Spheres - Math

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

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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}$

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

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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}$$

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

$A = 57,600 \pi$

Which is the greater quantity?

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

(b) 12

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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.

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

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(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 $\frac{3}{4\pi}$

$r =\sqrt[3]{ 729} = 9$ 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 = $\sqrt{100}$ = 10$ inches

(b) is greater.

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

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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.

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

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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$

Which is the greater quantity?

(a) The volume of a sphere with radius

(b) The volume of a cube with sidelength

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

Which is the greater quantity?

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

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

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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.

Which is the greater quantity?

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

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

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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}$$ .

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

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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.

Which is the greater quantity?

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

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

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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.

Find the diameter of a sphere if the volume is .

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Write the volume for the sphere.

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

Substitute the volume and solve for the radius.

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

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

$r=\sqrt[3]{$\frac{3}{4}$\pi}$

Double the radius to find diameter.

$d=2\sqrt[3]{$\frac{3}{4}$\pi}$

The surface area of a sphere is $100\pi ; $in^{2}$$ . What is the diameter of the sphere?

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The surface area of a sphere is given by $SA=4\pi $r^{2}$$

So the equation to sovle becomes $4\pi $r^{2}$=100\pi$ or so

To answer the question we need to find the diameter:

$d=2r=2\cdot 5=10; in$

A company wants to construct an advertising balloon spherical in shape. It can afford to buy 28,000 square meters of material to make the balloon. What is the largest possible diameter of this balloon (nearest whole meter)?

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This is equivalent to asking the diameter of a balloon with surface area 28,000 square meters.

The relationship between the surface area and the radius is:

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

To find the radius, substitute for the surface area, then solve:

$28,000=4\pi $r^{2}$$

$r=$\sqrt{\frac{28,000}{4\pi }$}\approx47$

To find the diameter , double the radius—this is 94.

If the volume of a sphere is , what is the sphere's diameter?

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Write the formula for the volume of a sphere:

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

Plug in the volume and find the radius by solving for :

$2=\frac{4}{3}$\pi $r^3$$

Start solving for by multiplying both sides of the equation by :

$2\cdot3=\frac{4}{3}$\pi $r^3$\cdot3$

$6=4\pi $r^3$$

Now, divide each side of the equation by $4\pi$ :

$\frac{6}{4\pi}$=\frac{4\pi $r^3$}{4\pi}$

Reduce the left side of the equation:

Finally, take the cubed root of both sides of the equation:

$\sqrt[3]{$\frac{3}{2\pi}$}=r$

Keep in mind that you've solved for the radius, not the diameter. The diameter is double the radius, which is: $2\sqrt[3]{$\frac{3}{2\pi}$}$ .

The circumference of a sphere is . Find the radius.

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If the circumference of a sphere is , that means we can easily solve for the diameter by using $C=\pi \cdot d$ . This is the equation to find the circumference.

By substituting in the value for circumference (), we can solve for the missing variable :

$84.823:in= \pi \cdot d$

$$\frac{84.823:in}{ \pi}$= d$

To find the radius of the sphere, we need to divide this value by :

$r=\frac{d}{2}$=\frac{27:in}{2}$=13.5:in$

Find the diameter of a sphere if the volume is $\pi$ .

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Write the formula for the volume of a sphere.

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

Plug in the given volume and solve for the radius.

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

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

$r=\sqrt[3]{$\frac{3}{4}$}$

The diameter is double the radius.

$d=2\sqrt[3]{$\frac{3}{4}$}$

Find the diameter of a sphere with the volume listed below.

$V=288\pi$

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In order to solve, we must the given volume into the volume formula for a sphere.

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

$288\pi =\frac{4}{3}\pi(r^{3})$

Divide both sides by pi.

$288 =\frac{4}{3}\(r^{3})$

Multiply both sides by 3.

$216 =(r^{3})$

Take the cube root of both sides to find the radius.

Double the radius to get the diameter, diameter is 12.

Find the diameter of a sphere if it has a volume of $16\pi$ .

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Recall how to find the volume of a sphere:

$\text{Volume of Sphere}$=\frac{4}{3}$\pi $r^3$ , where is the radius of the sphere.

Now, since the radius is half the diameter, the equation for the volume of a sphere can be rewritten as thus:

$\text{Volume of $Sphere}$=\frac{4}{3}$\pi($\frac{d}{2}$)^3$=\frac{1}{6}$\pi $d^3$ , where is the diameter of the sphere.

Rewrite the equation to solve for .

$d=\sqrt[3]{$\frac{6(\text{Volume of Sphere}$)}{\pi}}$

Now, plug in the volume of the sphere to find the diameter.

$d=\sqrt[3]{$\frac{6(16\pi)}{\pi}$}=4.58$

Find the diameter of a sphere if it has a volume of $24\pi$ .

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Recall how to find the volume of a sphere:

$\text{Volume of Sphere}$=\frac{4}{3}$\pi $r^3$ , where is the radius of the sphere.

Now, since the radius is half the diameter, the equation for the volume of a sphere can be rewritten as thus:

$\text{Volume of $Sphere}$=\frac{4}{3}$\pi($\frac{d}{2}$)^3$=\frac{1}{6}$\pi $d^3$ , where is the diameter of the sphere.

Rewrite the equation to solve for .

$d=\sqrt[3]{$\frac{6(\text{Volume of Sphere}$)}{\pi}}$

Now, plug in the volume of the sphere to find the diameter.

$d=\sqrt[3]{$\frac{6(24\pi)}{\pi}$}=5.24$