# Thread: polar enclossed area integral mmn 15 2b

1. ## polar enclossed area integral mmn 15 2b

i forgot to mention that y>=0 and theta is from 0 to pi

2. Drawing them would be easiest if you converted both equations to cartesians, using $\displaystyle \displaystyle x = r\cos{\theta}, y = r\sin{\theta}, x^2 + y^2 = r^2$.

3. was told to use polar by my prof

4. But you give one equation, $\displaystyle x^2+ y^2= 1$, in Cartesian coordinates! In polar coordinates, that is r= 1. That should be easy to draw. As for $\displaystyle r= 1+ cos(\theta)$, that is what is called a "cardioid". Look it up. As far as graphing is concerned, you could just use the "old fashioned" way- calculate r for a number of $\displaystyle \theta$ values. The are beween them will be the integral of $\displaystyle f_1(\theta)- f_2(\theta)= (1+ cos(\theta))- 1= cos(\theta)$. And the integration will be between values of $\displaystyle \theta$ where the graphs cross: solve $\displaystyle 1+ cos(\theta)= 1$.

5. Originally Posted by transgalactic
was told to use polar by my prof
Yes, use polars to evaluate the integral, but you can use Cartesians to aid in graphing...

6. i managed to draw it
but when i made 1=1+sin theta
theta=pi*k
but the drawing gives me pi/2 point
i dont know how to complerte the integral