frac{a}{\sqrt{a^2-x^2-y^2}} \ \text{d}x\text{d}y \qquad\text{on } x^2+y^2+z^2=a^2 \end{align*}

So the area is $$\iint_{x^2+y^2\le a^2}\frac{a}{\sqrt{a^2-x^2-y^2}} \ \text{d}x\text{d}y \text{.}$$ To evaluate this integral, we switch to polar coordinates, substituting $$x=r\cos\theta\text{,}$$ $$y=r\sin\theta\text{.}$$ This gives

\begin{align*} \text{area} &=\iint_{x^2+y^2\le a^2}\frac{a}{\sqrt{a^2-x^2-y^2}} \ \text{d}x\text{d}y =\int_0^a\text{d}r\ r\int_0^{2\pi}\text{d}\theta\ \frac{a}{\sqrt{a^2-r^2}}\\ &=2\pi a \int_0^a\text{d}r\ \frac{r}{\sqrt{a^2-r^2}} \end{align*}

We already showed, in Example 3.3.4, that the value of this integral is $$2\pi a^2\text{.}$$

Of course “integrating over a sphere” cries out for spherical coordinates. So this time we parametrize the hemisphere $$x^2+y^2+z^2=a^2\text{,}$$ $$z\ge 0\text{,}$$ using as parameters the angular coordinates $$\theta\text{,}$$ $$\varphi$$ of the spherical coordinates

\begin{align*} x&=\rho\sin\varphi\cos\theta\\ y&=\rho\sin\varphi\sin\theta\\ z&=\rho\cos\varphi \end{align*}

and then apply 3.3.1. In spherical coordinates the equation $$x^2+y^2+z^2=a^2$$ becomes just $$\rho^2=a^2\text{,}$$ and the condition $$z\ge 0$$ is $$0\le\varphi\le\frac{\pi}{2}\text{,}$$ $$0\le\theta \lt 2\pi\text{.}$$ So the hemisphere can be parametrized 4 by

\begin{align*} &\big(x(\theta,\varphi)\,,\, y(\theta,\varphi)\,,\, z(\theta,\varphi)\big) =\big(a\sin\varphi\cos\theta\,,\,a\sin\varphi\sin\theta\,,\,a\cos\varphi\,\big)\\ & 0\le\varphi\le\frac{\pi}{2},\ 0\le\theta \lt 2\pi \end{align*}

Since

\begin{align*} \Big(\frac{\partial x}{\partial\theta}\,,\, \frac{\partial y}{\partial\theta}\,,\, \frac{\partial z}{\partial\theta}\Big) &=\big(-a\sin\varphi\sin\theta\,,\, a\sin\varphi\cos\theta\,,\,0\big)\\ \Big(\frac{\partial x}{\partial\varphi}\,,\,\frac{\partial y}{\partial\varphi} \,,\, \frac{\partial z}{\partial\varphi}\Big) &=(a\cos\varphi\cos\theta\,,\,a\cos\varphi\sin\theta\,,\,-a\sin\varphi) \end{align*}

3.3.1 yields

\begin{align*} &\hat{\textbf{n}}\,\text{d}S =\pm\Big(\frac{\partial x}{\partial\theta}\,,\, \frac{\partial y}{\partial\theta}\,,\, \frac{\partial z}{\partial\theta}\Big) \times \Big(\frac{\partial x}{\partial\varphi}\,,\,\frac{\partial y}{\partial\varphi} \,,\, \frac{\partial z}{\partial\varphi}\Big) \ \text{d}\theta\text{d}\varphi\\ &=\pm \big(\!-a\sin\varphi\sin\theta, a\sin\varphi\cos\theta,0\big) \!\times\! (a\cos\varphi\cos\theta,a\cos\varphi\sin\theta,-a\sin\varphi) \,\text{d}\theta\text{d}\varphi\\ &=\pm\big(-a^2\sin^2\varphi\cos\theta\,,\, -a^2\sin^2\varphi\sin\theta\,,\, -a^2\sin\varphi\cos\varphi\Big)\ \text{d}\theta\text{d}\varphi\\ &=\mp a^2\sin\varphi \big(\sin\varphi\cos\theta\,,\, \sin\varphi\sin\theta\,,\, \cos\varphi\Big)\ \text{d}\theta\text{d}\varphi \end{align*}

and

\begin{align*} \text{d}S&=a^2\sin\varphi \sqrt{\sin^2\varphi\cos^2\theta +\sin^2\varphi\sin^2\theta +\cos^2\varphi}\ \text{d}\theta\text{d}\varphi\\ &=a^2\sin\varphi\ \text{d}\theta\text{d}\varphi \end{align*}

So the area of the hemisphere is

\begin{align*} a^2\int_0^{\frac{\pi}{2}} \text{d}\varphi \int_0^{2\pi}\text{d}\theta\ \sin\varphi &=2\pi a^2 \int_0^{\frac{\pi}{2}} \text{d}\varphi \ \sin\varphi =2\pi a^2\Big[-\cos\varphi\Big]_0^{\pi/2}\\ &=2\pi a^2 \end{align*}

There is an easier way to do this, using a little geometry.

We are now going to again compute the surface area of the hemisphere using spherical coordinates. But this time instead of determining $$\text{d}S$$ using the canned formula 3.3.1, we are going to read it off of a sketch.

Sketch the part of the hemisphere that is in the first octant, $$x\ge 0\text{,}$$ $$y\ge 0\text{,}$$ $$z\ge 0\text{.}$$ Slice it up into small pieces by drawing in curves of constant $$\theta$$ (the blue lines in the figure below) and curves of constant $$\varphi$$ (the red lines in the figure below).

Each piece is approximately a little rectangle. Concentrate on one of them, like the piece with the thick sides in the figure above. The area, $$\text{d}S\text{,}$$ of that piece is (essentially) the product of its height and its width. Each of the two sides of the piece is

• a segment of a circle of radius $$a$$ (a fat blue line in both the figure above and in the figure on the left below)
• that subtends an angle $$\text{d}\varphi$$
• and hence is the fraction $$\frac{\text{d}\varphi}{2\pi}$$ of a full circle of radius $$a$$ and hence is of length $$\frac{\text{d}\varphi}{2\pi} 2\pi a = a\text{d}\varphi\text{.}$$

The top of the piece is

• a segment of a circle of radius $$a\sin\varphi$$ (a fat red line in both the figure above and in the figure on the right below)
• that subtends an angle $$\text{d}\theta$$
• and hence is the fraction $$\frac{\text{d}\theta}{2\pi}$$ of a full circle of radius $$a\sin\varphi$$ and hence is of length $$\frac{\text{d}\theta}{2\pi} 2\pi a\sin\varphi = a\sin\varphi\text{d}\theta\text{.}$$

These are drawn in the figure below.

