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Spheres and Hemispheres: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A central circle

A sphere has diameter 7.2 cm. Find its radius and the exact circumference of a great circle. State why the great circle uses that radius.
1. Hint 1
  A great circle lies in a plane through the sphere center.
2. Hint 2
  Find the radius from the supplied diameter, then use the circle circumference formula.
Answer

Radius 3.6 cm; great-circle circumference $7.2\pi$ cm. The great circle has the sphere radius because its plane passes through the center.

Full solution

A radius is half a diameter, so the sphere radius is 3.6 cm.

A plane through the center cuts the sphere in a circle with the same radius.
Its circumference is
$C=7.2\pi\text{ cm}.$

Answer

Radius 3.6 cm; great-circle circumference $7.2\pi$ cm. The great circle has the sphere radius because its plane passes through the center.

Key idea

A great circle keeps the sphere radius, while its circumference remains a length.
Problem 2 Two measurements of one ball

A ball has radius 1.5 cm. Find its exact volume and the area of its boundary sphere.
1. Hint 1
  Volume measures the region inside; surface area measures the boundary.
2. Hint 2
  Cube the radius for volume and square it for surface area.
Answer

Volume $\frac{9\pi}{2}\text{ cm}^3$ ; boundary area $9\pi\text{ cm}^2$ .

Full solution

The volume formula uses three factors of the radius.
$\begin{aligned}V&=\frac43\pi\left(\frac32\right)^3\\V&=\frac{9\pi}{2}\text{ cm}^3.\end{aligned}$

The boundary sphere uses the square of the radius.
$\begin{aligned}S&=4\pi\left(\frac32\right)^2\\S&=9\pi\text{ cm}^2.\end{aligned}$

Answer

Volume $\frac{9\pi}{2}\text{ cm}^3$ ; boundary area $9\pi\text{ cm}^2$ .

Key idea

Keep square units for area and cubic units for volume.
Problem 3 Three hemisphere measurements

A solid hemisphere has radius 3.5 cm. Give exact answers for each case.

Case A: The curved boundary area, excluding the flat base.

Case B: The total boundary area, including the flat base.

Case C: The enclosed volume.

Text description of this figure

A hemisphere with base radius labeled 3.5 cm. The circular rim is drawn as an ellipse in perspective. The hidden back half of the rim is dashed. No computed area or volume is shown.
1. Hint 1
  Separate the curved cap from the flat disk before choosing the area formulas.
2. Hint 2
  The curved area is half a sphere area; the total adds one disk. The volume is half a ball volume.
Answer

$\begin{gathered}\text{A: }\frac{49\pi}{2}\text{ cm}^2\\\text{B: }\frac{147\pi}{4}\text{ cm}^2\\\text{C: }\frac{343\pi}{12}\text{ cm}^3.\end{gathered}$

Full solution

Case A: Half the sphere surface gives
$\begin{aligned}S_c&=2\pi\left(\frac72\right)^2\\S_c&=\frac{49\pi}{2}\text{ cm}^2.\end{aligned}$

Case B: Add the flat disk, giving three times the disk area.
$\begin{aligned}S_t&=3\pi\left(\frac72\right)^2\\S_t&=\frac{147\pi}{4}\text{ cm}^2.\end{aligned}$

Case C: Use the half-ball volume.
$\begin{aligned}V_H&=\frac23\pi\left(\frac72\right)^3\\V_H&=\frac{343\pi}{12}\text{ cm}^3.\end{aligned}$

Answer

$\begin{gathered}\text{A: }\frac{49\pi}{2}\text{ cm}^2\\\text{B: }\frac{147\pi}{4}\text{ cm}^2\\\text{C: }\frac{343\pi}{12}\text{ cm}^3.\end{gathered}$

Key idea

A hemisphere has half the sphere volume, but its whole boundary includes a new disk.
Problem 4 Information in the equator disk

The flat base of a solid hemisphere has area $81\pi\text{ cm}^2$ . Find the hemisphere radius and its exact volume.
1. Hint 1
  The flat base is a disk with the hemisphere radius.
2. Hint 2
  Use the base area to find the positive radius before calculating half-ball volume.
Answer

Radius 9 cm; volume $486\pi\text{ cm}^3$ .

Full solution

The base disk gives
$\begin{aligned}\pi r^2&=81\pi\\r^2&=81\\r&=9\text{ cm}.\end{aligned}$

The hemisphere volume is
$\begin{aligned}V_H&=\frac23\pi(9)^3\\V_H&=486\pi\text{ cm}^3.\end{aligned}$

Answer

Radius 9 cm; volume $486\pi\text{ cm}^3$ .

Key idea

The equator disk determines the radius, even when no length is supplied.
Problem 5 From a great circle to the whole sphere

A great circle of a sphere has circumference $11\pi$ cm. Find the sphere area and enclosed volume exactly.
1. Hint 1
  The great circle and the sphere share one radius.
2. Hint 2
  Recover that radius from circumference, then use it in both sphere formulas.
Answer

Area $121\pi\text{ cm}^2$ ; volume $\frac{1331\pi}{6}\text{ cm}^3$ .

