1. ## use cylindrical shells...

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

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

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

4. 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?

5. 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.

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

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

7. 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})
$
= $2\pi(\frac{27}{6})$ = $9\pi$

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