So the area of our piece is

$\text{d}S = \big(a\text{d}\varphi\big)\big(a\sin\varphi\text{d}\theta\big) = a^2\sin\varphi\,\text{d}\theta\text{d}\varphi \nonumber$

This is exactly the same formula that we found for $$\text{d}S$$ in Example 3.3.6 so that we will, yet again, get that the area of a hemisphere of radius $$a$$ is $$2\pi a^2\text{.}$$ (Phew!)

But wait! We can do it again, by yet another method!

We’ll compute the area of the hemisphere one last time 5. This time we’ll use that the equation of the hemisphere is

$z=f(x,y) = \sqrt{a^2-x^2-y^2} \qquad\text{with (x,y) running over x^2+y^2\le a^2} \nonumber$

So 3.3.2 yields

\begin{align*} \text{d}S&= \sqrt{1 + f_x(x,y)^2 + f_y(x,y)^2}\ \text{d}x\text{d}y\\ &=\sqrt{1+\Big(\frac{-x}{\sqrt{a^2-x^2-y^2}}\Big)^2 +\Big(\frac{-y}{\sqrt{a^2-x^2-y^2}}\Big)^2} \ \text{d}x\text{d}y\\ &=\sqrt{1+\frac{x^2+y^2}{a^2-x^2-y^2}} \ \text{d}x\text{d}y\\ &=\sqrt{\frac{a^2}{a^2-x^2-y^2}} \ \text{d}x\text{d}y \end{align*}

So the area is $$\iint_{x^2+y^2\le a^2}\frac{a}{\sqrt{a^2-x^2-y^2}} \ \text{d}x\text{d}y \text{.}$$ We already found, in Example 3.3.5, that the value of this integral in $$2\pi a^2\text{.}$$

Let’s do some more substantial examples, where the integrand is not 1.

Evaluate $$\ \iint_S x^2y^2z^2\ dS\$$ where $$S$$ is the part of the cone $$x^2+y^2=z^2$$ with $$0\le z\le 1\text{.}$$

Solution 1

We can express $$S$$ as

$z=f(x,y) = \sqrt{x^2+y^2}\qquad x^2+y^2\le 1 \nonumber$

Now since

$f_x(x,y) = \frac{x}{\sqrt{x^2+y^2}}\qquad f_y(x,y) = \frac{y}{\sqrt{x^2+y^2}} \nonumber$

3.3.2 gives 6

$\text{d}S = \Big[1 + \frac{x^2}{x^2+y^2} + \frac{y^2}{x^2+y^2}\Big]^{1/2} \ \text{d}x\text{d}y =\sqrt{2}\,\text{d}x\text{d}y \nonumber$

Our integral is then

\begin{align*} \iint_S x^2y^2z^2\ dS &=\sqrt{2}\iint_{x^2+y^2\le 1} x^2 y^2 (x^2+y^2)\ \text{d}x\text{d}y \end{align*}

Since we are integrating over a circular domain, let’s convert to polar coordinates.

\begin{align*} \iint_S x^2y^2z^2\ dS &=\sqrt{2}\int_0^{2\pi}\text{d}\theta\int_0^1 \text{d}r \ r(r\cos\theta)^2(r\sin\theta)^2 r^2\\ &=\sqrt{2} \left[\int_0^{2\pi}\text{d}\theta\ \cos^2\theta \sin^2\theta\right] \left[\int_0^1 \text{d}r \ r^7\right]\\ &=\frac{\sqrt{2}}{8} \int_0^{2\pi}\text{d}\theta\ \cos^2\theta \sin^2\theta =\frac{\sqrt{2}}{32} \int_0^{2\pi}\text{d}\theta\ \sin^2(2\theta)\\ &=\frac{\sqrt{2}}{64} \int_0^{2\pi}\text{d}\theta\ \big[1-\cos(4\theta)\big] \end{align*}

Remembering 7 that the integral of $$\cos(\theta)\text{,}$$ or $$\cos(4\theta)\text{,}$$ over a full period is $$0\text{,}$$ we end up with

$\iint_S x^2y^2z^2\ dS = \frac{\sqrt{2}}{64}(2\pi) =\frac{\pi\sqrt{2}}{32} \nonumber$

Solution 2

We may parametrize 8 the cone by

\begin{align*} \vecs{r} (z,\theta) &= z\cos\theta\,\hat{\pmb{\imath}} + z\sin\theta\,\hat{\pmb{\jmath}} + z\,\hat{\mathbf{k}} \qquad 0\le z\le 1,\ 0\le\theta\le 2\pi \end{align*}

Then because

$\begin{gather*} \frac{\partial\vecs{r} }{\partial z} = \cos\theta\,\hat{\pmb{\imath}} + \sin\theta\,\hat{\pmb{\jmath}} + \hat{\mathbf{k}} \qquad\text{and}\qquad \frac{\partial\vecs{r} }{\partial\theta} = -z\sin\theta\,\hat{\pmb{\imath}} + z\cos\theta\,\hat{\pmb{\jmath}} \end{gather*}$

3.3.1 yields 9

\begin{align*} \hat{\textbf{n}}\,\text{d}S &=\pm\det\left[\begin{matrix} \hat{\pmb{\imath}} & \hat{\pmb{\jmath}} & \hat{\mathbf{k}} \\ \cos\theta & \sin\theta & 1 \\ -z\sin\theta & z\cos\theta & 0 \end{matrix}\right]\,\text{d}z\text{d}\theta\\ &=\pm\big[-z\cos\theta\,\hat{\pmb{\imath}} – z\sin\theta\,\hat{\pmb{\jmath}} + z\,\hat{\mathbf{k}}\big]\ \text{d}z\text{d}\theta\\ \text{d}S&= \sqrt{2} z\ \text{d}z\text{d}\theta \end{align*}

So our integral becomes

\begin{align*} \iint_S x^2y^2z^2\ dS &= \sqrt{2}\int_0^{2\pi}\text{d}\theta \int_0^1\text{d}z\ z (z\cos\theta)^2 (z\sin\theta)^2 z^2\\ &= \sqrt{2}\int_0^{2\pi}\text{d}\theta \int_0^1\text{d}z\ z^7 \cos^2\theta \sin^2\theta\\ &=\frac{\sqrt{2}}{8} \int_0^{2\pi}\text{d}\theta\ \cos^2\theta \sin^2\theta \end{align*}

We evaluated this integral in Solution 1. So again

$\iint_S x^2y^2z^2\ dS =\frac{\pi\sqrt{2}}{32} \nonumber$

Let’s do something more celestial.

