use cylindrical shells...

**slapmaxwell1** · March 28th 2010, 09:44 PM

use cylindrical shells to find the volume of the solid generated when the region bounded by xy = 4, x + y = 5 is revolved about the x -axis. i did the work for this problem i was wondering if someone could check my work and let me know if my setup for this problem is right? thanks in advance...IMG.pdf

i cant add the second page, but my answer, the final answer for the problem was volume = (58/6)pi

**kaiser0792** · March 28th 2010, 10:00 PM

I came up with $9\pi$ but I worked it in a hurry!

**kaiser0792** · March 28th 2010, 10:03 PM

Is this the integral? $2\pi \int_{1}^{4} y(5y - y^2 - 4) dy$

**slapmaxwell1** · March 28th 2010, 10:36 PM

um not thats not quite what i got..one of of the functions xy = 4, so one of the functions should be divided by x or y, right?

**kaiser0792** · March 28th 2010, 10:53 PM

I'm not sure what your asking, but using the shell method and rotating the solid about the x-axis would require the equation:

$V = \int_{c}^{d} p(y)h(y) dy$ right?

And if so, p(y) = y, the distance from the axis of revolution to the representaive rectangle (incremental slice) and h(y) would represent the length of the representative rectangle which would be the difference of the two given functions x = 5 - y minus x = 4/y.

**kaiser0792** · March 28th 2010, 11:01 PM

I'm sorry, the integral that I gave still had "y" in it. That was an error, it should have been:

$\int_{1}^{4}(5y - y^2 -4) dy$

**kaiser0792** · March 28th 2010, 11:11 PM

The work you submitted is correct:

However, there may be a mistake in your algebra. Check out my solution
and compare.

$2\pi[\frac{5y^2}{2} - \frac{y^3}{3} - 4y]_{1}^{4}$

$2\pi[(40 - \frac{64}{3} - 16) - (\frac{5}{2} - \frac{1}{3} - 4)]$

$2\pi[(\frac{120 - 64 - 48}{3} - (\frac{15 - 2 - 24}{6})]$

$2\pi(\frac{8}{3} + \frac{11}{6})<br />$ = $2\pi(\frac{27}{6})$ = $9\pi$

**slapmaxwell1** · March 29th 2010, 04:45 AM

thanks im going to go back and check the algebra on that one, thanks again for the catch..

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