 58
 0
One of the problems I am working on is confusing me.
Given r=2cos(theta) and r=2sin(theta), you get a graph with two circles intersecting each other at pi/4.
So I worked it out, using the Area of a Polar Region formula.
[tex]\int\frac{1}{2}\pi*r^{2}[/tex]. When I got an answer, I checked and saw that it was incorrect. I noticed in the solutions guide that when they had started, they doubled the integral before working it out. That is the part the confused me. Why would you want to double the integral in this case? The enclosed region is all contained in one quadrant, so wouldn't just integrating from 0 to pi/4 be sufficient?
I'd like some help understanding this.
Given r=2cos(theta) and r=2sin(theta), you get a graph with two circles intersecting each other at pi/4.
So I worked it out, using the Area of a Polar Region formula.
[tex]\int\frac{1}{2}\pi*r^{2}[/tex]. When I got an answer, I checked and saw that it was incorrect. I noticed in the solutions guide that when they had started, they doubled the integral before working it out. That is the part the confused me. Why would you want to double the integral in this case? The enclosed region is all contained in one quadrant, so wouldn't just integrating from 0 to pi/4 be sufficient?
I'd like some help understanding this.
Attachments

1 KB Views: 295