Double Integral: Find Area Bounded by Two Circles with Polar Coordinates

[Sebs0r]

Homework Statement

Find area bounded by x^2 + y^2 = 1 and x^2 + y^2 = x + y

Homework Equations

The Attempt at a Solution

from the second circle, we can see r^2 >= r cos t + r sin t
so r >= cos t + sin t

Limits are:
cos t + sin t <= r <= 1
-pi/4 <= t <= 3pi/4

Doing the integrals however, I always seem to get zero:

$$\int^{3pi/4}_{-pi/4}\int^{1}_{cos t+ sin t} r dr dt$$
this gives me
$$\int^{3pi/4}_{-pi/4}-sin(2t)/2 dt$$
which is zero. What am I doing wrong?
Thanks

 

[Mark44]

If you integrate a periodic function over one complete period, you don't get the area between the graph and the horizontal axis - you get zero. You are integrating a multiple of sin(2t) over an interval of length pi, so naturally you'll get zero. If you want the area, you have to break up the integral into two integrals - one for the part where sin(2t) is above the horizontal axis, and the other where it is below. Then you'll get the area between the curve and the horizontal axis.

 

[Sebs0r]

Thanks, I just found that if I split the integral up into the first circle and two segments of the other circle, it ends up giving me a non-zero answer (not sure if right thought)
Area within first quadrant = pi/4

Area in 2nd and 4th quadrant =
$$\int^{0}_{-pi/4}\int^{cos t + sin t}_{0} r dr dt$$

which gives me
$$\int^{0}_{-pi/4} (1+sin2t)/2 dt$$
which is pi/8 + 1/4

so in total i have pi/4 + 2(pi/8 + 1/4)
= (pi + 1)/2

But yeah, what you said makes sense. Thanks