How to find the volume of a sphere - SAT Math

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Question

A cube with sides of 4” each contains a floating sphere with a radius of 1”. What is the volume of the space outside of the sphere, within the cube?

Answer

Volume of Cube = side3 = (4”)3 = 64 in3

Volume of Sphere = (4/3) * π * r3 = (4/3) * π * 13 = (4/3) * π * 13 = (4/3) * π = 4.187 in3

Difference = Volume of Cube – Volume of Sphere = 64 – 4.187 = 59.813 in3

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Question

The surface area of a sphere is $36\pi\ mm^2$ . Find the volume of the sphere in cubic millimeters.

Answer

$SA=4\pi r^2$

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

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

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

$V=\frac{4}{3}(27)\pi$

$V=4(9)\pi$

$V=36\pi$

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Question

If a sphere's diameter is doubled, by what factor is its volume increased?

Answer

The formula for the volume of a sphere is 4πr3/3. Because this formula is in terms of the radius, it would be easier for us to determine how the change in the radius affects the volume. Since we are told the diameter is doubled, we need to first determine how the change in the diameter affects the change in radius.

Let us call the sphere's original diameter d and its original radius r. We know that d = 2r.

Let's call the final diameter of the sphere D. Because the diameter is doubled, we know that D = 2d. We can substitute the value of d and obtain D = 2(2r) = 4r.

Let's call the final radius of the sphere R. We know that D = 2R, so we can now substituate this into the previous equation and write 2R = 4r. If we simplify this, we see that R = 2r. This means that the final radius is twice as large as the initial radius.

The initial volume of the sphere is 4πr3/3. The final volume of the sphere is 4πR3/3. Because R = 2r, we can substitute the value of R to obtain 4π(2r)3/3. When we simplify this, we get 4π(8r3)/3 = 32πr3/3.

In order to determine the factor by which the volume has increased, we need to find the ratio of the final volume to the initial volume.

32πr3/3 divided by 4πr3/3 = 8

The volume has increased by a factor of 8.

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Question

A sphere increases in volume by a factor of 8. By what factor does the radius change?

Answer

Volume of a sphere is 4Πr3/3. Setting that equation equal to the original volume, the new volume is given as 8*4Πr3/3, which can be rewritten as 4Π8r3/3, and can be added to the radius value by 4Π(2r)3/3 since 8 si the cube of 2. This means the radius goes up by a factor of 2

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Question

A foam ball has a volume of 2 units and has a diameter of x. If a second foam ball has a radius of 2x, what is its volume?

Answer

Careful not to mix up radius and diameter. First, we need to identify that the second ball has a radius that is 4 times as large as the first ball. The radius of the first ball is (1/2)x and the radius of the second ball is 2x. The volume of the second ball will be 43, or 64 times bigger than the first ball. So the second ball has a volume of 2 * 64 = 128.

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Question

A sphere is perfectly contained within a cube that has a volume of 216 units. What is the volume of the sphere?

Answer

To begin, we must determine the dimensions of the cube. This is done by solving the simple equation:

We know the volume is 216, allowing us to solve for the length of a side of the cube.

Taking the cube root of both sides, we get s = 6.

The diameter of the sphere will be equal to side of the cube, since the question states that the sphere is perfectly contained. The diameter of the sphere will be 6, and the radius will be 3.

We can plug this into the equation for volume of a sphere:

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

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

We can cancel out the 3 in the denominator.

$V=4\pi (3)^2$

Simplify.

$V=4\pi (9)$

$V=36\pi$

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Question

A cross-section is made at the center of a sphere. The area of this cross-section is 225_π_ square units. How many cubic units is the total volume of the sphere?

Answer

The solution to this is simple, though just take it step-by-step. First find the radius of the circular cross-section. This will give us the radius of the sphere (since this cross-section is at the center of the sphere). If the cross-section has an area of 225_π_, we know its area is defined by:

A = 225_π_ = _πr_2

Solving for r, we get r = 15.

From here, we merely need to use our formula for the volume of a sphere:

V = (4 / 3)_πr_3

For our data this is: (4 / 3)π * 153 = 4_π_ * 152 * 5 = 4500_π_

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Question

At x = 3, the line y = 4_x_ + 12 intersects the surface of a sphere that passes through the xy-plane. The sphere is centered at the point at which the line passes through the x-axis. What is the volume, in cubic units, of the sphere?

Answer

We need to ascertain two values: The center point and the point of intersection with the surface. Let's do that first:

The center is defined by the x-intercept. To find that, set the line equation equal to 0 (y = 0 at the x-intercept):

0 = 4x + 12; 4_x_ = –12; x = –3; Therefore, the center is at (–3,0)

Next, we need to find the point at which the line intersects with the sphere's surface. To do this, solve for the point with x-coordinate at 3:

y = 4 * 3 + 12; y = 12 + 12; y = 24; therefore, the point of intersection is at (3,24)

Reviewing our data so far, this means that the radius of the sphere runs from the center, (–3,0), to the edge, (3,24). If we find the distance between these two points, we can ascertain the length of the radius. From that, we will be able to calculate the volume of the sphere.