Full solution

The circumference determines the positive radius.
$\begin{aligned}2\pi r&=11\pi\\r&=\frac{11}{2}\text{ cm}.\end{aligned}$

Use the squared radius for area.
$\begin{aligned}S&=4\pi\left(\frac{11}{2}\right)^2\\S&=121\pi\text{ cm}^2.\end{aligned}$

Use the cubed radius for volume.
$\begin{aligned}V&=\frac43\pi\left(\frac{11}{2}\right)^3\\V&=\frac{1331\pi}{6}\text{ cm}^3.\end{aligned}$

Answer

Area $121\pi\text{ cm}^2$ ; volume $\frac{1331\pi}{6}\text{ cm}^3$ .

Key idea

A central circumference can supply the missing radius for both three-dimensional measurements.
Problem 6 Cover the curved dome

An ideal hemispherical dome has a circular rim of diameter 1.6 m. A covering goes over the outside curved cap only; it does not cover the circular opening. Ignore thickness, seams, and waste.

Find the exact area to cover. Use 3.14 for pi to estimate that area to the nearest hundredth of a square meter. Is a supply covering exactly 4 square meters sufficient under this estimate?

Text description of this figure

A hemisphere with base diameter labeled 1.6 m. The circular rim is drawn as an ellipse in perspective. The hidden back half of the rim is dashed. No computed area or volume is shown.
1. Hint 1
  The circular opening contributes no covered area.
2. Hint 2
  Halve the rim diameter before using the curved hemisphere area. Compare the estimated area with the supply.
Answer

Exact area $1.28\pi\text{ m}^2$ ; estimated area $4.02\text{ m}^2$ . No, a 4-square-meter supply is insufficient.

Full solution

The radius is half the 1.6 m diameter, or 0.8 m.

Only the curved cap is covered.
$\begin{aligned}S_c&=2\pi(0.8)^2\\S_c&=1.28\pi\text{ m}^2.\end{aligned}$

Using the specified approximation gives
$S_c\approx4.0192\text{ m}^2.$
This rounds to 4.02 square meters.

The estimate is greater than the 4 square meters available, so the supply is insufficient even with no allowance for waste.

Answer

Exact area $1.28\pi\text{ m}^2$ ; estimated area $4.02\text{ m}^2$ . No, a 4-square-meter supply is insufficient.

Key idea

Choose the actual covered boundary before comparing an area with available material.
Problem 7 Reconstruct the ball from a slice

A plane cuts a sphere in a circle of radius 12 cm. The perpendicular distance from the sphere center O to the plane is 5 cm. Let M be the perpendicular foot and P any point on the section circle.

Find the sphere radius and the exact volume of its ball.

Text description of this figure

Right triangle OMP has right angle at M. O is the sphere center, M is the perpendicular foot in the cutting plane, and P lies on the section circle. Vertical OM is 5 cm. Horizontal MP is 12 cm. The hypotenuse OP is labeled R, the unknown sphere radius.
1. Hint 1
  The section radius and perpendicular plane distance form legs of a right triangle.
2. Hint 2
  The sphere radius is its hypotenuse. Find that length before calculating the ball volume.
Answer

Radius 13 cm; ball volume $\frac{8788\pi}{3}\text{ cm}^3$ .

Full solution

M is the section circle center.
The perpendicular OM and the section radius MP meet at a right angle, while OP is a sphere radius.

Add the leg squares.
$\begin{aligned}R^2&=5^2+12^2\\R^2&=169\\R&=13\text{ cm}.\end{aligned}$

Then find the whole ball volume.
$\begin{aligned}V&=\frac43\pi(13)^3\\V&=\frac{8788\pi}{3}\text{ cm}^3.\end{aligned}$

Answer

Radius 13 cm; ball volume $\frac{8788\pi}{3}\text{ cm}^3$ .

Key idea

An off-center section carries enough right-triangle information to recover the sphere radius.
Problem 8 Why the comparison gives a half-ball

A hemisphere has any positive radius a. A right cylinder has radius a and height a. Remove the cone whose apex is the cylinder bottom center and whose base fills the cylinder top. Height z is measured upward from the common bottom level, from zero to a.

Explain why all corresponding horizontal section areas agree. Use that agreement to derive the hemisphere volume in terms of a.

Text description of this figure

Central vertical views of a hemisphere of radius a and a cylinder of radius a and height a. The cylinder has a cone removed whose apex is at the bottom center and whose circular base fills the top. Height z is measured upward from the common bottom level zero. The cone top is at height a.
1. Hint 1
  Find the hemisphere section radius squared using the central right triangle.
2. Hint 2
  Use similarity to express the cone section radius in terms of its distance above the bottom apex.
3. Hint 3
  Check the full height and every section area before applying Cavalieri.
Answer

The common section area is $\pi(a^2-z^2)$ . Cavalieri gives hemisphere volume $\frac23\pi a^3$ .

