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

1. 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.

2. 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:

$dV=2\pi rh\,dx$

where:

$r=x$

$h=8-x^3$

and so we have:

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

Evaluate this, and you should get:

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

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

3. 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?

4. 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.