<p>This is a problem from my calc. book...</p>
<p>Find the volume of the region bounded by the circle x^2 + y^2 = r^2 about the line x=R, where r is less than R. I set up the integral using the shell method, and I am pretty sure I set it up right, I just can't simplify it...</p>
<p>I set it up like this:</p>
<p>2pi times the definite integral from -r to r, of (R-x)(2((r^2 - x^2)^1/2))dx</p>
<p>hehe...I hate typing math...anyways, thanks!</p>
<p>is this bc?</p>
<p>yeah, it is bc...just started the class one week ago...</p>
<p>I tried to simplify the integral you got from the shell method. I also tried the washer method. Both ways I get the integral from -r to r of 4<em>pi</em>R*(r^2-x^2)^(1/2). I believe that's integrable, but not with a u-sub. Integration by parts may have to be used here.</p>
<p>yeah, and then integrating that is the area of a semi-circle...</p>
<p>Assuming what dualityim got was correct:
I = 4piR int{(r^2-x^2)^(1/2)}dx
Using the substitution x = rsinu (with dx/du = rcosu) you get:
I = 4piR int{r^2cos^2(u)}du = 4piRr^2 int{cos^2(u)}du
Now use the identity cos^2(u)=0.5(1+cos2u), and the rest should be easy.</p>