To solve for the volume of a sphere, you must first know the equation for the volume of a sphere.

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

In this equation, \(\displaystyle r\) is equal to the radius. We can plug the given radius from the question into the equation for \(\displaystyle r\).

\(\displaystyle V=\frac{4}{3}(\pi)(12^{3})\)

Now we simply solve for \(\displaystyle V\).

\(\displaystyle V=\frac{4}{3}(\pi)(1728)\)

\(\displaystyle V=(\pi)(2304)=2304\pi\)

The volume of the sphere is  \(\displaystyle 2304\pi\). 

To solve for the volume of a sphere you must first know the equation for the volume of a sphere.

The equation is

Then plug the radius into the equation for \(\displaystyle r\) yielding 

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

Then cube the radius to get

 \(\displaystyle V=\frac{4}{3}(64)\pi\)

Multiply the answer by \(\displaystyle \frac{4}{3}\) and  to yield \(\displaystyle 85.3\pi\).

The answer is \(\displaystyle 85.3\pi\).

Given the surface area, we can solve for the radius and then solve for the volume.

4πr² = 256π

4r² = 256

r² = 64

r = 8

Now solve the volume equation, substituting for r:

V = (4/3)π(8)³

V = (4/3)π*512

V = (2048/3)π

V = 683π

In order to find the volume of a sphere, use the formula

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

We were given the baseball's diameter, \(\displaystyle \dpi{100} D=76mm\), which must be converted to its radius.

\(\displaystyle D=2r\)

\(\displaystyle 76mm=2r\)

\(\displaystyle r=\frac{76mm}{2}\)

\(\displaystyle \rightarrow 38mm\)

Now we can solve for volume.

\(\displaystyle V=\frac{4}{3}\pi (38mm)^{3}\)

\(\displaystyle V=\frac{4}{3}\pi (54872mm^{3})\)

\(\displaystyle V=73162\frac{2}{3}mm^{3}*\pi\)

\(\displaystyle \rightarrow 229847.30mm^{3}\)

Convert to centimeters.

\(\displaystyle \dpi{100} \frac{229847.30mm^{3}}{1}*\frac{1cm^{3}}{1000mm^{3}}\approx 230cm^{3}\)

If you arrived at \(\displaystyle 1838cm^{3}\) then you did not convert the diameter to a radius.

In order to find the volume of a sphere, use the formula

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

We were given the radius of the sphere, \(\displaystyle r=1.6in\).Therefore, we can solve for volume.

\(\displaystyle \dpi{100} V=\frac{4}{3}\pi (1.6in)^{3}\)

\(\displaystyle \dpi{100} V=\frac{4}{3}\pi (4.096in^{3})\)

\(\displaystyle \dpi{100} V=17.16in^{3}\)

If you calculated the volume to be \(\displaystyle \dpi{100} 9.65in^{3}\) then you multiplied by \(\displaystyle \frac{3}{4}\) rather than by \(\displaystyle \frac{4}{3}\). 

The surface area of a sphere in terms of its radius \(\displaystyle r\) is 

\(\displaystyle A = 4\pi r^2\)

Substitute \(\displaystyle A = 1,000\) and solve for \(\displaystyle r\):

\(\displaystyle 4\pi r^2 =A\)

\(\displaystyle 4\pi r^2 =1,000\)

\(\displaystyle r^2 =\frac{1,000}{4\pi } = \frac{250}{\pi }\)

\(\displaystyle r =\sqrt{ \frac{250}{\pi }} \approx 8.92 \textrm{ cm }\)

Substitute for \(\displaystyle r\) in the formula for the volume of a sphere:

\(\displaystyle V = \frac{4\pi r^3}{3} \approx\frac{4\pi \cdot 8.92^3}{3} \approx 2,973.5 \textrm{ cm }^{3}\)

The formula for the volume of a sphere is:

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

where \(\displaystyle r\) is the radius of the sphere.

Plugging in our values, we get:

\(\displaystyle V=\frac{4}{3} \pi (6m)^3\)

\(\displaystyle V=\frac{4}{3} \pi 216m^3 = 288 \pi m^3\)

The formula for the volume of a sphere is:

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

Where \(\displaystyle r\) is the radius of the sphere

Plugging in our values, we get:

\(\displaystyle V = \frac{4 \pi (9m)^3}{3}\)

\(\displaystyle V = 972 \pi m^3\)

To find your answer, we would use the formula:  C=2πr. We are given that C = 29.5. Thus we can plug in to get  [29.5]=2πr and then multiply 2π to get 29.5=(6.28)r.  Lastly, we divide both sides by 6.28 to get 4.70=r. Then we would plug into the formula for volume V=(4π〖(4.7)〗³) / 3   (The information given of 22 ounces is useless) 

The formula for volume of a sphere is \(\displaystyle V=\frac{4}{3}\pi r^3\).

The problem gives us the diameter, however, and not the radius. Since the diameter is twice the radius, or \(\displaystyle d=2r\), we can find the radius.

\(\displaystyle 12=2r\)

\(\displaystyle \frac{12}{2}=r\)

\(\displaystyle 6=r\).

Now plug that into our initial equation.

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

\(\displaystyle V=\frac{4}{3}\pi (6)^3\)

\(\displaystyle V=\frac{4}{3}\pi(216)\)

\(\displaystyle V=288\pi\)