1. ## Volume of Revolution

y = x^3
y = 4x

Would this be right?...

integrate (from 2 to 0) [((8-x^3)^2)-(8-4x)^2] dx *pi

Can someone tell me if I am right, if not, could you tell me how it should be?
(I am using the washer method)

2. Originally Posted by Morgan82
y = x^3
y = 4x

Would this be right?...

integrate (from 2 to 0) [((8-x^3)^2)-(8-4x)^2] dx *pi

Can someone tell me if I am right, if not, could you tell me how it should be?
(I am using the washer method)
close enough ...

$V = \pi \int_0^2 (8-x^3)^2 - (8-4x)^2 \, dx$

3. Yes you integral is set up right, now simplify the expression and integrate.

4. Originally Posted by skeeter
close enough ...

$V = \pi \int_0^2 (8-x^3)^2 - (8-4x)^2 \, dx$
I figured it out... and I get -344pi/21

???

5. Volume should never be negative, so therefore check your integration. You may have made a mis-calculation somewhere.

6. Originally Posted by Morgan82
I figured it out... and I get -344pi/21

???
I get $\frac{832\pi}{21}$ ... check your algebra/antiderivative/arithmetic.

7. Same as skeeter, work it out spaced out neatly, the difference sign in between the two can often make one lose track.

8. Originally Posted by skeeter
I get $\frac{832\pi}{21}$ ... check your algebra/antiderivative/arithmetic.

Are you VERY sure about this? My friend didn't get that.. and it seems like she is always right.

9. We are both positive, just because she seems like she is always right, doesn't mean she is right.