The volume of a sphere = \(\displaystyle \frac{4}{3}\pi r^{3}\)

Radius is \(\displaystyle \frac{1}{2}\) of the diameter so the radius = 5.

\(\displaystyle V=\frac{4}{3}\pi5^{3}\)

or \(\displaystyle V=\frac{500}{3\pi}\)

which is approximately \(\displaystyle 167\pi\)

To find the volume of a sphere we use the sphere volume formula:

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

First we need to find the radius of the sphere. A sphere has a radius of half the diameter. So we see that \(\displaystyle 12/2 =6=r\).

Next we plug 6 in for our radius and get
\(\displaystyle V = \frac{4}{3}\pi 6^3=288\pi in^3\) 

(don't forget your units).

The formula for the volume of a sphere is:
\(\displaystyle v = \frac{4}{3}\pi * r^3\), thus we need to just determine the radius and plug it into the equation. Remember that \(\displaystyle 2r = d\) and so 

\(\displaystyle 6 = 2r \newline 3 = r\)

And plugging in we get \(\displaystyle \frac{108}{3}\pi \textup{in}^3\)

First find the radius from the surface area set the given surface area equal to the surface area formula and solve for the radius.
\(\displaystyle 576\pi = 4\pi r^2\)

\(\displaystyle 144\pi = \pi r^2 \newline 144 = r^2 \newline 12 = r\)

Now plug the radius into the volume formula:
\(\displaystyle V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi 12^3 = 2304 \pi\textup{ ft}^3\)

Define the radius of Autumn’s orange as r. The volume of her orange is \(\displaystyle \pi r^3\). Ariana’s orange has twice the radius of Autumn’s, so the radius of her orange is \(\displaystyle 2r\), and the volume is \(\displaystyle \pi(2r)^3 = 8\pi r^3\), which is 8 times larger than Autumn’s orange.

To solve, simply use the formula for the volume of a sphere. Thus,

\(\displaystyle V=\frac{4}{3}\pi{r^3}=\frac{4}{3}*\pi*2^3=\frac{4}{3}*8*\pi=\frac{32}{3}\pi\)

The volume of a cylinder is:

\(\displaystyle V=\pi r^{2}h\)

You can think of the volume as the area of the base times the height. Since it is given that the diameter is twice the length of the height, the radius (half the diameter) equals the height. If it helps to visualize these dimensions, draw the cylinder described.

The equation can be rewritten, using the height in terms of the radius.

\(\displaystyle h=r\)

\(\displaystyle V=\pi r^2(r)\)

\(\displaystyle V=\pi r^{3}\)

Plug in the given volume to solve for the radius.

\(\displaystyle 64\pi=\pi r^{3}\)

\(\displaystyle 64=r^{3}\)

\(\displaystyle 4=r\)

To solve this question, you must remember that the formula for volume is the product of the area of the base and the height. The area of the base of this cylinder is \(\displaystyle \pi r^2\).

\(\displaystyle V=A_b*h=\pi r^2h\)

Plug in the given radius and height to solve.

\(\displaystyle V=\pi(5)^2(9)\)

\(\displaystyle V=\pi(25)(9)\)

\(\displaystyle V=225\pi\)

The volume of a cylinder is given by the formula: \(\displaystyle V=\pi r^2h\).

For a shape with a hole through the center, the final volume is equal to the total volume of the shape minus the volume of the inner hole. In this question, we are looking for the volume given by the larger radius minus the volume given by the smaller radius. The height is equal to the thickness of the washer.

\(\displaystyle V=\pi r_1^2h-\pi r_2^2h\)

\(\displaystyle V=\pi(8)^2(0.5)-\pi(2)^2(0.5)\)

\(\displaystyle V=\pi(64)(0.5)-\pi(4)(0.5)\)

\(\displaystyle V=32\pi-2\pi\)

\(\displaystyle V=30\pi\ in^3\)

Cylinder:  \(\displaystyle V=Bh\)          Cone:  \(\displaystyle V=\frac{1}{3}Bh\)

Thus the difference is 2/3Bh and that means a cylinder can hold 2/3Bh more given the same radius and height.