Triple integral change of variable examples

Multivariable Calculus | Triple integral with spherical coordinates: Example.
Multivariable Calculus | Triple integral with spherical coordinates: Example.

Triple integral change of variable examples

Spherical coordinates example

For spherical coordinates, the change of variables function is \begin{align*} (x,y,z) &= \cvarf(\rho,\theta,\phi) \end{align*} where the components of $\cvarf$ are given by \begin{align*} x &= \rho \sin\phi \cos\theta\\ y &= \rho \sin\phi \sin\theta\\ z &= \rho \cos\phi. \end{align*} We can compute that \begin{align*} \jacm{\cvarf}(\rho,\theta,\phi) &= \left| \begin{array}{ccc} \displaystyle \pdiff{x}{\rho} & \displaystyle \pdiff{x}{\theta} & \displaystyle \pdiff{x}{\phi}\\ \displaystyle \pdiff{y}{\rho} & \displaystyle \pdiff{y}{\theta} & \displaystyle \pdiff{y}{\phi}\\ \displaystyle \pdiff{z}{\rho} & \displaystyle \pdiff{z}{\theta} & \displaystyle \pdiff{z}{\phi}\\ \end{array} \right|\\ &= \left| \begin{array}{ccc} \sin\phi \cos\theta & – \rho\sin\phi \sin\theta & \rho\cos\phi\cos\theta\\ \sin\phi \sin\theta & \rho \sin\phi \cos\theta & \rho\cos\phi\sin\theta\\ \cos\phi & 0 & -\rho \sin\phi \end{array} \right|\\ &= -\rho^2 \sin\phi. \end{align*}

The change of variable factor is the absolute value of the determinant \begin{align*} \left| \jacm{\cvarf}(\rho,\theta,\phi) \right| = \rho^2 \sin\phi. \end{align*} This means that the map from spherical coordinates to rectangular coordinates changes volume by the factor $\rho^2 \sin\phi$. For this reason, you need to do the above calculation only once. Now, you can just remember that the factor for spherical coordinates is $\rho^2 \sin\phi$.

Now we’re ready for the example: find the mass of a star $\dlv$ that is a ball of radius 3 centered at the origin if the density of the star is $g(x,y,z) = 10 – x^2-y^2-z^2$.

If we try to compute the integral directly in rectangular coordinates, it isn’t so easy: \begin{align*} &\iiint_\dlv g(x,y,z) dx\,dy\,dz\\ &= \int_{-3}^3 \int_{-\sqrt{9-z^2}}^{\sqrt{9-z^2}} \int_{-\sqrt{9-z^2-y^2}}^{\sqrt{9-z^2-y^2}} (10 – x^2 -y^2-z^2) dx\,dy\,dz \end{align*} This would lead to a mess.

We can find the mass of the star more easily in spherical coordinates. The star density $g(x,y,z) = 10 – x^2-y^2-z^2$ becomes $g(\cvarf(\rho,\theta,\phi)) = 10-\rho^2$.

In spherical coordinates, the integral over ball of radius 3 is the integral over the region \begin{align*} 0 \le \rho \le 3, \quad 0 \le \theta \le 2\pi, \quad 0 \le \phi \le \pi. \end{align*} The volume element is $\rho^2 \sin\phi \,d\rho\,d\theta\,d\phi$.

Therefore, the mass of the star is \begin{align*} &\int_0^3 \int_0^{2\pi} \int_0^{\pi} (10-\rho^2) \rho^2 \sin\phi \,d\phi\,d\theta\,d\rho\\ &=\int_0^3 \int_0^{2\pi} \left.(10-\rho^2) \rho^2 (-\cos\phi ) \right|_{\phi=0}^{\phi=\pi} d\theta\,d\rho\\ &=\int_0^3 \int_0^{2\pi} (10-\rho^2) \rho^2 2 d\theta\,d\rho\\ &=\int_0^3 4\pi (10-\rho^2) \rho^2 d\rho = \frac{828 \pi}{5} \approx 520. \end{align*}

