To solve this question, you need to know the surface area of a cone equation:  \(\displaystyle \textup{surface area}=\pi r(r+\sqrt{h^2 +r^2}),\) where \(\displaystyle r\) is the radius of the cone \(\displaystyle (3\textup{cm})\), and \(\displaystyle h\) is the height of the cone.  We must substitute all of the values that we know, and solve for the height of the cone to get the answer.   

\(\displaystyle 24\textup{cm}^2=\pi *3\textup{cm}(3\textup{cm}+\sqrt{h^2 +(3\textup{cm})^2}),\)  Therefore, the height of the cone is \(\displaystyle 4\textup{cm.}\)

Cylinder Volume = \(\displaystyle \pi*r^{2}*h\)

Cone Volume = \(\displaystyle \frac{1}{3}*\pi*r^{2}*h\)

Cylinder Diameter = 6, therefore Cylinder Radius = 3

Cone Radius = 3

Empty Volume = Cylinder Volume – Cone Volume

\(\displaystyle =\pi*r^{2}*h-\frac{1}{3}*\pi*r^{2}*h\) 

\(\displaystyle =\pi*3^{2}*10-\frac{1}{3}*\pi*3^{2}*10\)

\(\displaystyle =\pi*90-\pi*30\)

\(\displaystyle =\pi*60\)

1. Find the volume of each cone:

\(\displaystyle Cone Volume=\frac{1}{3}\pi r^{2}h\)

Cone 1:

Since the diameter is \(\displaystyle 6\), the radius is \(\displaystyle 3\).

\(\displaystyle Volume=\frac{1}{3}\pi (3^{2})(10)=30\pi\)

Cone 2:

Since the diameter is \(\displaystyle 3\), the radius is \(\displaystyle 1.5\).

\(\displaystyle Volume=\frac{1}{3}\pi (1.5^{2})(8)=6\pi\)

2. Subtract the smaller volume from the larger volume:

\(\displaystyle 30\pi-6\pi=24\pi\)

1. Find the volumes of the cone and cylinder:

Cone:

Since the diameter is \(\displaystyle 8\), the radius is \(\displaystyle 4\).

\(\displaystyle Volume=\frac{1}{3}\pi r^{2}h\)

\(\displaystyle Volume=\frac{1}{3}\pi (4^{2})(9)=48\pi\)

Cylinder:

Since the diameter is \(\displaystyle 10\), the radius is \(\displaystyle 5\).

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

\(\displaystyle Volume=\pi (5^{2})(12)=300\pi\)

2. Subtract the cone's volume from the cylinder's volume:

\(\displaystyle 300\pi-48\pi=252\pi\)

To find the volume of a cone plug the radius and height into the formula for the volume of a cone.

\(\displaystyle V = \frac{1}{3}\pi r^2\cdot h\)

\(\displaystyle \\=\frac{1}{3}\pi 4^2\cdot 7\\ \\=\frac{112}{3}\pi cm^3\)

Bearing in mind the volume formula for a cone:

\(\displaystyle V_{cone} = \frac{1}{3}\pi r^2h\)

Because the volume varies by the square of the radius, doubling the radius will quadruple the volume (since \(\displaystyle 2^2 = 4\).) Because the volume also varies linearly by the height, quadrupling the height will quadruple the volume.

To find the volume of a cone with radius \(\displaystyle r\), and height \(\displaystyle h\) use the formula:

\(\displaystyle V = \frac{1}{3}\pi *r^2*h\).

We plug in our given radius and height to find:

\(\displaystyle V = \frac{1}{3}\pi * 6^2*2 = \frac{72}{3}\pi = 24\pi\)

The formula for the volume of a cone is:

\(\displaystyle V = \frac{\pi r^2h}{3}\)

With the information we have, we can simply plug values into this equation and solve.

\(\displaystyle V = \frac{\pi r^2h}{3} = \frac{(100^2)(70)\pi}{3} \approx 233333\pi\)

So our cone can hold approximately \(\displaystyle 233333\pi\) cubic millimeters of motor oil.

To find the volume of a cone, simply use the following formula. Thus,

\(\displaystyle V=\frac{1}{3}\pi{r^2}h=\frac{1}{3}*\pi*3^2*2=\frac{1}{3}*9*2*\pi=6\pi\)

To solve simply use the formula for the voluma of a cone. Thus,

\(\displaystyle V=\frac{1}{3}\pi{r^2}h=\frac{1}{3}*\pi*3^2*4=12\pi\)