Consider a spherical shell of radius $$a$$ with mass density $$\mu$$ per unit area. Think of it as a hollow planet 10. We are going to determine the gravitational force that it exerts on a particle of mass $$m$$ a distance $$b$$ away from its centre. This particle can be either outside the shell ($$b \gt a$$) or inside the shell ($$b \lt a$$). We can choose the coordinate system so that the centre of the shell is at the origin and the particle is at $$(0,0,b)\text{.}$$ By Newton’s law of gravitation, the force exerted on the particle by a tiny piece of the shell of surface area $$\text{d}S$$ located at $$\vecs{r}$$ is

$\frac{G\,(\mu\text{d}S)\,m}{|\vecs{r} -(0,0,b)|^3}(\vecs{r} -(0,0,b)) \nonumber$

Here $$G$$ is the gravitational constant, $$\mu\text{d}S$$ is the mass of the tiny piece of shell, $$m$$ is the mass of the particle

and $$\vecs{r} -(0,0,b)$$ is the vector from the particle to the piece of shell. If we work in spherical coordinates, as we did in Example 3.3.6,

\begin{align*} \text{d}S &= a^2\sin\varphi\,\text{d}\varphi\text{d}\theta \end{align*}

and

\begin{align*} \vecs{r} &= a\sin\varphi\cos\theta\,\hat{\pmb{\imath}} + a\sin\varphi\sin\theta\,\hat{\pmb{\jmath}} + a\cos\varphi\,\hat{\mathbf{k}}\\ \vecs{r} -(0,0,b) &= a\sin\varphi\cos\theta\,\hat{\pmb{\imath}} + a\sin\varphi\sin\theta\,\hat{\pmb{\jmath}} + (a\cos\varphi-b)\,\hat{\mathbf{k}}\\ |\vecs{r} -(0,0,b)|^2 &= a^2+b^2 -2ab\cos\varphi \end{align*}

The total force is then

$\begin{gather*} \vecs{F} = G\mu m a^2\int_0^\pi\!\!\text{d}\varphi \int_0^{2\pi}\!\!\text{d}\theta\ \sin\varphi\ \frac{a\sin\varphi\cos\theta\,\hat{\pmb{\imath}} \!+\! a\sin\varphi\sin\theta\,\hat{\pmb{\jmath}} \!+\! (a\cos\varphi\!-\!b)\,\hat{\mathbf{k}}} {\big[a^2+b^2 -2ab\cos\varphi\big]^{3/2}} \end{gather*}$

Note for future reference that the square root in $$[a^2+b^2 -2ab\cos\varphi]^{3/2}$$ is the positive square root because $$[b^2+a^2 -2ab\cos\varphi]^{1/2}$$ is the length of $$\vecs{r} -(0,0,b)\text{,}$$ which is positive.

This integral is a little different than other integrals that we have encountered so far in that the integrand is a vector. By definition 11,

$\iint_S \big[G_1\,\hat{\pmb{\imath}} + G_2\,\hat{\pmb{\jmath}} + G_3\,\hat{\mathbf{k}} \big]\ \text{d}S =\hat{\pmb{\imath}} \iint_S G_1\ \text{d}S +\hat{\pmb{\jmath}} \iint_S G_2\ \text{d}S +\hat{\mathbf{k}} \iint_S G_3\ \text{d}S \nonumber$

so we just have to compute the three components separately.

In our case, the $$\hat{\pmb{\imath}}$$ and $$\hat{\pmb{\jmath}}$$ components

\begin{align*} \vecs{F} \cdot\hat{\pmb{\imath}} & = G\mu m a^2 \int_0^\pi\text{d}\varphi \left[ \sin\varphi\ \frac{a\sin\varphi} {\big[a^2+b^2 -2ab\cos\varphi\big]^{3/2}} \int_0^{2\pi}\text{d}\theta\ \cos\theta \right]\\ \vecs{F} \cdot\hat{\pmb{\jmath}} & = G\mu m a^2\int_0^\pi\text{d}\varphi\left[ \sin\varphi\ \frac{a\sin\varphi} {\big[a^2+b^2 -2ab\cos\varphi\big]^{3/2}} \int_0^{2\pi}\text{d}\theta\ \sin\theta \right] \end{align*}

are both zero 12 because $$\int_0^{2\pi}\cos\theta\ \text{d}\theta =\int_0^{2\pi}\sin\theta\ \text{d}\theta =0$$ so that

\begin{align*} \vecs{F} &= G\mu m a^2 \hat{\mathbf{k}}\int_0^\pi\text{d}\varphi \int_0^{2\pi}\text{d}\theta\ \sin\varphi\ \frac{a\cos\varphi-b} {\big[a^2+b^2 -2ab\cos\varphi\big]^{3/2}}\\ &= 2\pi G\mu m a^2 \hat{\mathbf{k}}\int_0^\pi\text{d}\varphi\ \sin\varphi\ \frac{a\cos\varphi-b} {\big[a^2+b^2 -2ab\cos\varphi\big]^{3/2}} \end{align*}

To evaluate this integral we substitute

$\begin{gather*} u=a^2+b^2 -2ab\cos\varphi\qquad \text{d}u = 2ab\sin\varphi\,\text{d}\varphi\qquad \cos\varphi = \frac{a^2+b^2-u}{2ab} \end{gather*}$

When $$\varphi=0\text{,}$$ $$u=(a-b)^2$$ and when $$\varphi =\pi\text{,}$$ $$u=(a+b)^2\text{,}$$ so

\begin{align*} \vecs{F} &= \frac{\pi G\mu m a}{b} \hat{\mathbf{k}}\int_{(a-b)^2}^{(a+b)^2} \text{d}u\ \frac{\frac{a^2+b^2-u}{2b}-b} {u^{3/2}}\\ &= \frac{\pi G\mu m a}{b} \hat{\mathbf{k}}\int_{(a-b)^2}^{(a+b)^2} \text{d}u\ \frac{\frac{a^2-b^2-u}{2b}} {u^{3/2}}\\ &= \frac{\pi G\mu m a}{b} \hat{\mathbf{k}}\Big[ \Big(\frac{a^2-b^2}{2b}\Big)\frac{u^{-1/2}}{-1/2} -\Big(\frac{1}{2b}\Big)\frac{u^{1/2}}{1/2} \Big]_{(a-b)^2}^{(a+b)^2} \end{align*}

Recalling that $$u^{1/2}$$ is the positive square root,

\begin{align*} \vecs{F} &=\frac{\pi G\mu m a}{b} \hat{\mathbf{k}}\Big[ \Big(\frac{b^2-a^2}{b}\Big)\frac{1}{a+b} -\frac{a+b}{b} -\Big(\frac{b^2-a^2}{b}\Big)\frac{1}{|a-b|} +\frac{|a-b|}{b} \Big] \end{align*}

If $$b \gt a\text{,}$$ so that $$|a-b|=b-a$$

\begin{align*} \vecs{F} &=\frac{\pi G\mu m a}{b} \hat{\mathbf{k}}\Big[ \frac{b-a}{b} -\frac{a+b}{b} -\frac{a+b}{b} +\frac{b-a}{b} \Big] =-\frac{G(4\pi a^2\mu)m}{b^2} \hat{\mathbf{k}} \end{align*}

If $$b \lt a\text{,}$$ so that $$|a-b|=a-b$$

\begin{align*} \vecs{F} &=\frac{\pi G\mu m a}{b} \hat{\mathbf{k}}\Big[ \frac{b-a}{b} -\frac{a+b}{b} +\frac{a+b}{b} +\frac{a-b}{b} \Big] =0 \end{align*}

The moral 13 is

• if the particle is inside the shell, it feels no gravitational force at all, and
• if the particle is outside the shell, it feels the same gravitational force as it would if the entire mass of the shell ($$4\pi a^2\mu$$) were concentrated at the centre of the shell.