The distance between these two points is defined by the distance formula:

d = √( (_x_1 – _x_0)2 + (_y_1 – _y_0)2 )

For our data, that is:

√( (3 + 3)2 + (24 – 0)2 ) = √( 62 + 242 ) = √(36 + 576) = √612 = √(2 * 2 * 3 * 3 * 17) = 6√(17)

Now, the volume of a sphere is defined by: V = (4/3)_πr_3

For our data, that would be: (4/3)π * (6√(17))3 = (4/3) * 63 * 17√(17) * π = 4 * 2 * 62 * 17√(17) * π = 4896_π_√(17)

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Question

The surface area of a sphere is 676_π_ in2. How many cubic inches is the volume of the same sphere?

Answer

To begin, we must solve for the radius of our sphere. To do this, recall the equation for the surface area of a sphere: A = 4_πr_2

For our data, that is: 676_π_ = 4_πr_2; 169 = _r_2; r = 13

From this, it is easy to solve for the volume of the sphere. Recall the equation:

V = (4/3)_πr_3

For our data, this is: V = (4/3)π * 133 = (4_π_ * 2197)/3 = (8788_π)_/3

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Question

What is the difference between the volume and surface area of a sphere with a radius of 6?

Answer

Surface Area = 4_πr_2 = 4 * π * 62 = 144_π_

Volume = 4_πr_3/3 = 4 * π * 63 / 3 = 288_π_

Volume – Surface Area = 288_π_ – 144_π_ = 144_π_

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Question

A solid hemisphere has a radius of length r. Let S be the number of square units, in terms of r, of the hemisphere's surface area. Let V be the number of cubic units, in terms of r, of the hemisphere's volume. What is the ratio of S to V?

Answer

First, let's find the surface area of the hemisphere. Because the hemisphere is basically a full sphere cut in half, we need to find half of the surface area of a full sphere. However, because the hemisphere also has a circular base, we must then add the area of the base.

$S = \frac{1}{2}\cdot$ (surface area of sphere) + (surface area of base)

The surface area of a sphere with radius r is equal to $4\pi r^2$ . The surface area of the base is just equal to the surface area of a circle, which is $\pi r^2$ .

$S=\frac{1}{2}\cdot 4\pi r^2+\pi r^2=2\pi r^2+\pi r^2=3\pi r^2$

The volume of the hemisphere is going to be half of the volume of an entire sphere. The volume for a full sphere is $\frac{4}{3}\pi r^3$ .

$V = \frac{1}{2}\cdot$ (volume of sphere)

$V = \frac{1}{2}\cdot \frac{4}{3}\pi r^3=\frac{2}{3}\pi r^3$

Ultimately, the question asks us to find the ratio of S to V. To do this, we can write S to V as a fraction.

$\frac{S}{V}=\frac{3\pi r^2}{\frac{2}{3}\pi r^3}$

In order to simplify this, let's multiply the numerator and denominator both by 3.

$\frac{S}{V}=\frac{3\pi r^2}{\frac{2}{3}\pi r^3}$ = $\frac{9\pi r^2}{2\pi r^3}=\frac{9}{2r}$

The answer is $\frac{9}{2r}$ .

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Question

Find the volume of a sphere with a radius of $\frac{3}{5}$ .

Answer

Write the formula to find the volume of the sphere.

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

Substitute the radius and solve for the volume.

$V=\frac{4}{3}\pi (\frac{3}{5})^3=\frac{4}{3}\pi (\frac{27}{125})=\frac{36}{125}\pi$

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Question

Find the volume of a sphere whose diameter is 10cm.

Answer

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

Radius is half of the diameter. Half of 10cm is 5cm.

$Volume=\frac{4}{3}\pi 5^3$

$Volume=\frac{500}{3}\pi$

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Question

Six spheres have volumes that form an arithmetic sequence. The two smallest spheres have radii 4 and 6. Give the volume of the largest sphere.

Answer

The volume of a sphere with radius can be determined using the formula

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

The smallest sphere, with radius , has volume

$V_{1} = \frac{4}{3} \pi \cdot 4 ^{3} = \frac{4}{3} \pi \cdot 64 = \frac{256}{3} \pi$ .

The second-smallest sphere, with radius , has volume

$V_{2} = \frac{4}{3} \pi \cdot 6 ^{3} = \frac{4}{3} \pi \cdot 216= \frac{864}{3} \pi$ .

The volumes are in an arithmetic sequence; their common difference is the difference of these two volumes, or

$d = V_{2} - V_{1}= \frac{864}{3} \pi - \frac{256}{3} \pi = \frac{608}{3} \pi$

Since the six volumes are in an arithmetic sequence, the volume of the largest of the six spheres - that is, the sixth-smallest sphere - is

$V_{6}= V_{1}+ 5d$

$= \frac{256}{3} \pi + 5 \cdot \frac{608}{3} \pi$

$= \frac{256}{3} \pi + \frac{3,040}{3} \pi$

$= \frac{3,296}{3} \pi$

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Question

The radii of six spheres form an arithmetic sequence. The smallest and largest spheres have radii 10 and 30, respectively. Give the volume of the second-largest sphere.

