# Volume of region bounded by two graphs?

• May 21st 2009, 04:57 PM
xvaliant
Volume of region bounded by two graphs?
The region is bounded by y=x^3 and y=x in the first quadrant.

I already found the area, but I'm having trouble with the volume.

First:
A)find the volume rotated around y=2
B) find volume rotated around the y axis
C)the area (1/4) is the base of a solid. There are cross sections perpendicular to the x-axis that are semi circles. Find the volume of the solid.

Help?
• May 21st 2009, 05:59 PM
csawyer1109
The formula to find the volume is:

$\displaystyle \pi \int (outer - line going around)^2 - (inner - line going around)^2$

That means "a" would be setup as follows:

$\displaystyle 2 \times \pi \int (x-2)^2 - (x^3-2)^2$

The integral will be from 0 to 1.

The reason we multiply it by 2 is because the two lines actually form a solid twice. Therefore we will take the integral from a point of intersection to the next point of intersection on one solid and multiply times 2 in order to get the volume of both solids.

b)To find the area going around the y-axis we need to place all of our $\displaystyle x$'s in terms of $\displaystyle y$.

We come out with $\displaystyle x=\sqrt[3]{y}$ and $\displaystyle x=y$. Remember the graph looks the same, but the equations are just in terms of $\displaystyle y$.

Then just use the formula to find the volume.

Your setup should look as follows:

$\displaystyle 2 \times \pi \int (\sqrt[3]{y})^2 - (y)^2$

The integral will be from 0 to 1 again because the graphs intersect at those points.

Question "c" is beyond my knowledge. Maybe someone else here can enlighten me.

If you have any questions, just ask!
• May 21st 2009, 06:17 PM
xvaliant
So, would the graph be considered dx or dy? Because oringally I thought it was dx, but then it asked for it to be rotated around y=2 and the y-axis. So, in order to do that, don't I need to make the y=2 volume change from x to y like in (b)?
• May 21st 2009, 06:20 PM
csawyer1109
No, for "a" your terms need to be in terms of $\displaystyle x$.

This is because $\displaystyle y=2$ is virtually the same as the $\displaystyle x-axis$, but shifted up 2.
• May 21st 2009, 06:28 PM
xvaliant
ok. And it's the washer method because of the formula used? Btw, I appreciate your help!
• May 21st 2009, 06:30 PM
csawyer1109
You got it!

No problem and good luck!
• May 21st 2009, 06:31 PM
xvaliant
Ok. That's probably what I was doing wrong. I had thought it was the shell method because it made a "cup" shape. Haha.