The “Gravity Train” 14 refers to the following curious, though admittedly not very practical, thought experiment.

• Pretend that the Earth is a perfect sphere of radius $$R$$ and that it has a constant mass density $$\rho\text{.}$$
• Pick any two distinct points on the surface of the Earth. Call them $$V$$ and $$M\text{.}$$
• Bore a tunnel straight through the Earth from $$V$$ to $$M\text{.}$$
• Place a train in the tunnel at $$V\text{.}$$ Assume that the only forces acting on the train are gravity, $$\textbf{G}\text{,}$$ and a normal force, $$\textbf{N}\text{,}$$ that the tunnel imposes on the train to keep it in the tunnel. In particular, there are no frictional forces, like air resistance, and the train does not have an engine. Release the train and assume that it does not melt as it passes through the centre of the Earth.

What happens?

We’ll simplify our analysis of the motion of the train by picking a convenient coordinate system.

• First translate our coordinate system so that the centre of the Earth, call it $$O\text{,}$$ is at the origin, $$(0,0,0)\text{.}$$
• Then rotate our coordinate system about the origin so that the origin, $$V$$ and $$M$$ all lie in the $$xz$$-plane.
• Then rotate our coordinate system about the $$y$$-axis so that $$V$$ and $$M$$ have the same $$z$$-coordinate $$Z\ge 0\text{.}$$ So the coordinates of $$V$$ and $$M$$ are $$\big(\pm\sqrt{R^2-Z^2}\,,\,0\,,Z\big)\text{.}$$ Let’s suppose that $$V$$ is at $$\big(\sqrt{R^2-Z^2}\,,\,0\,,Z\big)$$ and $$M$$ is at $$\big(-\sqrt{R^2-Z^2}\,,\,0\,,Z\big)\text{.}$$ It really doesn’t matter which is which, but we can always arrange that it is $$V$$ at $$\big(+\sqrt{R^2-Z^2}\,,\,0\,,Z\big)$$ by rotating around the $$z$$-axis by $$180^\circ$$ if necessary.

The $$y$$- and $$z$$-coordinates of the train are always fixed at $$0$$ and $$Z\text{,}$$ respectively. So let’s call the $$x$$-coordinate at time $$t$$ $$x(t)\text{,}$$ and look at the $$x$$-component of Newton’s law of motion.

$\begin{gather*} m\textbf{a} = \textbf{G} +\textbf{N} \end{gather*}$

It is

$m x”(t) = \textbf{G}\cdot\hat{\pmb{\imath}} \nonumber$

because the normal force $$\textbf{N}$$ has no $$\hat{\pmb{\imath}}$$ component. Recall that Newton’s law of gravity says that

$\textbf{G} = -\frac{GMm}{|\vecs{r} |^3}\vecs{r} \nonumber$

where $$G$$ is the gravitational constant, $$\vecs{r}$$ is the vector from $$O$$ to the train, and $$m$$ is the mass of the train. In this case, because of our computation in Example 3.3.10, the train only feels gravity from shells of the Earth that are inside the train, so that $$M$$ is the mass of the

part of the Earth whose distance to the centre of the Earth is no more than $$|\vecs{r} |\text{.}$$ So

$M = \frac{4}{3}\pi |\vecs{r} |^3\rho \nonumber$

and

$mx”(t) = -\frac{Gm}{|\vecs{r} |^3}\ \frac{4}{3}\pi |\vecs{r} |^3\rho\ \vecs{r} \cdot\hat{\pmb{\imath}} \nonumber$

so that

$x”(t) + \frac{4\pi G\rho}{3} x(t) = 0 \nonumber$

This is exactly the differential equation of simple harmonic motion. We have seen it before in Example 2.2.7. Except for the constant $$\frac{4\pi G\rho}{3}\text{,}$$ it is identical to the equation solved in Example A.9.4 of the Appendix A.9, entitled “Review of Linear Ordinary Differential Equations”. The general solution is

$x(t) = C_1 \cos\left(\sqrt{\frac{4\pi G\rho}{3}}\ t\right) + C_2 \sin\left(\sqrt{\frac{4\pi G\rho}{3}}\ t\right) \nonumber$

with $$C_1$$ and $$C_2$$ being arbitrary constants. If we release the train, from rest, at $$t=0\text{,}$$ then $$x(0) = \sqrt{R^2-Z^2}$$ and $$x'(0)=0$$ so that $$C_1= \sqrt{R^2-Z^2}\text{,}$$ $$C_2=0$$ and

$x(t) = \sqrt{R^2-Z^2} \cos\left(\sqrt{\frac{4\pi G\rho}{3}}\ t\right) \nonumber$

The train reaches $$M$$ when $$x(t) = -\sqrt{R^2-Z^2}\text{.}$$ That is, when $$\cos\left(\sqrt{\frac{4\pi G\rho}{3}}\ t\right)=-1\text{.}$$ So the transit time, $$T\text{,}$$ from $$V$$ to $$M$$ obeys

$\sqrt{\frac{4\pi G\rho}{3}}\ T=\pi \implies T= \pi \sqrt{\frac{3}{4\pi G\rho}} = \sqrt{\frac{3\pi}{4 G\rho}} \nonumber$

Notice that this transit time depends only on the gravitational constant $$G$$ and the density of the Earth $$\rho\text{.}$$ In particular it is completely independent of

• where $$V$$ and $$M$$ are and, in particular,
• how close together $$V$$ and $$M$$ are, and also of
• the radius of the Earth.

In the case of the Earth, the transit time is about 42 minutes.