Cylindrical coordinates example

For cylindrical coordinates, the change of variables function is \begin{align*} (x,y,z) &= \cvarf(r,\theta,z) \end{align*} where the components of $\cvarf$ are given by \begin{align*} x &= r \cos\theta\\ y &= r \sin\theta\\ z &= z. \end{align*} We can compute that \begin{align*} \jacm{\cvarf}(\rho,\theta,\phi) &= \left| \begin{array}{ccc} \displaystyle \pdiff{x}{r} & \displaystyle \pdiff{x}{\theta} & \displaystyle \pdiff{x}{z}\\ \displaystyle \pdiff{y}{r} & \displaystyle \pdiff{y}{\theta} & \displaystyle \pdiff{y}{z}\\ \displaystyle \pdiff{z}{r} & \displaystyle \pdiff{z}{\theta} & \displaystyle \pdiff{z}{z}\\ \end{array} \right|\\ &= \left| \begin{array}{ccc} \cos\theta & -r\sin\theta &0\\ \sin\theta & r \cos\theta &0\\ 0 & 0 & 1 \end{array} \right|\\ &= r \cos^2\theta + r\sin^2\theta = r. \end{align*} Volume changes by $|r|=r$ (just like area change for polar coordinates). Replace $dx\,dy\,dz$ by $r \,dr\,d\theta\,dz$.

We can calculate the following example problem. Find the volume of cone of height 1 and radius one. It is bounded by surface $z=\sqrt{x^2+y^2}$ and plane $z=1$.

The volume is \begin{align*} \int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{\sqrt{x^2+y^2}}^{1} dz \, dy\, dz. \end{align*}

The integral is easier to compute in cylindrical coordinates. In cylindrical coordinates, the cone is described by \begin{align*} 0 \le \theta \le 2\pi, \quad 0 \le r \le 1,\quad r \le z \le 1. \end{align*}

The volume of cone is \begin{align*} &\int_0^1 \int_0^{2\pi} \int_r^{1} r \, dz\,d\theta\,dr\\ &=\int_0^1 \int_0^{2\pi} (1 -r)r\, d\theta\,dr\\ &=\int_0^1 2\pi (1-r)r\, dr = \frac{\pi}{3}. \end{align*}

Ice cream cone revisited

Armed with the knowledge of how to change to spherical coordinates, we can now revisit the ice cream cone example.

Applet: Ice cream cone region

Applet loading

Ice cream cone region. The ice cream cone region is bounded above by the half-sphere $z=\sqrt{1-x^2-y^2}$ and bounded below by the cone $z=\sqrt{x^2+y^2}$.

On the triple integral examples page, we tried to find the volume of an ice cream cone $\dlv$ and discovered the volume was \begin{align*} \iiint_\dlv dV = \int_{-1/\sqrt{2}}^{1/\sqrt{2}} \int_{-\sqrt{1/2-x^2}}^{\sqrt{1/2-x^2}} \int_{\sqrt{x^2+y^2}}^{\sqrt{1-x^2-y^2}} dz\,dy\,dx. \end{align*}

We need to describe the bounds in terms of spherical coordinates. Since the cone is symmetrical around the $z$ axis, $\theta$ is easy. $0 \le \theta \le 2\pi$ in the cone and the ranges of the other variables don’t depend on $\theta$. Also, we see that $0 \le r \le 1$, since a given line from the origin extends until it hits the sphere $z=\sqrt{1-x^2-y^2}$, which is a sphere of radius 1. Lastly, $\phi$ is determined by requiring that we are in the cone $z \ge \sqrt{x^2+y^2}$. (To see this is exactly a condition on $\phi$, look at the page on spherical coordinates.) On the cone, $z=\sqrt{x^2+y^2}$, $\phi = \pi/4$. Consquently, the condition on $\phi$ is $0 \le \phi \le \pi/4$.

In summary, the ice cream cone is described by \begin{align*} 0 \le \theta \le 2\pi, \quad 0 \le r \le 1, \quad 0 \le \phi \le \pi/4. \end{align*}

Changing to spherical coordinates, we calculate that the volume of the ice cream cone is \begin{align*} \int_0^1 \int_0^{2\pi} \int_0^{\pi/4} \rho^2 \sin \phi d\phi \,d\theta\,d\rho &=\int_0^1 \int_0^{2\pi} [\cos(0)-\cos(\pi/4)] \rho^2 d\theta\,d\rho\\ &=\int_0^1 \int_0^{2\pi} \frac{2-\sqrt{2}}{2} \rho^2 d\theta\,d\rho\\ &=\int_0^1 \pi (2-\sqrt{2}) \rho^2 d\rho\\ &= \pi \frac{2-\sqrt{2}}{3}. \end{align*}

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