Question

Hi, I need some help with volume >.<.
Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disc or washer.

Answer 1

\[x = y-y^2, x =0\] about the y axis, I'm having a hard time figuring this stuff out. I understand how areas between curves work, but this stuff is sort of confusing.

Answer 2

@Psymon

Answer 3

Any method required or whichever?

Answer 4

Which ever

Answer 5

I'm just trying to understand this before my midterm which is not to far away, just hard for me to grasp. :\

Answer 6

Im just drawing it out for myself real quick :P

Answer 7

=P

Answer 8

Alright, well this one seems like it'll be easy to do with just disk method



So just by factoring the function into y(1-y), we can see we have y-intercepts of 0 and 1. So I choose the disk method mainly because the disk method uses limits of integration parallel to the axis that the graph is being rotated around. This graph is about y, so I use y-limits. That and my function is already in terms of x = f(y), so very easy. So since I'm dealing with everything in terms of y, factoring and solving for y showed me that I need my limitsofintegration to be from 0 to 1. The next thing I look at is if I revolve my graph about the axis of revolution, is a hole created. In this case absolutely not. Because no hole is created, I just have the simple disk method formula of

\[\pi \int\limits_{a}^{b}[R(y)]^{2}dy\]Where the R(y) radius, is represented by the graph which covers the bounded area. Because the bounded area is merely the area under the curve of y-y^2, y-y^2 is our radius. So this means we have this:

\[\pi \int\limits_{0}^{1}(y-y^{2})^{2}dy\] which shouldnt be too bad to integrate and do from there. The set up is the part that would beeasy to mess up, so let me know what ya think/