## Optional — Dropping Higher Order Terms in $$\text{d}u,\text{d}v$$

In the course of deriving 3.3.1, that is, $$\hat{\textbf{n}}\text{d}S$$ and $$\text{d}S$$ formulae for

we approximated, for example, the vectors

\begin{alignat*}{2} \overrightarrow{P_0P_1} &=\vecs{r} (u_0+\text{d}u, v_0) -\vecs{r} (u_0\,,\,v_0) &= \frac{\partial\vecs{r} }{\partial u}(u_0\,,\,v_0)\,\text{d}u + E_1 &\approx \frac{\partial\vecs{r} }{\partial u}(u_0\,,\,v_0)\,\text{d}u\\ \overrightarrow{P_0P_2} &=\vecs{r} (u_0, v_0+\text{d}v)-\vecs{r} (u_0\,,\,v_0) &= \frac{\partial\vecs{r} }{\partial v}(u_0\,,\,v_0)\,\text{d}v + E_2 &\approx \frac{\partial\vecs{r} }{\partial v}(u_0\,,\,v_0)\,\text{d}v \end{alignat*}

where $$E_1$$ is bounded 15 by a constant times $$\text{d}u^2$$ and $$E_2$$ is bounded by a constant times $$\text{d}v^2\text{.}$$ That is, we assumed that we could just drop $$E_1$$ and $$E_2\text{.}$$

So we approximated

\begin{align*} \big|\overrightarrow{P_0P_1}\times\overrightarrow{P_0P_2}\big| &=\Big|\Big[\frac{\partial\vecs{r} }{\partial u}(u_0\,,\,v_0)\,\text{d}u + E_1\Big] \times\Big[\frac{\partial\vecs{r} }{\partial v}(u_0\,,\,v_0)\,\text{d}v + E_2\Big] \Big|\\ &=\Big|\frac{\partial\vecs{r} }{\partial u}(u_0\,,\,v_0)\,\text{d}u \times\frac{\partial\vecs{r} }{\partial v}(u_0\,,\,v_0)\,\text{d}v + E_3 \Big|\\ &\approx \Big|\frac{\partial\vecs{r} }{\partial u}(u_0\,,\,v_0)\,\text{d}u \times\frac{\partial\vecs{r} }{\partial v}(u_0\,,\,v_0)\,\text{d}v \Big| \end{align*}

where the length of the vector $$E_3$$ is bounded by a constant times $$\text{d}u^2\,\text{d}v+\text{d}u\,\text{d}v^2\text{.}$$ We’ll now see why dropping terms like $$E_3$$ does not change the value of the integral at all 16.

Suppose that our domain of integration consists of all $$(u,v)$$’s in a rectangle of width $$A$$ and height $$B\text{,}$$ as in the figure below.

Subdivide the rectangle into a grid of $$n\times n$$ small subrectangles by drawing lines of constant $$v$$ (the red lines in the figure) and lines of constant $$v$$ (the blue lines in the figure). Each subrectangle has width $$\text{d}u = \frac{A}{n}$$ and height $$\text{d}v = \frac{B}{n}\text{.}$$ Now suppose that in setting up the integral we make, for each subrectangle, an error that is bounded by some constant times

$\text{d}u^2\,\text{d}v+\text{d}u\,\text{d}v^2 =\Big(\frac{A}{n}\Big)^2 \frac{B}{n} + \frac{A}{n}\Big(\frac{B}{n}\Big)^2 =\frac{AB(A+B)}{n^3} \nonumber$

Because there are a total of $$n^2$$ subrectangles, the total error that we have introduced, for all of these subrectangles, is no larger than a constant times

$n^2 \times \frac{AB(A+B)}{n^3} = \frac{AB(A+B)}{n} \nonumber$

When we define our integral by taking the limit $$n\rightarrow 0$$ of the Riemann sums, this error converges to exactly $$0\text{.}$$

## Exercises

### Stage 1

Let $$0 \lt \theta \lt \frac{\pi}{2}\text{,}$$ and $$a,b \gt 0\text{.}$$ Denote by $$S$$ the part of the surface $$z=y\,\tan\theta$$ with $$0\le x\le a\text{,}$$ $$0\le y\le b\text{.}$$

1. Find the surface area of $$S$$ without using any calculus.
2. Find the surface area of $$S$$ by using (3.3.2).

Let $$a,b,c \gt 0\text{.}$$ Denote by $$S$$ the triangle with vertices $$(a,0,0)\text{,}$$ $$(0,b,0)$$ and $$(0,0,c)\text{.}$$

1. Find the surface area of $$S$$ in three different ways, each using (3.3.2).
2. Denote by $$T_{xy}$$ the projection of $$S$$ onto the $$xy$$-plane. (It is the triangle with vertices $$(0,0,0)$$ $$(a,0,0)$$ and $$(0,b,0)\text{.}$$) Similarly use $$T_{xz}$$ to denote the projection of $$S$$ onto the $$xz$$-plane and $$T_{yz}$$ to denote the projection of $$S$$ onto the $$yz$$-plane. Show that

$\text{Area}(S) =\sqrt{\text{Area}(T_{xy})^2 +\text{Area}(T_{xz})^2 +\text{Area}(T_{yz})^2 } \nonumber$

Let $$a,h \gt 0\text{.}$$ Denote by $$S$$ the part of the cylinder $$x^2+z^2=a^2$$ with $$x\ge 0\text{,}$$ $$0\le y\le h$$ and $$z\ge 0\text{.}$$

1. Find the surface area of $$S$$ without using any calculus.
2. Parametrize $$S$$ by

$\vecs{r} (\theta,y) = a\,\cos\theta\ \hat{\pmb{\imath}} +y\,\hat{\pmb{\jmath}} +a\sin\theta\ \hat{\mathbf{k}}\quad 0\le\theta\le\frac{\pi}{2},\ 0\le y\le h \nonumber$

Find the surface area of $$S$$ by using (3.3.1).

### Stage 2

Let $$S$$ be the part of the surface $$z = xy$$ lying inside the cylinder $$x^2 + y^2 = 3\text{.}$$ Find the moment of inertia of $$S$$ about the $$z$$-axis, that is,

$I = \iint_S (x^2 + y^2)\ \text{d}S \nonumber$

Find the surface area of the part of the paraboloid $$z = a^2 – x^2 – y^2$$ which lies above the $$xy$$–plane.

