y = x^3

y = 4x

in the first quadrant about line y = 8

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)

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- December 16th 2009, 03:59 PMMorgan82Volume of Revolution
y = x^3

y = 4x

in the first quadrant about line y = 8

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) - December 16th 2009, 04:03 PMskeeter
- December 16th 2009, 04:10 PMRockHard
Yes you integral is set up right, now simplify the expression and integrate.

- December 16th 2009, 04:13 PMMorgan82
- December 16th 2009, 04:15 PMRockHard
Volume should never be negative, so therefore check your integration. You may have made a mis-calculation somewhere.

- December 16th 2009, 04:17 PMskeeter
- December 16th 2009, 04:24 PMRockHard
Same as skeeter, work it out spaced out neatly, the difference sign in between the two can often make one lose track.

- December 16th 2009, 05:38 PMMorgan82
- December 17th 2009, 09:03 PMRockHard
We are both positive, just because she seems like she is always right, doesn't mean she is right. (Rock)