Volume of a Sphere

An informal argument using Cavalieri's Principle to derive the formula \( V = \frac{4}{3}\pi r^3 \) by comparing cross-sectional areas of solids.

Geometry

What is Volume?

Volume measures the amount of three-dimensional space enclosed by a solid figure. For familiar shapes like prisms and cylinders, we compute volume by multiplying a base area by a height. But what about curved solids like spheres?

Cavalieri's Principle

Cavalieri's Principle states: if two solids have the same height and the same cross-sectional area at every level, then they have the same volume.

The Key Idea: Compare Cross-Sections

To find the volume of a sphere of radius \( r \), we will compare it to a simpler solid whose volume we already know. We need a solid where, at every height, the cross-sectional area exactly matches the sphere's cross-sectional area.

Sphere

Unknown volume. Cross-sections are circles whose radius depends on height.

Comparison Solid

A cylinder with an inverted cone removed. Known volume, matching cross-sections.

Cross-Section of a Sphere

The Comparison Solid

Consider a cylinder of radius \( r \) and height \( r \), with an inverted cone of the same radius and height removed from it. We call this a "cylinder minus cone" or anti-cone.

Sphere vs. Cylinder Minus Cone

Computing the Volume

1

Volume of cylinder: \( V_{\text{cyl}} = \pi r^2 \cdot r = \pi r^3 \)

2

Volume of cone: \( V_{\text{cone}} = \frac{1}{3}\pi r^2 \cdot r = \frac{1}{3}\pi r^3 \)

3

Volume of hemisphere: \( V_{\text{hemi}} = \pi r^3 - \frac{1}{3}\pi r^3 = \frac{2}{3}\pi r^3 \)

4

Volume of full sphere: \( V = 2 \times \frac{2}{3}\pi r^3 = \frac{4}{3}\pi r^3 \)

The Sphere Volume Formula

Volume vs. Surface Area

	Volume	Surface Area
Formula	\( \frac{4}{3}\pi r^3 \)	\( 4\pi r^2 \)
Units	Cubic (e.g., cm³)	Square (e.g., cm²)
Measures	Space inside	Area of the skin
Exponent on r	3	2
Coefficient	\( \frac{4}{3} \)	4

How to Apply the Formula

1

Identify the radius. If given the diameter, divide by 2. If given circumference, use \( r = \frac{C}{2\pi} \).

2

Cube the radius. Compute \( r^3 \). Be careful with fractions and decimals.

3

Multiply by \( \frac{4}{3}\pi \). Either leave the answer in terms of \( \pi \) or multiply by \( \approx 3.14159 \).

4

Include units. Volume is always in cubic units (cm³, m³, in³, etc.).

Example 1: Radius Given

Find the volume of a sphere with radius \( r = 6 \) cm.

1

Radius: \( r = 6 \) cm

2

Cube it: \( 6^3 = 216 \)

3

Multiply: \( V = \frac{4}{3}\pi(216) = 288\pi \approx 904.78 \text{ cm}^3 \)

Example 2: Diameter Given

A basketball has a diameter of 24 cm. Find its volume.

1

Find radius: \( r = \frac{d}{2} = \frac{24}{2} = 12 \) cm

2

Cube it: \( 12^3 = 1,728 \)

3

Multiply: \( V = \frac{4}{3}\pi(1,728) = 2,304\pi \approx 7,238.23 \text{ cm}^3 \)

Example 3: Working Backwards

A sphere has a volume of \( 36\pi \) cubic inches. What is its radius?

1

Set up equation: \( \frac{4}{3}\pi r^3 = 36\pi \)

2

Divide both sides by \( \pi \): \( \frac{4}{3} r^3 = 36 \)

3

Multiply both sides by \( \frac{3}{4} \): \( r^3 = 27 \)

4

Take the cube root: \( r = \sqrt[3]{27} = 3 \) inches

Common Mistakes

✕ Using Diameter for Radius

Plugging diameter into \( \frac{4}{3}\pi r^3 \) gives a volume 8 times too large, since \( d^3 = 8r^3 \).

✕ Forgetting the Fraction

Writing \( 4\pi r^3 \) instead of \( \frac{4}{3}\pi r^3 \). The \( \frac{1}{3} \) comes from the cone subtraction.

✕ Confusing with Surface Area

Surface area is \( 4\pi r^2 \). Volume uses \( r^3 \) and has the coefficient \( \frac{4}{3} \), not 4.

Summary of the Argument

Step	What We Did	Result
1	Sliced the hemisphere at height \( h \)	Area \( = \pi(r^2 - h^2) \)
2	Sliced the cylinder minus cone at height \( h \)	Area \( = \pi(r^2 - h^2) \)
3	Applied Cavalieri's Principle	Hemisphere = Cyl. − Cone
4	Computed Cyl. − Cone volume	\( \pi r^3 - \frac{1}{3}\pi r^3 = \frac{2}{3}\pi r^3 \)
5	Doubled for full sphere	\( V = \frac{4}{3}\pi r^3 \)

Question 1 of 5

What is the volume of a sphere with radius 5 cm?

\( V = \frac{4}{3}\pi(5)^3 = \frac{4}{3}\pi(125) = \frac{500}{3}\pi \text{ cm}^3 \). Remember to cube the radius, then multiply by \( \frac{4}{3}\pi \).

Question 2 of 5

Cavalieri's Principle states that two solids have the same volume if:

Cavalieri's Principle requires two conditions: same height and identical cross-sectional area at every horizontal level. The shapes of the cross-sections can differ, but their areas must match.

Question 3 of 5

In the informal argument for sphere volume, the hemisphere is compared to which solid?

The comparison solid is a cylinder (radius r, height r) with a cone of the same dimensions removed. At each height, the annular cross-section of this solid matches the circular cross-section of the hemisphere.

Question 4 of 5

A sphere has a volume of \( 288\pi \) cubic meters. What is its radius?

Set \( \frac{4}{3}\pi r^3 = 288\pi \). Divide by \( \pi \): \( \frac{4}{3}r^3 = 288 \). Multiply by \( \frac{3}{4} \): \( r^3 = 216 \). Take the cube root: \( r = 6 \) m.

Question 5 of 5

If you double the radius of a sphere, by what factor does the volume increase?

Since \( V = \frac{4}{3}\pi r^3 \), replacing \( r \) with \( 2r \) gives \( \frac{4}{3}\pi(2r)^3 = \frac{4}{3}\pi \cdot 8r^3 = 8V \). Volume scales with the cube of the radius, so doubling the radius multiplies volume by \( 2^3 = 8 \).