Evaluating this double integral with polar coordinates

Question

Find $\int_\gamma x^2y^3dx+y\,dy$ where $\gamma$ is the positively oriented border of $\{(x,y) \in R^2 : 1\le x^2+y^2 \le4\} $

If I convert to polar coordinates and sketch the domain, I get that $$\gamma :=\{(\rho,\theta) \in R^2 : 1\le \rho\le2, \theta \in [0, \pi]\} $$

The original integral is confusing me, though. I don't know how to convert it in polar coordinates.

Answer 1

Apply Green's theorem

$\int_R udx+vdy = \int\int_S(\frac{\partial v}{\partial x}-\frac{\partial u }{\partial y})dydx$

Here $u=x^2y^3$ and $v=y$

So, $A = \int\int_S(0-3x^2y^2)dxdy$

Now use the fact that if $x=r\cos\theta$ and $y=r\sin\theta$, $dxdy = r\ dr \ d\theta$

Answer 2

First, you mean $\theta\in [0,2\pi]$. You don't want to do the original line integral. You want to apply Green's Theorem (or Stokes's Theorem, if you prefer), so you will be doing the double integral in polar coordinates. You should get $$-\int_S 3x^2y^2\,dA.$$ Now put this in polar coordinates. Some double-angle formulas might help when you do the actual integration.