Math Help - Another volume integral

1. Another volume integral

Let R be the region in the first quadrant bounded by the curves y = x^3 and y = 2x-x^2. Calculate the following quantities.

(a) The area of R
(I did this already and got 37/12 units squared)

(b) The volume obtained by rotating R about the x-axis
(Having trouble with this one. I feel like I know what I'm doing, but can't seem to get a satisfactory answer. I split it into two integrals:
$V = \pi \int_0^1((2x-x^2)^2 -(x^3)^2)~dx + \pi\int_{-2}^0((x^3)^2 -(2x-x^2)^2) ~dx$

But the second integral keeps returning negative volume, and I don't understand why. I am pretty sure I put the correct function on top, and that I took it over the correct interval. :/ I could just switch the sign (and hope that I did everything else correctly, it's an even problem so don't know the answer) but I wouldn't have any idea why I was doing that other than the method I thought should work didn't.

(c) The volume obtained by rotating R about the y-axis
(haven't even attempted this yet, can't get past b)

2. Ok

Originally Posted by angel.white
Let R be the region in the first quadrant bounded by the curves y = x^3 and y = 2x-x^2. Calculate the following quantities.

(a) The area of R
(I did this already and got 37/12 units squared)

(b) The volume obtained by rotating R about the x-axis
(Having trouble with this one. I feel like I know what I'm doing, but can't seem to get a satisfactory answer. I split it into two integrals:
$V = \pi \int_0^1((2x-x^2)^2 -(x^3)^2)~dx + \pi\int_{-2}^0((x^3)^2 -(2x-x^2)^2) ~dx$

But the second integral keeps returning negative volume, and I don't understand why. I am pretty sure I put the correct function on top, and that I took it over the correct interval. :/ I could just switch the sign (and hope that I did everything else correctly, it's an even problem so don't know the answer) but I wouldn't have any idea why I was doing that other than the method I thought should work didn't.

(c) The volume obtained by rotating R about the y-axis
(haven't even attempted this yet, can't get past b)
you should have that $V=\pi\int_{-2}^{0}x^6-(2x-x^2)^2dx$ $+\pi\int_0^1{(2x-x^2)^2-x^6dx}$...is the volume $\frac{531\pi}{35}$?

3. Originally Posted by Mathstud28
you should have that $V=\pi\int_{-2}^{0}x^6-(2x-x^2)^2dx$ $+\pi\int_0^1{(2x-x^2)^2-x^6dx}$...is the volume $\frac{531\pi}{35}$?
I don't know, it's an even problem so the book doesn't say.

I have a theory, though: Since when x <1, y<1. This means that even though the function y=x^3 appears to be on top, it is actually the inner function (once they are flipped around the x-axis, it will be on the bottom. This means that I should be integrating outer - inner all the way along, and so I can do it with a single integral $\int_{-2}^1 ((2x-x^2)^2-(x^3)^2)~dx$

I am feeling pretty confident in this answer, it makes sense to me, and explains why the integral returned negative volume earlier (because I was taking the inner radius - the outer radius). I think I got confused because I was mixing concepts for areas with concepts for volumes.

EDIT: For the last one, rotated about the y-axis, I got (63/10)*pi Does anyone have any way of checking whether this is correct?

4. Since they're asking for the first quadrant, we only need to integrate from 0 to 1.

Here are both methods for both axes.

Shells: $2{\pi}\int_{0}^{1}y(y^{\frac{1}{3}}-(1-\sqrt{1-y}))dy=\frac{41{\pi}}{105}$

Washers: ${\pi}\int_{0}^{1}\left[(2x-x^{2})^{2}-x^{6}\right]=\frac{41{\pi}}{105}dx$

Shells: $2{\pi}\int_{0}^{1}x(2x-x^{2}-x^{3})dx=\frac{13\pi}{30}$

Washers: ${\pi}\int_{0}^{1}\left[y^{\frac{2}{3}}-(1-\sqrt{1-y})^{2}\right]dy=\frac{13\pi}{30}$

5. Originally Posted by galactus
Since they're asking for the first quadrant, we only need to integrate from 0 to 1.
well . Why don't I ever read the instructions?

Thank you so much, you saved me from getting -2 points (we get negative points if we do the wrong problem)

Going back over it, I was able to get the same answers as you ^_^ I appreciate it, you've been very helpful today.