Volume of Rotation

1. Feb 2, 2015

tree.lee

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

Find volume when the area rotates around a line.

2. Relevant equations

x=y^2
x=1-y^2

3. The attempt at a solution

Could someone please check my formula for Area? I have pi[ (3-(y^2))^2 - (2-(y^2))^2 ] which gives me the integral pi[5y-(2/3)y^3]. I think it's wrong though because I plugged it into the wolfram alpha program and it gives me the wrong answer, the answer according to the textbook is (10)(sqrt2)(pi)/3 and the program says according to my integral between the intersection points (-1/sqrt2, 1/sqrt2) the volume would be (14)(sqrt2)(pi)/3. I'm really confused. Any help would be greatly appreciated! Thank you.

2. Feb 2, 2015

Staff: Mentor

You are missing either $\Delta x$ or $\Delta y$, which makes it harder for me to tell whether you are using disks or shells. I'm sure I could figure out what you're doing with some effort, but I shouldn't have to.

3. Feb 2, 2015

tree.lee

Oh, I think I'm using delta y. Like my widths are on the y axis? Is that what is meant? Sorry I'm not very good with terminology at all, haha.

4. Feb 2, 2015

Brian T

The radius of your circles run parallel to the x-axis, and the summing of the areas occurs along the y-axis axis (the circles are vertically stacked), which gives us
$∫\pi(x(y))^2Δy$
In this case, x(y) is not directly explicit. You have to use the given functions and the fact that it revolves around the x=3 axis to determine the real form of x(y)

5. Feb 2, 2015

Staff: Mentor

I get the same number you got -- $\frac{14 \pi \sqrt{2}}{3}$. Double-check the problem information and answer. If everything is as you reported, it seems that the book has a typo.

Last edited: Feb 2, 2015
6. Feb 2, 2015

Staff: Mentor

The above is incorrect. tree.lee is using "washers", or disks with holes in them.
Please read the problem more carefully. Both functions give x explicitly in terms of y.

7. Feb 2, 2015

Dick

3-(1-y^2) does not give 2-y^2.

8. Feb 2, 2015

tree.lee

Oh. Wouldn't subtracting the functions from x=3 give the distance from it and thus the radii desired? Then subtracting the inner function from the outer function should give the area I want to integrate, right? That gives me pi[5y-(2/3)y^3], doesn't it?
But the textbook says (10)(sqrt2)(pi)/3.

9. Feb 2, 2015

Staff: Mentor

Yes, I know. It might be a typo in the book. That has been known to happen.

What book are you using?

10. Feb 2, 2015

Brian T

I believe I got the answer. Let me type out my work and someone correct me if I'm wrong

11. Feb 2, 2015

tree.lee

OH I
Sorry I posted replies that I didn't mean to I'm new to this forum and there are a lot of brackets saying quotes and stuff and it confuses me and I'm not sure what I"m inputting when I hit post reply. Sorry. Thanks for the help A LOT! I did double check everything.

12. Feb 2, 2015

Dick

No. See my post #7.

13. Feb 2, 2015

Staff: Mentor

Yes, good eye, Dick -- you're right. The inner radius is (3 - (1 - y2)2, which is (2 + y2)2. Using that gives the textbook's answer.

14. Feb 2, 2015

tree.lee

Oh. Wow, this community is so cool. Thanks a billion. It's seriously appreciated. Hope you're all having a wonderful day. Thanks a lot Dick.