Now, $ x = rcos\theta \\ \; \\ diff. \; partially \; w.r.t. \; r, \\ \; \\ \therefore \frac{\partial x }{\partial r} \; = \; cos\theta \\ \; \\ y = rsin\theta \\ \; \\ diff. \; partially \; w.r.t. \; r, \\ \; \\ \therefore \frac{\partial y }{\partial r} \; = \; sin\theta \\ \; \\ \; \\ diff. \;x \; partially \; w.r.t. \; \theta, \\ \; \\ \therefore \frac{\partial x }{\partial \theta} \; = \; -r \;sin\theta \\ \; \\ diff. \; y \; partially \; w.r.t. \; \theta, \\ \; \\ \therefore \frac{\partial y }{\partial \theta} \; = \; r \;cos\theta \\ \; \\ \; \\ Jocobian|J| \; = \; \frac{\partial (x,y)}{\partial (r,\theta)} \;= \; \left| \begin{array}{ccc} \frac{\partial x }{\partial r} & \frac{\partial x }{\partial \theta} \\ \frac{\partial y }{\partial r} & \frac{\partial y }{\partial \theta} \end{array} \right| \;= \; \left| \begin{array}{ccc} cos\theta & -r \; sin\theta \\ sin\theta & r \; cos\theta \end{array} \right| \\ \; \\ \therefore \frac{\partial (x,y)}{\partial (r,\theta)} \;= \; cos\theta (r \; cos\theta) \; - \; (-r sin\theta) sin\theta \\ \; \\ = rcos^2\theta \; + \; r \; sin^2 \theta \\ = r \; (cos^2\theta \; + \; sin^2 \theta ) \; = \; r $