Applying Green's Theorem 3

Question

When using green's theorem to evaluate a line integral $$\oint_C P\;dx + Q\;dy = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)\;dx\,dy$$ where $C$ is a circle, I know we have to use the cylindrical coordinates $(r,\theta,z)$. My question is how do we change the $dx\,dy$ on the right hand side of the equation to $r\,dr\,d\theta$.

Answer 1

You change the coordinates using a Jacobian determinant. \begin{gathered} x = r\cos \theta \hfill \\ y = r\sin \theta \hfill \\ \left| J \right|dxdy = \left| {\begin{array}{*{20}{c}} {\frac{{\partial x}}{{\partial r}}}&{\frac{{\partial x}}{{\partial \theta }}} \\ {\frac{{\partial y}}{{\partial r}}}&{\frac{{\partial y}}{{\partial \theta }}} \end{array}} \right| = \left( {\frac{{\partial x}}{{\partial r}}\frac{{\partial y}}{{\partial \theta }} - \frac{{\partial x}}{{\partial \theta }}\frac{{\partial y}}{{\partial r}}} \right)drd\theta = \left( {r{{\cos }^2}\theta + r{{\sin }^2}\theta } \right)drd\theta = rdrd\theta \hfill \\ \end{gathered}