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find the volume of the solid obtained by rotating the region bounded by..........

find the volume of the solid obtained by rotating the region bounded by: y=x^3, y=8, x=0 about the y-axis

I drew the graph up it the attachments. Im using tube method to solve this ∫ 2*pi*r*h, so for me it will be integral from 2 to 0 ∫ 2*pi*x*8 from that i get 32pi. And the answer is (96pi)/5.

Can some please help out im not sure what the problem is really, or i must be missing some small, please look at the attachment of the graph for it.

Thanks in advance

Re: find the volume of the solid obtained by rotating the region bounded by..........

If I were going to use the tube (or shell as I know it as) method, I would begin by finding the volume of an arbitrary tube:

$\displaystyle dV=2\pi rh\,dx$

where:

$\displaystyle r=x$

$\displaystyle h=8-x^3$

and so we have:

$\displaystyle V=2\pi\int_0^2 8x-x^4\,dx$

Evaluate this, and you should get:

$\displaystyle V=\frac{96\pi}{5}$

which disagrees with the result you cite. I get the same result I gave above using the disk method.

Re: find the volume of the solid obtained by rotating the region bounded by..........

if you anti diff 8x-x^4 you get 4*x^2-x^5/5 and if you sub 2 in to that it will give you 9.6*2pi. Whats the problem with my work?

Re: find the volume of the solid obtained by rotating the region bounded by..........

i just relised that its the answer im such an ediot.