Find the area of the portion of the cone $$z^2 = x^2 + y^2$$ lying between the planes $$z = 2$$ and $$z = 3\text{.}$$

Determine the surface area of the surface given by $$z = \frac{2}{3}\big(x^{3/2} + y^{3/2}\big)\text{,}$$ over the square $$0 \le x \le 1\text{,}$$ $$0 \le y \le 1\text{.}$$

1. To find the surface area of the surface $$z = f (x,y)$$ above the region $$D\text{,}$$ we integrate $$\iint_D F(x,y)\ \text{d}A\text{.}$$ What is $$F(x,y)\text{?}$$
2. Consider a “Death Star”, a ball of radius $$2$$ centred at the origin with another ball of radius $$2$$ centred at $$(0, 0, 2\sqrt{3})$$ cut out of it. The diagram below shows the slice where $$y = 0\text{.}$$
1. The Rebels want to paint part of the surface of Death Star hot pink; specifically, the concave part (indicated with a thick line in the diagram). To help them determine how much paint is needed, carefully fill in the missing parts of this integral:

$\text{surface area} = \int_{\fbox{\vphantom{L}\qquad}}^{\fbox{\vphantom{L}\qquad}} \int_{\fbox{\vphantom{L}\qquad}}^{\fbox{\vphantom{L}\qquad}} \fbox{\vphantom{L}\qquad\qquad\qquad}\ \text{d}r\,\text{d}\theta \nonumber$

2. What is the total surface area of the Death Star?

Find the area of the cone $$z^2=x^2+y^2$$ between $$z=1$$ and $$z=16\text{.}$$

Find the surface area of that part of the hemisphere $$z=\sqrt{a^2-x^2-y^2}$$ which lies within the cylinder $$\big(x-\frac{a}{2}\big)^2+y^2=\big(\frac{a}{2}\big)^2\text{.}$$

The cylinder $$\ x^2+y^2=2x\$$ cuts out a portion $$S$$ of the upper half of the cone $$\ x^2+y^2=z^2\text{.}$$ Compute

$\iint_S(x^4-y^4+y^2z^2-z^2x^2+1)\,\text{d}S \nonumber$

Find the surface area of the torus obtained by rotating the circle $$(x-R)^2+z^2=r^2$$ (the circle is contained in the $$xz$$-plane) about the $$z$$-axis.

A spherical shell of radius $$a$$ is centred at the origin. Find the centroid (i.e. the centre of mass with constant density) of the part of the sphere that lies in the first octant.

Find the area of that part of the cylinder $$x^2+y^2=2ay$$ lying outside $$z^2=x^2+y^2\text{.}$$

Let $$a$$ and $$b$$ be positive constants, and let $$\mathcal{S}$$ be the part of the conical surface

$a^2 z^2 = b^2(x^2+y^2) \nonumber$

where $$0\le z\le b\text{.}$$ Consider the surface integral

$I = \iint_{\mathcal{S}} (x^2+y^2)\,\text{d}\mathcal{S}. \nonumber$

1. Express $$I$$ as a double integral over a disk in the $$xy$$-plane.
2. Use the parametrization $$x=t\cos\theta\text{,}$$ $$y=t\sin\theta\text{,}$$ etc., to express $$I$$ as a double integral over a suitable region in the $$t\theta$$-plane.
3. Evaluate $$I$$ using the method of your choice.

Evaluate, for each of the following, the flux $$\ \iint_S\vecs{F} \cdot\hat{\textbf{n}} \,\text{d}S\$$ where $$\hat{\textbf{n}}$$ is the outward normal to the surface $$S\text{.}$$

1. $$\ \vecs{F} ={(x^2+y^2+z^2)}^n(x\,\hat{\pmb{\imath}}+y\,\hat{\pmb{\jmath}}+z\,\hat{\mathbf{k}})\$$ and the surface $$S$$ is the sphere $$x^2+y^2+z^2=a^2\text{.}$$
2. $$\ \vecs{F} =x\,\hat{\pmb{\imath}}+y\,\hat{\pmb{\jmath}}+z\,\hat{\mathbf{k}}\$$ and $$S$$ is the surface of the rectangular box $$0\le x\le a\text{,}$$ $$0\le y\le b\text{,}$$ $$0\le z\le c\text{.}$$
3. $$\ \vecs{F} =y\,\hat{\pmb{\imath}}+z\,\hat{\mathbf{k}}\$$ and $$S$$ is the surface of the solid cone $$0\le z\le 1-\sqrt{x^2+y^2}\text{.}$$

Let $$\mathcal{S}$$ be the part of the surface $$x^2+y^2+2z=2$$ that lies above the square $$-1\le x\le 1\text{,}$$ $$-1\le y\le 1\text{.}$$

1. Find $$\displaystyle\iint_{\mathcal{S}} \frac{x^2+y^2}{\sqrt{1+x^2+y^2}}\ \text{d}S\text{.}$$
2. Find the flux of $$\vecs{F} =x\hat{\pmb{\imath}}+y\hat{\pmb{\jmath}}+z\hat{\mathbf{k}}$$ upward through $$\mathcal{S}\text{.}$$

Let $$\mathcal{S}$$ be the part of the surface $$z=xy$$ that lies above the square $$0\le x\le 1\text{,}$$ $$0\le y\le 1$$ in the $$xy$$-plane.

1. Find $$\displaystyle\iint_{\mathcal{S}} \frac{x^2y}{\sqrt{1+x^2+y^2}}\ \text{d}S\text{.}$$
2. Find the flux of $$\vecs{F} =x\hat{\pmb{\imath}}+y\hat{\pmb{\jmath}}+\hat{\mathbf{k}}$$ upward through $$\mathcal{S}\text{.}$$

Find the area of the part of the surface $$z=y^{3/2}$$ that lies above $$0\le x,y\le 1\text{.}$$

Let $$\mathcal{S}$$ be spherical cap which consists of the part of the sphere $$x^2+y^2+(z-2)^2=4$$ which lies under the plane $$z=1\text{.}$$ Let $$f(x,y,z)=(2-z)(x^2+y^2)\text{.}$$ Calculate

$\iint_{\mathcal{S}} f(x,y,z)\,\text{d}S \nonumber$

1. Find a parametrization of the surface $$S$$ of the cone whose vertex is at the point $$(0, 0, 3)\text{,}$$ and whose base is the circle $$x^2 + y^2 = 4$$ in the $$xy$$-plane. Only the cone surface belongs to $$S\text{,}$$ not the base. Be careful to include the domain for the parameters.
2. Find the $$z$$-coordinate of the centre of mass of the surface $$S$$ from (a).

Let $$S$$ be the surface of a cone of height $$a$$ and base radius $$a\text{.}$$ The surface $$S$$ does not include the base of the cone or the interiour of the cone. Find the centre of mass of $$S\text{.}$$

Locate the cone in a coordinate system so that its base is in the $$xy$$-plane, and its vertex on the $$z$$-axis. So the vertex will be the point $$(0, 0, a)\text{.}$$ The base is a circle of radius $$a$$ in the $$xy$$-plane with centre at the origin. The cone surface is characterized by the fact that for every point of $$S\text{,}$$ the distance from the $$z$$-axis and the distance from the $$xy$$-plane add up to $$a\text{.}$$

Let $$S$$ be the portion of the elliptical cylinder $$x^2+\frac{1}{4}y^2=1$$ lying between the planes $$z=0$$ and $$z=1$$ and let $$\hat{\textbf{n}}$$ denote the outward normal to $$S\text{.}$$ Let $$\vecs{F} =x\,\hat{\pmb{\imath}}+xyz\,\hat{\pmb{\jmath}}+zy^4\,\hat{\mathbf{k}}\text{.}$$ Calculate the flux integral $$\iint_S \vecs{F} \cdot\hat{\textbf{n}}\, \text{d}S$$ directly, using an appropriate parameterization of $$S\text{.}$$

Evaluate the flux integral

$\iint_S \vecs{F} \cdot\hat{\textbf{n}}\, \text{d}S \nonumber$

where $$\vecs{F} (x, y, z) = (x+1)\,\hat{\pmb{\imath}}+(y +1)\,\hat{\pmb{\jmath}}+2z\,\hat{\mathbf{k}}\text{,}$$ and $$S$$ is the part of the paraboloid $$z = 4-x^2 -y^2$$ that lies above the triangle $$0 \le x \le 1\text{,}$$ $$0 \le y \le 1 – x\text{.}$$ $$S$$ is oriented so that its unit normal has a negative $$z$$-component.

