Spheres - ISEE Upper Level (grades 9-12) Quantitative Reasoning

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Question

The volume of a sphere is one cubic yard. Give its radius in inches.

Answer

The volume of a sphere with radius is

$V = \frac{4}{3} \pi r^{3}$ .

To find the radius in yards, we set and solve for .

$1 = \frac{4}{3} \pi r^{3}$

$\frac{3}{4} \cdot 1 =\frac{3}{4} \cdot \frac{4}{3} \pi r^{3}$

$\frac{3}{4} = \pi r^{3}$

$\frac{1}{\pi} \cdot \frac{3}{4} =\frac{1}{\pi} \cdot \pi r^{3}$

$\frac{3}{4\pi} = r^{3}$

$r = \sqrt[3]{ \frac{3}{4\pi} } = \sqrt[3]{ \frac{3\cdot 2 \pi^{2}}{4\pi \cdot 2 \pi^{2}} }= \sqrt[3]{ \frac{6 \pi^{2}}{8 \pi^{3}} } = \frac{ \sqrt[3]{6 \pi^{2}} } {2\pi}$ yards.

Since the problem requests the radius in inches, multiply by 36:

$36 \times \frac{ \sqrt[3]{6 \pi^{2}} } {2\pi} = \frac{ 18\sqrt[3]{6 \pi^{2}} } {\pi}$

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Question

In terms of $\pi$ , give the surface area, in square inches, of a spherical water tank with a diameter of 20 feet.

Answer

feet = $20 \times 12 = 240$ inches, the diameter of the tank; half of this, or 120 inches, is the radius. Set , substitute in the surface area formula, and solve for :

$A =4\pi r^{2}$

$A =4\pi \cdot 120^{2}$

$A =4\pi \cdot 14,400$

$A = 57,600 \pi$

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Question

Which is the greater quantity?

(a) The surface area of a sphere with radius 1

(b) 12

Answer

The surface area of a sphere can be found using the formula

$A = 4\pi r^{2}$ .

The surface area of the given sphere can be found by substituting :

$A = 4\pi \cdot 1^{2} = 4\pi$

$\pi > 3$ so $4 \times \pi >4 \times 3$ , or $4\pi > 12$

This makes (a) greater.

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Question

Sphere A has volume $972 \pi \textrm{ in}^{3}$ . Sphere B has surface area $400 \pi \textrm{ in}^{2}$ . Which is the greater quantity?

(a) The radius of Sphere A

(b) The radius of Sphere B

Answer

(a) Substitute $V = 972 \pi$ in the formula for the volume of a sphere:

$\frac{4}{3}\pi r^{3}= V$

$\frac{4}{3}\pi r^{3}= 972 \pi$

$\frac{4}{3}\pi r^{3} \div \frac{4}{3}\pi = 972 \pi \div \frac{4}{3}\pi$

$\frac{4}{3}\pi r^{3} \cdot \frac{3}{4\pi} = 972 \pi\cdot \frac{3}{4\pi}$

$r^{3} = 729$

$r =\sqrt[3]{ 729} = 9$ inches

(b) Substitute $A = 400 \pi$ in the formula for the surface area of a sphere:

$4\pi r^{2}= A$

$4\pi r^{2}= 400 \pi$

$4\pi r^{2} \div 4\pi = 400 \pi \div 4\pi$

$r^{2} = 100$

$r = \sqrt{100} = 10$ inches

(b) is greater.

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Question

is a positive number. Which is the greater quantity?

(A) The surface area of a sphere with radius

(B) The surface area of a cube with edges of length

Answer

The surface area of a sphere is $4 \pi$ times the square of its radius, which here is ; the surface area of the sphere in (A) is $4 \pi t^{2}$ .

The area of one face of a cube is the square of the length of an edge, which here is , so the area of one face of the cube in (B) is $(2t) ^{2} = 4t^{2}$ . The cube has six faces so the total surface area is $6 \cdot 4t^{2} = 24t^{2}$ .

$\pi < 4$ , so $4 \pi t^{2} < 16 t^{2} < 24 t^{2}$ , giving the sphere less surface area. (B) is greater.

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Question

In terms of $\pi$ , give the volume, in cubic feet, of a spherical tank with diameter 36 inches.

Answer

36 inches = $36 \div 12 = 3$ feet, the diameter of the tank. Half of this, or $\frac{3}{2}$ feet, is the radius. Set $r = \frac{3}{2}$ , substitute in the volume formula, and solve for :

$V = \frac{4}{3} \pi r^{3}$

$V = \frac{4}{3} \pi\left \cdot \left( \frac{3}{2} \right ) ^{3}$

$V = \frac{4}{3}\cdot \frac{3}{2} \cdot \frac{3}{2} \cdot \frac{3}{2} \cdot \pi\left$

$V = \frac{1}{1}\cdot \frac{1}{1} \cdot \frac{3}{1} \cdot \frac{3}{2} \cdot \pi\left$

$V = \frac{9}{2} \pi$

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Question

Which is the greater quantity?

(a) The volume of a sphere with radius

(b) The volume of a cube with sidelength

Answer

A sphere with radius has diameter and can be inscribed inside a cube of sidelength . Therefore, the cube in (b) has the greater volume.

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Question

Which is the greater quantity?

(a) The volume of a cube with sidelength inches.

(b) The volume of a sphere with radius inches.

Answer

You do not need to calculate the volumes of the figures. All you need to do is observe that a sphere with radius inches has diameter inches, and can therefore be inscribed inside the cube with sidelength inches. This give the cube larger volume, making (a) the greater quantity.

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Question

Which is the greater quantity?

(a) The volume of a sphere with diameter one foot

(b) $864 \textrm{ in}^{3}$

Answer

The radius of the sphere is one half of its diameter of one foot, which is six inches, so substitute :

$V = \frac{4}{3} \pi r ^{3}$

$V = \frac{4}{3} \pi \cdot 6 ^{3}$

$V = \frac{4}{3} \pi \cdot 216$

$V = 288 \pi \approx 288 \cdot 3.14 = 904.32$ cubic inches,

which is greater than $864 \textrm{ in}^{3}$ .

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Question

is a positive number. Which is the greater quantity?

(A) The volume of a cube with edges of length

(B) The volume of a sphere with radius

Answer

No calculation is really needed here, as a sphere with radius - and, subsequently, diameter - can be inscribed inside a cube of sidelength . This makes (A), the volume of the cube, the greater.

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Question

Which is the greater quantity?

(a) The radius of a sphere with surface area $36 \pi$

(b) The radius of a sphere with volume $36 \pi$

Answer

The formula for the surface area of a sphere, given its radius , is

$A = 4 \pi r^{2}$

The sphere in (a) has surface area $36 \pi$ , so

$36 \pi = 4 \pi r^{2}$

$36 \pi\div 4 \pi = 4 \pi r^{2} \div 4 \pi$

$9 = r^{2}$

$r = \sqrt{9} = 3$

The formula for the volume of a sphere, given its radius , is

$V = \frac{4}{3} \pi r^{3}$

The sphere in (b) has volume $36 \pi$ , so

$36 \pi = \frac{4}{3} \pi r^{3}$

$\frac{3} {4} \cdot 36 \pi = \frac{3} {4} \cdot \frac{4}{3} \pi r^{3}$

$27 \pi = \pi r^{3}$

$27 \pi\div \pi = \pi r^{3} \div \pi$

$27 = r^{3}$

$r = \sqrt[3]{27} = 3$

The radius of both spheres is 3.

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