Find the area of the region shared by the curves

[apiwowar]

find the area of the region shared by the curves r=6cosx and r=6sinx

i know that if you graph those two functions you get two circles, both with radius 3, the first one has its center on the cartesian x-axis and the other has its center on the y axis.

i also know that if you draw a line through the center of where they meet the angle of that line is pi/4

what i don't know is which function to integrate and what would be the lower limit of integration.

i know that the formula is 1/2 integral from a to b of (g(x)^2 - f(x)^2)

making 6sinx the second function makes sense since that's the bottom part of the area that i want to find but you can't make the first function 6cosx and integrate from 0 to pi/4 because that would give you more than just where the two functions overlap

any help or ideas about how to solve this would be appreciated.

 

[Mark44]

The formula that you "know" is incorrect. A ray going out from the origin should extend to only one or the other of the two circles, so you won't have g(x)^2 - f(x)^2 in your integral. If you have drawn a sketch of the region you will see this.

An easy way to do this is to exploit the symmetry of the two curves, and integrate along the circle r = 6sint from t = 0 to t = pi/4, then double that value. Note that I changed the x you had to t (for theta), since the curves are (I believe) curves in polar coordinates, in which r and theta appear, but x and y don't.

Another way is one that doesn't exploit symmetry. To do it this way you need two integrals, one in which you integrate along the sine curve from t = 0 to pi/4, and the other in which you integrate along the cosine curve back to the origin, as t runs from pi/4 to pi/2.

Both techniques should give you the same area.

 

[apiwowar]

yea i used x instead of theta

but i completely see that now. i just wasnt thinking straight.

thanks