# Volume of Revolution

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• December 16th 2009, 03:59 PM
Morgan82
Volume 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 PM
skeeter
Quote:

Originally Posted by Morgan82
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)

close enough ...

$V = \pi \int_0^2 (8-x^3)^2 - (8-4x)^2 \, dx$
• December 16th 2009, 04:10 PM
RockHard
Yes you integral is set up right, now simplify the expression and integrate.
• December 16th 2009, 04:13 PM
Morgan82
Quote:

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

???
• December 16th 2009, 04:15 PM
RockHard
Volume should never be negative, so therefore check your integration. You may have made a mis-calculation somewhere.
• December 16th 2009, 04:17 PM
skeeter
Quote:

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

???

I get $\frac{832\pi}{21}$ ... check your algebra/antiderivative/arithmetic.
• December 16th 2009, 04:24 PM
RockHard
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 PM
Morgan82
Quote:

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.
• December 17th 2009, 09:03 PM
RockHard
We are both positive, just because she seems like she is always right, doesn't mean she is right. (Rock)