Answer

The radii of the spheres form an arithmetic sequence, with

$r_{1} =1 0$ and $r_{6} =30$

The common difference can be computed as follows:

$r_{6} = (6-1) d + r_{1}$

The second-largest - or fifth-smallest - sphere has radius:

$r_{5} = (5-1) d + r_{1} = 4 \cdot 4 + 10 = 16 + 10 = 26$

The volume of a sphere with radius can be determined using the formula

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

Set :

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

$= \frac{4}{3} \pi \cdot 26^{3}$

$= \frac{4}{3} \pi \cdot 17,576$

$= \frac{70, 304}{3} \pi$

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Question

A sphere fits inside of a cube so that its surface barely touches each side of the cube at any given time. If the volume of the box is 27 cubic centimeters, then what is the volume of the sphere?

Answer

If the volume of the cube is 27 cubic centimeters, then its height, width, and depth are all 3cm. Since the sphere fits perfectly in the cube, then the sphere's diameter is also 3. This means that its radius is $\frac{3}{2}$ .

Substitute this radius value into the equation for the volume of a sphere:

$V=\frac{4}{3}\pi r^3=\frac{4}{3}\cdot\pi \cdot(\frac{3}{2})^3=\frac{4}{3}\cdot\pi\cdot\frac{27}{8}$

$=\frac{108}{24}\pi = \frac{27}{6}\pi$

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Question

A sphere with radius fits perfectly inside of a cube so that the sides of the cube are barely touching the sphere. What is the volume of the cube that is not occupied by the sphere?

Answer

Because the sides of the interior of the cube are tangent to the sphere, we know that the length of each side is equal to the diameter of the sphere. Since the radius of this sphere is , then its diameter is .

To find the volume that is not occupied by the sphere, we will subtract the sphere's volume from the volume of the cube.

The volume of the cube is:

$V=lwh=14\cdot14\cdot14=2744$

The volume of the sphere is:

$V=\frac{4}{3}\pi r^3 = \frac{4}{3}\cdot\pi\cdot7^3 = \frac{1372}{3}\pi$

Therefore, with these values, the volume of the cube not occupied by the sphere is:

$2744-\frac{1372}{3}\pi$

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Question

The surface area of a sphere is 100. Give its volume.

Answer

The surface area of a sphere is equal to

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

where is the radius.

Setting and solving for :

$4 \pi r ^{2} = 100$

$\frac{4 \pi r ^{2}}{4 \pi } = \frac{100}{4 \pi }$

$r ^{2} = \frac{25 }{\pi }$

$r = \sqrt{\frac{25 }{\pi } }$

Applying the Quotient of Radicals Rule:

$r = \frac{\sqrt{25 } }{\sqrt{\pi}}$

$r = \frac{5 }{\sqrt{\pi}}$

The volume of the sphere can be determined from the radius using the formula

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

so, setting $r = \frac{5 }{\sqrt{\pi}}$ :

$V = \frac{4}{3} \pi \cdot \left ( \frac{5 }{\sqrt{\pi}} \right ) ^{3}$

Applying the Quotient of Radicals Rule:

$V = \frac{4}{3} \pi \cdot \frac{5^{3} }{\left ( \sqrt{\pi}\right ) ^{3}}$

$= \frac{4}{3} \pi \cdot \frac{125 }{\pi \sqrt{\pi} }$

$= \frac{500}{3\sqrt{\pi}}$

Rationalizing the denominator:

$V = \frac{500 \cdot \sqrt{\pi}}{3\sqrt{\pi} \cdot \sqrt{\pi}} = \frac{500 \sqrt{\pi}}{3 \pi}$

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Question

is a point on a sphere, and is the point on the sphere farthest from . The shortest curve from to that is completely on the sphere is $24 \pi$ in length.

Give the volume of the sphere.

Answer

The figure referenced is below. is the center of the sphere.

The shortest curve connecting to its opposite point is a semicircle. Also, $\overline{AB}$ is a diameter of the sphere, and $\overline{OA}$ and $\overline{OB}$ are radii.

Given radius , a semicircle has length

$l = \frac{1}{2} C= \frac{1}{2} \cdot 2 \pi r = \pi r$

Setting $l = 24\pi$ and solving for :

$\pi r = 24 \pi$

$\frac{\pi r }{\pi}= \frac{24 \pi}{\pi}$

.

The volume of a sphere, given its radius , is

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

Setting :

$V = \frac{4}{3} \pi \cdot 24 ^{3} = \frac{4}{3} \cdot 13,824 \cdot \pi = 18,432 \pi$

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Question

The radius of the sphere shown here is long. What is its volume?

Answer

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

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

The radius is stated to be long; hence, we can calculate the sphere's volume by substituting this value for into the formula, as shown:

$V=\frac{4\pi (5)^3}{3}=\frac{4\pi(125)}{3}=\frac{500\pi}{3}$

Hence, the volume of the sphere is $\frac{500\pi}{3} in.^3$

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