SAT Math - How to find the volume of a sphere

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?

128 units

16 units

8 units

4 units

2 units

Explanation

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.

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?

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

$1000-\frac{4}{3}\pi$

$2744-\frac{4}{3}\pi$

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

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

Explanation

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:

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

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.

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

$\frac{10, 976 }{3}\pi$

$464 \pi$

$784 \pi$

$196 \pi$

Explanation

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

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

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

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

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

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

$36\pi\ mm^3$

$4\pi\ mm^3$

$9\pi\ mm^3$

$3\pi\ mm^3$

$6\pi\ mm^3$

Explanation

$SA=4\pi r^2$

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

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

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

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

$V=4(9)\pi$

$V=36\pi$

Spheres element76

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

$\frac{500\pi}{3} in.^3$

$100\pi$

$\frac{100\pi}{3}$

$\frac{400\pi}{3}in.^3$

$\frac{125\pi}{3}in.^3$

Explanation

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 $\frac{500\pi}{3} in.^3$

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?

$\frac{9}{2r}$

$3r$

$\frac{4r}{3}$

$\frac{9r}{2}$

$\frac{3}{r}$

Explanation

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 $\frac{9}{2r}$ .

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

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

$\frac{25}{3}\pi$

$100\pi$

$\frac{400}{3}\pi$

$400\pi$

Explanation

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

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

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

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

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?

None of the other answers

2040_π_√(7)

816_π_√(11)

4896_π_

4896_π_√(17)

Explanation

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)

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?

59.813 in3

64 in3

4.187 in3

11.813 in3

Explanation

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

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?

$\frac{27}{6}\pi$

$6\pi$

$\frac{27}{12}\pi$

$3\pi$

None of the given answers.

Explanation

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:

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

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

How to find the volume of a sphere

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