Surface Area of a Sphere

A sphere is a perfectly round geometrical 3-dimensional object. It can be characterized as the set of all points located distance \(r\) (radius) away from a given point (center). It is perfectly symmetrical, and has no edges or vertices.

A sphere with radius \(r\) has a volume of \( \frac{4}{3} \pi r^3 \) and a surface area of \( 4 \pi r^2 \).

A sphere has several interesting properties, one of which is that, of all shapes with the same surface area, the sphere has the largest volume.

## Proof

To prove that the surface area of a sphere of radius \(r\) is \(4 \pi r^2 \), one straightforward method we can use is calculus. We first have to realize that for a curve parameterized by \(x(t)\) and \(y(t\)), the arc length is

\[ S = \int_a^b \sqrt{ \left(\frac{dy}{dt}\right)^2 + \left( \frac{dx}{dt}\right)^2 } \, dt. \]

From this we can derive the formula for the surface area of the solid obtained by rotating this about the \(x\)-axis. This turns out to be

\[ A = 2\pi \int_a^b y\sqrt{ \left(\frac{dy}{dt}\right)^2 + \left( \frac{dx}{dt}\right)^2 } \, dt .\]

We can obtain a sphere by revolving half a circle about the \(x\)-axis. This circle can be parameterized as \(x(t)=r\cos(t) \) and \(y(t) = r\sin(t) \) for \(0 \leq t \leq \pi \). From this, we get

\[ \frac{dx}{dt} = -r\sin(t), \quad \frac{dy}{dt} = r\cos(t) .\]

Substituting in our equations for surface area gives

\[\begin{align} A &= 2\pi \int_0^\pi r\sin(t)\sqrt{ \big(-r\sin(t)\big)^2 + \big( r\cos(t) \big)^2 } \ dt \\ &= 2\pi \int_0^\pi r\sin(t)\sqrt{ r^2\big(\sin(t)^2 + \cos(t)^2 \big) } \ dt \\ &= 2\pi \int_0^\pi r^2 \sin(t) \ dt \\ &= 2\pi r^2 \int_0^\pi\sin(t) \ dt \\ &= 4 \pi r^2. \ _\square \end{align} \]

## Archimedes’ Hat-Box Theorem

Archimedes’ Hat-Box Theorem

Archimedes’ hat-box theorem states that for any sphere section, its lateral surface will equal that of the cylinder with the same height as the section and the same radius of the sphere.

Let us recall our last proof section. After revolving the semicircle around the \(x\)-axis, we will obtain a sphere’s surface area, and if we cut just a partial section with parallel bases, the new surface area will be demonstrated in the image below:

From the image, the section’s lateral surface area is colored light blue with 2 circular bases of different radii. In order to visualize the section’s height better, this section will be rotated by 90 degrees, as shown below:

Now inside the section, there are 2 variable angles, \(\angle a\) and \(\angle b\), which appear as the integral borders of the cut section.

From the proof’s conclusion, the surface area of the section \((A’)\) can be calculated as

\[\begin{align} A’ &= 2\pi r^2 \int_a^b\sin(t) \ dt \\ &= 2\pi r^2 \left[\left.-\cos(t) \right|_a^b\right] \\ &= (2\pi r)r\big[\cos (a) – \cos (b)\big] . \end{align}\]

Considering the right triangles with radius \(r\) (thick red) in the image, it is obvious that \(r\) is the hypotenuse side for both. As a result, the vertical sides can be calculated as \(r\times \cos (a)\) and \(r\times \cos (b)\) for the left and right triangles, respectively.

Hence, the height of the section is \(h = \big(r\times \cos (a)\big) – \big(r\times \cos (b)\big) = r\big[\cos (a) – \cos (b)\big]\).

Substituting this term to the previous equation gives

\[A’ = (2\pi r)r\big[\cos (a) – \cos (b)\big] = 2\pi rh. \]

Clearly, this is the formula for the cylinder’s lateral surface with radius \(r\) and height \(h\)!

That means the lateral surface area of the sphere section equals the lateral surface area of the cylinder with radius \(r\) and height \(h,\) as shown in the image, and this holds true for any level of the sphere involved. \(_\square\)

A spherical tomato and a cylindrical portion of a cucumber have the same height and radius. Then they are chopped into slices of equal thickness, as shown above.

Comparing each slice of both kinds, which slice will have more lateral surface area of the peel?

A small green circle is inscribed within the section of a bigger blue circle, touching the mid-chord, as shown above left. Then the graphs are revolved around the \(y\)-axis to generate three figures: a blue cover dome, a green spherical melon, and a red serving plate.

Which of the following options will have more surface area?

I. The blue dome

II. The melon plus the plate

A sweets shop sells candies in 2 different styles: a spherical ball and a dome. The dome-like shape is a spherical section of a larger sphere with height \(h\) and base radius \(R,\) as shown above, while the candy ball has radius \(r\) with \(2r = R + h\).

If both shapes have the same total surface area, what is the ratio \(\frac{R}{h}\)?

## Practice Problems

What is the surface area of a sphere of radius 3? The surface area is \( 4 \pi \times 3^2 = 36 \pi \). \( _\square \)

If the volume of a sphere is \(36\pi,\) what is the surface area of the sphere? Observe that the volume of the sphere can be rewritten as \[36\pi=\frac{4}{3}\pi \times 3^3.\] Then, since the volume of a sphere with radius \(r\) is \( \frac{4}{3} \pi r^3 ,\) it follows that the radius of the sphere in this problem is \(r=3.\) Hence, its the surface area is \[4 \pi r^2 =4\pi \times 3^2 =36\pi. \ _\square\]

The volume of a sphere has grown 8 times. Then how many times has the surface area grown in the meanwhile? Observe that the volume of the sphere is \( \frac{4}{3} \pi r^3.\) This implies that it is proportional to \(r^3,\) that is\( \frac{4}{3} \pi r^3 \propto r^3\). Then 8 times growth in the volume of the sphere implies 2 times growth in the radius of sphere. Then, since the surface area of sphere is \( 4 \pi r^2 \propto r^2, \) the surface area of the sphere has grown \(2^2 = 4\) times. \(\ _\square\)

You have a watermelon whose volume is \(288 \text{ cm}^3.\) If you cut the watermelon into halves, what is the surface area of one half of the watermelon? (Assume that the watermelon is a perfect sphere.) From the formula \( V=\frac{4}{3} \pi r^3 \) for the volume of a sphere with radius \(r,\) you know that the radius of the watermelon is \(r=6 \text{ cm}.\) Since you cut the watermelon into two exact halves, you may think that the surface area of a half watermelon is also exactly half the surface area of the whole watermelon. However, this thinking is wrong.

As shown int the above diagram, the surface area of a half watermelon is bigger than half the surface area of a whole watermelon, by the area the cross section \(A.\) Thus, the surface area of a half watermelon is \[\text{(Half the surface area of the watermelon)} + \text{(Area of A)}. \] Since \(A\) is a circle whose radius is the same as the radius of the watermelon, our answer is \[\frac{1}{2} \times 4\pi \times 6^2 + \pi \times 6^2 = 108 \pi. \ _\square\]

The diameter of a solid metallic right circular cylinder is equal to its height. After cutting out the largest possible solid sphere \(S\) from this cylinder, the remaining material is recast to form a solid sphere \(Q.\)

What is the ratio of the radius of sphere \(S\) to that of sphere \(Q?\)