Full solution

For example, with radius and height 3 cm, a section 1 cm above the base has hemisphere area $8\pi\text{ cm}^2$ .
The comparison section has cylinder area $9\pi\text{ cm}^2$ minus cone area $\pi\text{ cm}^2$ , also $8\pi\text{ cm}^2$ .
The cylinder volume is $27\pi\text{ cm}^3$ , and the cone volume is $9\pi\text{ cm}^3$ .
Their difference is $18\pi\text{ cm}^3$ .

For arbitrary positive a, Pythagoras gives the hemisphere section radius squared as a squared minus z squared.
Thus
$A_H(z)=\pi(a^2-z^2).$

The removed cone widens upward from the base-level apex.
At positive heights, similarity gives section radius z, since its full radius and height are equal.
At height zero the section is the apex, of radius zero.
Therefore
$\begin{aligned}A_C(z)&=\pi a^2-\pi z^2\\A_C(z)&=\pi(a^2-z^2).\end{aligned}$

Both regions have height a, and the sections agree at every corresponding height.
Cavalieri gives the hemisphere the cylinder volume minus the cone volume.
$\begin{aligned}V_H&=\pi a^3-\frac13\pi a^3\\V_H&=\frac23\pi a^3.\end{aligned}$
The argument did not depend on the example radius 3.

Answer

The common section area is $\pi(a^2-z^2)$ . Cavalieri gives hemisphere volume $\frac23\pi a^3$ .

Key idea

Matching every section area over a common height transfers the known volume difference.
Problem 9 What the radial picture approximates

A sphere of positive radius r is divided into many small curved patches. Each patch is joined to the center. To use the ordinary pyramid volume formula, each patch is replaced by a small flat polygon through nearby points of the sphere.

Explain two reasons the sum of these ordinary pyramid volumes is an approximation for a finite partition. State what refinement does and whether this sketch by itself is a complete rigorous proof of the sphere-area formula.
1. Hint 1
  Compare a curved surface patch with a flat polygon.
2. Hint 2
  Compare the radius to a curved point with the perpendicular distance to a flat polygon.
3. Hint 3
  Describe what improves when the patches become smaller.
Answer

The flat bases approximate curved patch areas, and their heights are near r. Refinement improves both approximations. The sketch is an informal limiting argument, not a complete rigorous proof.

Full solution

First, a curved patch is not an ordinary pyramid base.
Replacing it by a flat polygon changes its area, so summing the flat base areas only approximates the sphere area.

Second, an ordinary pyramid height is the perpendicular distance from the center to its flat base plane.
That distance is not exactly the radius to a curved surface point.

As the patches become finer, the flat areas better approximate the curved areas, the perpendicular heights approach r, and the approximating volume sum approaches the ball volume.

This motivates the sphere-area relation, but a complete rigorous proof must control the approximation errors.
The finite sketch is an informal limiting explanation.

Answer

The flat bases approximate curved patch areas, and their heights are near r. Refinement improves both approximations. The sketch is an informal limiting argument, not a complete rigorous proof.

Key idea

A useful limiting explanation must identify which finite quantities are only approximations.
Problem 10 Recover dimensions and the visible boundary

A solid hemisphere is reported to have volume $\frac{686\pi}{3}\text{ cm}^3$ and total boundary area $147\pi\text{ cm}^2$ , including its flat base. All dimensions are positive.

Find its radius and diameter. Check that both reported measurements agree with those dimensions.

The flat base is then placed flush against an opaque table, leaving the whole curved cap uncovered. Find the exposed area.
1. Hint 1
  Either supplied measurement can determine the radius; the other must then check it.
2. Hint 2
  Total hemisphere area includes three disk areas. Use the positive root.
3. Hint 3
  The table hides one flat disk, so the exposed area is the curved cap.
Answer

Radius 7 cm; diameter 14 cm. Both reports agree. Exposed area $98\pi\text{ cm}^2$ .

Full solution

The total area gives
$\begin{aligned}3\pi r^2&=147\pi\\r^2&=49\\r&=7\text{ cm}.\end{aligned}$
The diameter is 14 cm.

Check the reported volume directly.
$\frac23\pi(7)^3=\frac{686\pi}{3}\text{ cm}^3.$

Check the total area separately.
$3\pi(7)^2=147\pi\text{ cm}^2.$
Both measurements agree with the recovered radius.

The table hides the flat base.
Count only the curved cap.
$\begin{aligned}S_e&=2\pi(7)^2\\S_e&=98\pi\text{ cm}^2.\end{aligned}$

Answer

Radius 7 cm; diameter 14 cm. Both reports agree. Exposed area $98\pi\text{ cm}^2$ .

Key idea

Recover and verify dimensions before choosing which part of the boundary is exposed.

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