It can be shown that the volume of the space between a cone and a cylinder with radius \begin{align*}r\end{align*} and height \begin{align*}r\end{align*} is the same as the volume of half a sphere (a hemisphere) with radius \begin{align*}r\end{align*}. Given this, what's the formula for the volume of a sphere?

#### Guidance

Recall that a **sphere** is the set of all points in space that are equidistant from a given point. The distance from the center of a sphere to any point on the sphere is the radius.

You have seen the formula for the volume of a sphere before.

\begin{align*}Sphere: V=\frac{4}{3} \pi r^3\end{align*}

The key to understanding the formula for the volume of a sphere is to compare a sphere inside of a cylinder with radius \begin{align*}r\end{align*} and height \begin{align*}2r\end{align*} with a double cone inside of a cylinder with radius \begin{align*}r\end{align*} and height \begin{align*}2r\end{align*}.

The volume of the sphere is the same as the volume of the **space between** the cylinder and the double cone.

**Example A**

Consider half of the double cone inside the cylinder and half the sphere (a hemisphere). If the radius of the sphere is 5 inches, label all the dimensions that you can. What is the area of the top surface of the hemisphere? What is the area of the top of the cylinder?

**Solution:**

The area of the top of the cylinder is the same as the area of the top of the hemisphere. In each case the area is \begin{align*}\pi (5^2)=25 \pi \ in^2\end{align*}.

**Example B**

Imagine a slice is made at height \begin{align*}h\end{align*}. Draw the horizontal and vertical cross sections of each and label all dimensions in terms of \begin{align*}h\end{align*}.

**Solution:** First consider the vertical cross section through the center of the cone in the cylinder.

Note that because the height and the radius of the cone are each 5 inches, isosceles right triangles are formed. Next consider the vertical cross section through the center of the sphere.

The radius of the circular horizontal cross section is \begin{align*}x\end{align*}. You can find the length of \begin{align*}x\end{align*} by using the Pythagorean Theorem:

\begin{align*}x^2+(5-h)^2 &= 5^2\\ x^2+25-10h+h^2 &= 25\\ x &= \sqrt{10h-h^2}\end{align*}

Now you can consider the horizontal cross sections.

Note that the shaded areas are the cross sections of the solids you are interested in. You are looking for the volume of the space **between** the cylinder and the cone and the volume of the sphere.

**Example C**

Confirm that the area of each shaded region below is the same. What does this tell you about the volume of the space between the cylinder and the cone compared to the volume of the sphere?

**Solution:** The area of the shaded region on the left is:

\begin{align*}A &= 25 \pi - (5-h)^2 \pi\\ &= 10 h \pi - h^2 \pi\end{align*}

The area of the shaded region on the right is:

\begin{align*}A &=\left(\sqrt{10h-h^2}\right)^2 \pi \\ &= 10 h \pi - h^2 \pi\end{align*}

The areas are the same. Because the two solids lie between parallel planes, have the same heights, and have equal cross sectional areas, **their volumes must be the same.**

**Concept Problem Revisited**

The volume of the cylinder is \begin{align*}\pi r^3\end{align*}. The volume of the cone is \begin{align*}\frac{\pi r^3}{3}\end{align*}. Therefore, the volume of the space between the cone and the cylinder is:

\begin{align*}\pi r^3-\frac{\pi r^3}{3}=\frac{2 \pi r^3}{3}\end{align*}

If this is also the volume of a hemisphere, then the volume of a sphere must be twice as big. The volume of a sphere is:

\begin{align*}\frac{2 (2\pi r^3)}{3}=\frac{4 \pi r^3}{3}\end{align*}

#### Vocabulary

*Cavalieri's*** principle** states that if two solids lying between parallel planes have equal heights and all cross sections at equal distances from their bases have equal areas, then the solids have equal volumes.

A ** sphere** is the set of all points in space that are equidistant from a given point. The distance from the center of a sphere to any point on the sphere is the radius.

#### Guided Practice

1. Find the volume of the space between the cylinder and the cone below.

2. Describe what portion of a sphere has the same volume as the volume calculated in #1.

3. Find the volume of a sphere with a diameter of 15 *cm*.

**Answers:**

1. The volume of the cylinder is \begin{align*}1728 \pi \ in^3\end{align*} and the volume of the cone is \begin{align*}576 \pi \ in^3\end{align*}. The volume of the space between the cylinder and the cone is \begin{align*}1152 \pi \ in^3\end{align*}.

2. A hemisphere with radius 12 *in* would have the same volume.

3. \begin{align*}V=\frac{4 \pi (7.5)^3}{3}=562.5 \pi \ cm^3\end{align*}

#### Practice

1. Find the volume of a sphere with a radius of 3 inches.

2. Find the volume of a sphere with a diameter of 12 inches.

3. Find the volume of a sphere with a diameter of 8 inches.

4. In your own words, explain where the formula for the volume of a sphere came from. How does it relate to a double cone within a cylinder?

A bead is created from a sphere by drilling a cylinder through the sphere. The original sphere has a radius of 8 mm. The cylinder drilled through the center has a radius of 4 mm.

5. What is the height of the bead? (Hint: Draw a right triangle and use the Pythagorean Theorem.)

6. What is the volume of the original sphere? What is the volume of the cylinder?

7. Due to Cavalieri's principle, the volume of the space above the cylinder is the same as the volume between a cone and a cylinder (see picture below). What is the approximate volume of the space above and below the cylinder that was cut off when making the bead?

8. What is the approximate volume of the bead?

A cylindrical container holds three tennis balls. The diameter of the cylinder is 4 inches, which is approximately the same as the diameter of each tennis ball. The height of the cylinder is 12 inches.

9. What is the volume of one tennis ball?

10. What is the volume of the space between the tennis balls and the cylinder?

11. If one cup of water has a volume of approximately \begin{align*}14.44 \ in^3\end{align*}, how many cups of water would fit in the cylinder with the tennis balls?

Think of a sphere of radius \begin{align*}r\end{align*} as being made up of a large number *\begin{align*}k\end{align*}* of congruent small square based pyramids. Let the area of each square base be \begin{align*}B\end{align*}.

12. What is the volume of one pyramid in terms of \begin{align*}r\end{align*}, \begin{align*}k\end{align*}, and \begin{align*}B\end{align*}?

13. What is the surface area of the sphere in terms of \begin{align*}k\end{align*} and \begin{align*}B\end{align*}?

14. What is the volume of the sphere in terms of \begin{align*}r\end{align*}, \begin{align*}k\end{align*}, and \begin{align*}B\end{align*}?

15. Use your answers to #13 and #14 and the formula for the volume of a sphere \begin{align*}\left(V=\frac{4 \pi r^3}{3} \right)\end{align*} to find the formula for the surface area of a sphere.