# Double Integrals - Volume of a Cylinder

1. Apr 30, 2013

### Illania

1. The problem statement, all variables and given/known data

A cylinder has a diameter of 2 inches. One end is cut perpendicular to the side of the cylinder and the other side is cut at an angle of 40 degrees to the side. The length at the longest point is 10 inches. Find the volume of the sample.

I believe this is what it would look like:

2. Relevant equations

3. The attempt at a solution

I began by dividing it into two regions, D1 and D2.

D1 should have a height of $10 - \frac{2}{tan(40°)}$
To find the volume of D1, I did
$\int^{2}_{0}\int^{{10}_{\frac{2}{tan(40°)}}} (x^2-y^2-1) dy dx$
I did this because I think x should go from 0 to 2, the diameter, and y should go from $\frac{2}{tan(40°)}$ to 10, the bottom of the region to the top.

When I attempted this, I got a negative answer so it cannot be right. Can anyone help me find the fault in my logic here?

2. Apr 30, 2013

### HallsofIvy

Staff Emeritus
I presume that you know that the volume of the entire cylider is $\pi r^2h= \pi(1)(10)= 10\pi$.

So all you really need to do is subtract off the volume of the section cut off. If you set up a coordinate system so that (0, 0, 0) is in the center of the end of the cylinder that is cut off, with the axis of the cylinder extending up the z- axis, the surface of the cylinder is given by $x^2+ y^2= 1$. Taking the line of cut parallel to the x-axis, at 45 degrees, so that z= x, we will have $x^2+ y^2+ z^2= 2x^2+ y^2= 1$. It will cross the xz-plane at z= 2 so you will want to integrate, with respect to z, from 0 to 2 with, at each z, integrating, with respect to y from $-\sqrt{1- 2x^2}$ to $\sqrt{1- 2x^2}$.

3. Apr 30, 2013

### Illania

I'm not quite sure I understand your explanation. Also, if I integrate using those values, won't I be left with x's in my final answer? It looks as if I'll end up with complex numbers as well.

4. Apr 30, 2013

### Staff: Mentor

HallsOfIvy worked a slightly different problem, in which the cutting plane makes an angle of 45°, not 40° as in your problem. The only thing different would be the equation of the plane.