The formula for the volume of a partial cylinder is:

\(\displaystyle V = (base)(height)(part)\)

\(\displaystyle V = (\pi r^2)(h)\left(\frac{part}{360}\right)\)

where \(\displaystyle r\) is the radius of the cylinder, \(\displaystyle h\) is the height of the cylinder, and \(\displaystyle part\) is the degrees of the sector.

Plugging in our values, we get:

\(\displaystyle V=(\pi (6m)^2)(20m)\left(\frac{270}{360}\right)\)

\(\displaystyle V=540 \pi m^3\)

The total surface area will be equal to the area of the two bases added to the area of the outer surface of the cylinder. If "unwrapped" the area of the outer surface is simply a rectangle with the height of the cylinder and a base equal to the circumference of the cylinder base. We can use these relationships to find a formula for the total area of the cylinder.

\(\displaystyle A=2A_{base}+A_{rectangle}\)

\(\displaystyle A=2(\pi r^2)+(2\pi r)(h)\)

Use the given radius and height to solve for the final area.

\(\displaystyle \small 2\pi(2)^{2} + 2\pi (2)(5)\)

\(\displaystyle \small 8\pi + 20\pi\)

\(\displaystyle \small 28\pi\)

The volume of a cylinder is equal to:

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

Use this formula and the given radius to solve for the height.

\(\displaystyle 81\pi=\pi(3)^2h\)

\(\displaystyle 81\pi=9\pi(h)\)

\(\displaystyle h=9\)

Now that we know the height, we can solve for the surface area. The surface area of a cylinder is equal to the area of the two bases plus the area of the outer surface. The outer surface can be "unwrapped" to form a rectangle with a height equal to the cylinder height and a base equal to the circumference of the cylinder base. Add the areas of the two bases and this rectangle to find the total area.

\(\displaystyle A=2A_{base}+A_{rec}\)

\(\displaystyle A=2\pi r^2+(2\pi r)(h)\)

Use the radius and height to solve.

\(\displaystyle A=2\pi (3)^2+2\pi (3)(9)\)

\(\displaystyle A=18\pi+54\pi\)

\(\displaystyle A=72\pi\)

Area of a circle \(\displaystyle A=\pi r^{2}=\pi(\frac{d}{2})^{2}=\pi(\frac{6}{2})^{2}=\pi(3)^{2}=9\pi in^{2}\)

Circumference of a circle \(\displaystyle C=\pi d=6\pi in^{2}\)

Surface area of a cylinder \(\displaystyle SA=2A+hC=2(9\pi)+8(6\pi)=18\pi+48\pi=66\pi in^{2}\)

The surface area of a sphere is given by the formula

\(\displaystyle S (sphere) = 4\pi*r^2\) , which here equals \(\displaystyle 144\pi\) .   So

\(\displaystyle 4\pi*r^2 = 144\pi\)

\(\displaystyle r^2 = 36\)

\(\displaystyle r = 6\)

To find out what the water level would be, we need to know how much water is in the balloon.  So we need to find the volume of the balloon.  

The amount of water that is in the balloon is

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

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

\(\displaystyle 288\pi\)

The cylindrical column of water will have this volume, and will have base radius 4. Using the formula for volume of a cylinder, we can solve for the height of the column of water.

\(\displaystyle V(cylinder) = \pi*r^2*h\), where \(\displaystyle r\) is the base radius 4 in this case.

\(\displaystyle 288\pi = \pi*(4^2)*h\)

\(\displaystyle 18 = h\)

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 \(\displaystyle \frac{4}{3}\).