Evaluate the surface integral

$\iint_S xy^2\ \text{d}S \nonumber$

where $$S$$ is the part of the sphere $$x^2 + y^2 + z^2 = 2$$ for which $$x \ge \sqrt{y^2 + z^2}\text{.}$$

Let $$S$$ be the surface given by the equation

$\begin{gather*} x^2 + z^2 = \sin^2y \end{gather*}$

lying between the planes $$y = 0$$ and $$y = \pi\text{.}$$ Evaluate the integral

$\begin{gather*} \iint_S\sqrt{1 + \cos^2y}\, \text{d}S \end{gather*}$

Let $$S$$ be the part of the paraboloid $$z=1-x^2-y^2$$ lying above the $$xy$$-plane. At $$(x,y,z)$$ $$S$$ has density

$\begin{gather*} \rho(x,y,z) = \frac{z}{\sqrt{5-4z}} \end{gather*}$

Find the centre of mass of $$S\text{.}$$

Let $$S$$ be the part of the plane

$x + y + z =2 \nonumber$

that lies in the first octant oriented so that $$\hat{\textbf{n}}$$ has a positive $$\hat{\mathbf{k}}$$ component. Let

$\vecs{F} = x\,\hat{\pmb{\imath}} + y\,\hat{\pmb{\jmath}} + z\,\hat{\mathbf{k}} \nonumber$

Evaluate the flux integral

$\iint_S \vecs{F} \cdot\hat{\textbf{n}}\, \text{d}S \nonumber$

Find the net flux $$\iint_S \vecs{F} \cdot\hat{\textbf{n}}\,\text{d}S$$ of the vector field $$\vecs{F} (x,y,z) = (x,y,z)$$ upwards (with respect to the $$z$$-axis) through the surface $$S$$ parametrized $$\vecs{r} = \big(uv^2\,,\,u^2v\,,\,uv\big)$$ for $$0\le u\le 1\text{,}$$ $$0\le v\le 3\text{.}$$

Let $$S$$ be the surface obtained by revolving the curve $$z = e^y$$, $$0 \le y \le 1\text{,}$$ around the $$y$$-axis, with the orientation of $$S$$ having $$\hat{\textbf{n}}$$ pointing toward the $$y$$-axis.

1. Draw a picture of $$S$$ and find a parameterization of $$S\text{.}$$
2. Compute the integral $$\iint_S e^y\, \text{d}S\text{.}$$
3. Compute the flux integral $$\iint_S\vecs{F} \cdot\hat{\textbf{n}}\,\text{d}S$$ where $$\vecs{F} = (x, 0, z)\text{.}$$

Compute the net outward flux of the vector field

$\vecs{F} = \frac{\vecs{r} }{|\vecs{r} |} = \frac{x\,\hat{\pmb{\imath}}+y\,\hat{\pmb{\jmath}}+z\,\hat{\mathbf{k}}}{\sqrt{x^2+y^2+z^2}} \nonumber$

across the boundary of the region between the spheres of radius $$1$$ and radius $$2$$ centred at the origin.

Evaluate the surface integral $$\iint_S z^2\,\text{d}S$$ where $$S$$ is the part of the cone $$x^2+y^2=4z^2$$ where $$0 \le x \le y$$ and $$0 \le z \le 1\text{.}$$

Compute the flux integral $$\iint_S \vecs{F} \cdot\hat{\textbf{n}}\,\text{d}S\text{,}$$ where

$\vecs{F} = \Big(-\frac{1}{2}x^3 – xy^2 \,,\, -\frac{1}{2} y^3 \,,\, z^2\Big) \nonumber$

and $$S$$ is the part of the paraboloid $$z = 5 – x^2 – y^2$$ lying inside the cylinder $$x^2 + y^2 \le 4\text{,}$$ with orientation pointing downwards.

Let the thin shell $$S$$ consist of the part of the surface $$z^2=2xy$$ with $$x\ge 1\text{,}$$ $$y\ge 1$$ and $$z\le 2\text{.}$$ Find the mass of $$S$$ if it has surface density given by $$\rho(x,y,z)=3z$$ kg per unit area.

Let $$S$$ be the portion of the paraboloid $$x=y^2 + z^2$$ that satisfies $$x \le 2 y \text{.}$$ Its unit normal vector $$\hat{\textbf{n}}$$ is so chosen that $$\hat{\textbf{n}} \cdot \hat{\pmb{\imath}} \gt 0\text{.}$$ Find the flux of $$\vecs{F} = 2\,\hat{\pmb{\imath}} + z\,\hat{\pmb{\jmath}} + y\,\hat{\mathbf{k}}$$ out of $$S\text{.}$$

Let $$S$$ denote the portion of the paraboloid $$z=1-\frac{1}{4}x^2-y^2$$ for which $$z\ge 0\text{.}$$ Orient $$S$$ so that its unit normal has a positive $$\hat k$$ component. Let

$\vecs{F} (x,y,z) = (3y^2+z)\,\hat{\pmb{\imath}}+(x-x^2)\,\hat{\pmb{\jmath}}+\hat{\mathbf{k}} \nonumber$

Evaluate the surface integral $$\iint_S\nabla\times\vecs{F} \cdot\hat{\textbf{n}}\,\text{d}S\text{.}$$

Let $$S$$ be the boundary of the apple core bounded by the sphere $$x^2+y^2+z^2=16$$ and the hyperboloid $$x^2+y^2-z^2=8\text{.}$$ Find the flux integral $$\ \iint_S\vecs{F} \cdot\hat{\textbf{n}} \,dS\$$ where $$\vecs{F} =x\,\hat{\pmb{\imath}}+y\,\hat{\pmb{\jmath}}+z\,\hat{\mathbf{k}}$$ and $$\hat{\textbf{n}}$$ is the outward normal to the surface $$S\text{.}$$

