# Thread: One more question :P

1. ## One more question :P

How do you SET UP this question? I can probably do it if I set it up, but I just don't know how to set it up. Also, please EXPLAIN how to set it up so I can set up other problems like this.

Let V be the region bounded by the $y-axis$, $y = 8$, $y = 1$, and $y = x^{.75}$. Find the exact volume by rotating V around the y-axis.

My attempt:

$\pi{\int^8}[(8-x^{.75}+1)^2]dx$
. . $^0$

2. You can do it with shells or disks. But don't forget to add on the rectangular region from 0 to 1. See it?. It is $7{\pi}$

${\pi}\int_{1}^{8}[(y^{\frac{4}{3}})^{2}-1]dy$

or

$2{\pi}\int_{1}^{16}[x(8-x^{\frac{3}{4}})]dx$

This is one way to go about it.

3. Thanks for clearing that up. It makes a lot more sense now.

One more question though...

Let's say that you had $y = x^{.75}$, but this time the region is bounded by the the $x-axis$ and the lines $x = 1$ and $x = 8$. Of course you still rotate the region around the y-axis, but how would the problem change when you set it up?

4. Then, since with shells the cross-sections are parallel to the axis you are revolving around you would have:

$2{\pi}\int_{1}^{8}x(x^{\frac{3}{4}})dx$