# using polar coordinates to find area of cardioid

• May 5th 2010, 06:24 PM
Dr Zoidburg
using polar coordinates to find area of cardioid
Use polar coordinates to calculate $\int \int_R \frac{1}{\sqrt{x^2+y^2}}dA$where R is the region inside thecardioid r = 1 + sinq and above the x-axis.
In polar form, the equation is just 1/r, right?
I'm having a problem working out the limits here. For the inner integral are the limits $0\leq r \leq (1+sin\theta)$.
What are the limits for the outer integral? Am i right in thinking the lower limit is also 0 as it's above the x-axis? What about the upper limit? Because it's a cardioid is it $2\pi$?
• May 5th 2010, 09:47 PM
matheagle
yes $r=\sqrt{x^2+y^2}$ but don't forget the Jacobian.

you also have $dxdy=rdrd\theta$

above the x-axis means $0<\theta <\pi$

$\int_0^{\pi}\int_0^{1+\sin\theta}drd\theta$
• May 5th 2010, 11:55 PM
Dr Zoidburg
D'oh! I always forget about that r! I was hopelessly stuck because I was doing 1/r which of course integrates to log(r) & log(0) is undefined. Hence my total loss at what to do.
That makes it a lot easier to do. Thanks for that.