### Stage 3

1. Consider the surface $$S$$ given by the equation

$x^2 + z^2 = \cos^2 y \nonumber$

Find an equation for the tangent plane to $$S$$ at the point $$\big(\frac{1}{2} , \frac{\pi}{4} , \frac{1}{2} \big)\text{.}$$

2. Compute the integral

$\iint_S \sin y\ \text{d}S \nonumber$

where $$S$$ is the part of the surface from (a) lying between the planes $$y = 0$$ and $$y = \frac{1}{2}\pi\text{.}$$

Let $$f$$ be a function on $$\mathbb{R}^3$$ such that all its first order partial derivatives are continuous. Let $$S$$ be the surface $$\left \{(x, y, z)|f(x,y,z) = c \right \}$$ for some $$c \in \mathbb{R}\text{.}$$ Assume that $$\vecs{ \nabla} f \ne \vecs{0}$$ on $$S\text{.}$$ Let $$\vecs{F}$$ be the gradient field $$\vecs{F} = \vecs{ \nabla} f\text{.}$$

1. Let $$C$$ be a piecewise smooth curve contained in $$S$$ (not necessarily closed). Must it be true that $$\int_C \vecs{F} \cdot \text{d}\vecs{r} = 0\text{?}$$ Explain why.
2. Prove that for any vector field $$\textbf{G}\text{,}$$

$\iint_S (\vecs{F} \times \textbf{G}) \cdot\hat{\textbf{n}}\, \text{d}S = 0. \nonumber$

1. Give parametric descriptions of the form $$\vecs{r} (u, v) = \big(x(u,v)\,,\, y(u,v)\,,\, z(u,v)\big)$$ for the following surfaces. Be sure to state the domains of your parametrizations.
1. The part of the plane $$2x + 4y + 3z = 16$$ in the first octant

$\left \{(x, y, z)| x \ge 0,\ y \ge 0,\ z \ge 0 \right \} \nonumber$

2. The cap of the sphere $$x^2 + y^2 + z^2 = 16$$ for $$4/\sqrt{2} \le z \le 4\text{.}$$
3. The hyperboloid $$z^2 = 1 + x^2 + y^2$$ for $$1 \le z \le 10\text{.}$$
2. The part of the plane $$2x + 4y + 3z = 16$$ in the first octant
3. Use your parametrization from part (a) to compute the surface area of the cap of the sphere $$x^2 + y^2 + z^2 = 16$$ for $$4/\sqrt{2} \le z \le 4\text{.}$$

Let $$S$$ be the part of the sphere $$x^2+y^2+z^2=2$$ where $$y\ge 1\text{,}$$ oriented away from the origin.

1. Compute

$\iint_S y^3\,\text{d}S \nonumber$

2. Compute

$\iint_S \big(xy\,\hat{\pmb{\imath}} + xz\,\hat{\pmb{\jmath}} + zy\,\hat{\mathbf{k}}\big)\cdot\hat{\textbf{n}}\,\text{d}S \nonumber$

Let $$\mathcal{S}$$ be the part of the surface $$(x+y+1)^2+z^2=4$$ which lies in the first octant. Find the flux of $$\vecs{F}$$ downwards through $$\mathcal{S}$$ where

$\vecs{F} =xy\,\hat{\pmb{\imath}}+(z-xy)\,\hat{\pmb{\jmath}} \nonumber$

1. As we mentioned above, the approximation below becomes exact when the limit $$\text{d}u,\text{d}v\rightarrow 0$$ is taken in the definition of the integral. See the optional §3.3.5.
2. This is called the implicit function theorem. We will not prove it. But it is not so hard to understand why it is true, if one thinks in terms of the Taylor expansion of $$G$$ about the point. For simplicity, let’s suppose that the point is $$(0,0,0)$$ and $$G$$ happens to be exactly equal to its first order Taylor expansion about $$(0,0,0)\text{.}$$ That is, $$G(x,y,z) = A +Bx +Cy +Dz\text{,}$$ for some constants $$A\text{,}$$ $$B\text{,}$$ $$C\text{,}$$ $$D\text{.}$$ Since $$(0,0,0)$$ is on the surface, $$A=K\text{.}$$ As $$\frac{\partial G}{\partial z}=D \ne 0$$ we can easily solve $$G(x,y,z)=K$$ for $$z$$ as a function of $$x$$ and $$y\text{.}$$ Namely $$z=\frac{1}{D}(-Bx-Cy)\text{.}$$ The general proof is based on the fact that, under reasonable hypotheses, the first order Taylor expansion is a good approximation to $$G$$ near $$(0,0,0)\text{.}$$
3. The symbols $$r\text{,}$$ $$\theta\text{,}$$ $$z$$ are the standard mathematics symbols for the cylindrical coordinates. Appendix A.7 gives another set of symbols that is commonly used in the physical sciences and engineering.
4. As we have noted before, the spherical coordinate system really breaks down at $$\varphi=0\text{,}$$ because $$\rho=1\text{,}$$ $$\varphi=0$$ gives the same point, namely the north pole $$(0,0,1)\text{,}$$ for all values of $$\theta\text{.}$$ We should really treat our integral like an improper integral, first integrating over $$\varepsilon \lt \varphi\le\frac{\pi}{2}$$ and then taking the limit $$\varepsilon\rightarrow 0^+\text{.}$$ However the breakdown of the spherical coordinate system at $$\varphi=0\text{,}$$ just like the breakdown of polar coordinates at $$r=0\text{,}$$ rarely causes problem and it is routine to skip the “improper integral” step.
5. We promise!
6. This answer for $$\text{d}S$$ is a very clean. Think about why. Hint: review the discussion following 3.3.2.
7. If you have forgotten why, sketch the graph.
8. We did so previously, with different variable names, in Example 3.2.2.
9. Again the formula for $$\text{d}S$$ is very neat. Think about why.
10. A favourite of science fiction and fantasy writers. Plug “subterranean fiction” into your favourite search engine. While you’re at it, also try “gravity train”. We’ll look at it in the optional Example 3.3.11.
11. Under this definition we still have $$\iint (\textbf{A}+\textbf{B})\,\text{d}S = \iint \textbf{A}\,\text{d}S +\iint\textbf{B}\,\text{d}S\text{.}$$
12. Think about why the $$\hat{\pmb{\imath}}$$ and $$\hat{\pmb{\jmath}}$$ components should both be zero. Think symmetry.
13. These two results appeared in Isaac Newton’s Principia Mathematica (1687). They are known as Newton’s “superb theorems”.
14. The British physicist and architect (he was Surveyor to the City of London and chief assistant to Christopher Wren) Robert Hooke (1635–1703) wrote about the gravity train idea in a letter to Isaac Newton. A gravity train was used in the 2012 movie Total Recall.
15. Remember the error in the Taylor polynomial approximations.
16. See the optional §1.1.6 of the CLP-2 text for an analogous argument concerning Riemann sums.

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