Exercises: Explain volume formulas - Common Core Math - HS Geometry

A

Recall / Warm-Up

1.

A cylinder has radius $r = 4$ cm and height $h = 5$ cm. What is its volume?

A.

$80\pi$ cm³

B.

$\frac{80\pi}{3}$ cm³

C.

$40\pi$ cm³

D.

$160\pi$ cm³

2.

A cone has radius $r = 3$ cm and height $h = 6$ cm. What is its volume?

A.

$18\pi$ cm³

B.

$54\pi$ cm³

C.

$12\pi$ cm³

D.

$6\pi$ cm³

3.

A right triangle has legs $a = 5$ and $h = 12$ . Which expression gives the length of the hypotenuse?

A.

$\sqrt{25 + 144} = 13$

B.

$5 + 12 = 17$

C.

$\sqrt{144 - 25} = \sqrt{119}$

D.

$5 \times 12 = 60$

B

Fluency Practice

1.

Two solids each have height 8 cm. At every height $h$ between 0 and 8 cm, the cross-sectional area of Solid A equals the cross-sectional area of Solid B. By Cavalieri's principle, if Solid A has volume $320\pi$ cm³, what is the volume of Solid B in cm³?

2.

A hemisphere of radius $r = 6$ cm is compared to a cylinder-minus-cone solid, where both the cylinder and the cone have radius 6 cm and height 6 cm. What is the volume of the cylinder-minus-cone solid in terms of $\pi$ ?

$Hemisphere cross-section showing the right triangle formed by radius r = 5, height h = 3, and cross-sectional radius x$

3.

A hemisphere of radius $r = 5$ cm is sliced horizontally at height $h = 3$ cm above its flat base. Using the Pythagorean Theorem, find the cross-sectional area at that height. Express your answer in terms of $\pi$ .

4.

For a hemisphere of radius $r$ , at height $h$ above the flat base, the cross-sectional radius $x$ satisfies ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ . So the cross-sectional area is $A_{\text{hemisphere}}(h) =$ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ .

equation relating x, h, r:

area formula in terms of r and h:

5.

For the cylinder-minus-cone comparison solid (cylinder radius $r$ , height $r$ ; cone apex at base, base radius $r$ at top), at height $h$ the cone's inner radius equals ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ . The annular cross-sectional area is $A_{\text{annulus}}(h) =$ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ .

cone's inner radius at height h:

annular area formula:

6.

Using the hemisphere argument: a hemisphere of radius $r$ has the same volume as a cylinder-minus-cone (both with radius $r$ and height $r$ ). Compute the hemisphere volume when $r = 3$ cm. Express your answer in terms of $\pi$ .

C

Varied Practice

D

Word Problems

1.

A spherical water tank at a summer camp has a radius of 9 m. Camp staff need to know how much water it can hold.

Using the sphere volume formula $V = \frac{4}{3}\pi r^3$ , find the exact volume of the tank in terms of $\pi$ .

2.

Tomás is verifying the sphere volume formula by using the Cavalieri argument. He uses a hemisphere of radius $r = 6$ cm compared to a cylinder (radius 6 cm, height 6 cm) with a cone (radius 6 cm, height 6 cm) removed from the interior.

1.

Compute the volume of the cylinder in terms of $\pi$ .

2.

Compute the volume of the cone in terms of $\pi$ .

3.

By Cavalieri's principle, the hemisphere volume equals the cylinder volume minus the cone volume. Use this to find the full sphere volume in terms of $\pi$ .

3.

Two stacks of books have the same height. Each book in the first stack is rectangular with a base of 100 cm². Each book at the same level in the second stack is circular with an area of 100 cm². Both stacks have 20 books.

By Cavalieri's principle, if the first stack has total volume 2000 cm³, what is the volume of the second stack?

E

Error Analysis

1.

A student is computing the annular cross-sectional area of the cylinder-minus-cone comparison solid at height $h$ .

Student's work:

"The cylinder has outer radius $r$ , and the cone's cross-section at height $h$ has inner radius $(r - h)$ (since the cone gets smaller as you go up). So the annulus area is:
$A_{\text{annulus}}(h) = \pi r^2 - \pi(r-h)^2$
This is not equal to $\pi(r^2 - h^2)$ , so Cavalieri's principle cannot apply."

What error did the student make? Which choice correctly identifies the mistake and gives the right annulus area?

A.

The student used the wrong hemisphere cross-section formula.

B.

The student oriented the cone incorrectly. The cone's apex is at the base ( $h = 0$ ), so its radius grows as $h$ increases: inner radius $= h$ . The correct annulus area is $\pi r^2 - \pi h^2 = \pi(r^2 - h^2)$ .

C.

The student should use $h^2$ for the outer radius, not $r^2$ .

D.

The Pythagorean Theorem cannot be applied here, so the cross-section formula is different.

2.

A student is deriving the sphere volume from the Cavalieri argument.

Student's work:

"By Cavalieri's principle, the hemisphere has the same volume as the cylinder minus the cone.
$V_{\text{cylinder}} = \pi r^3$
$V_{\text{cone}} = \frac{1}{3}\pi r^3$
So the sphere volume is $\pi r^3 - \frac{1}{3}\pi r^3 = \frac{2}{3}\pi r^3$ ."

What error did the student make?

A.

The student computed the cylinder volume incorrectly.

B.

The student forgot to double the hemisphere to get the full sphere. The result $\frac{2}{3}\pi r^3$ is the hemisphere volume. The sphere volume is $2 \times \frac{2}{3}\pi r^3 = \frac{4}{3}\pi r^3$ .

C.

The student used the wrong formula for the cone; the cone factor should be $\frac{1}{2}$ , not $\frac{1}{3}$ .

D.

Cavalieri's principle cannot be applied here because the hemisphere and the cylinder-minus-cone have different shapes.

F

Challenge

1.

The sphere volume formula is $V = \frac{4}{3}\pi r^3$ . Trace the arithmetic of the Cavalieri derivation to explain, in your own words, where the fraction $\frac{4}{3}$ comes from. Your explanation should identify the two steps that produce this factor.

2.

An oblique prism and a right rectangular prism both have a rectangular base of length 8 cm and width 5 cm, and both have vertical height 12 cm. The oblique prism's lateral faces are tilted at an angle.