Completing Proof of Jacobian Determinant

Question

I am actually trying to Prove the formula: $$dxdy=rdrd\theta$$ When we change Cartesian System to Polar Coordinates:

We have: $$x=r\cos \theta$$ So

$$dx=-r\sin \theta d\theta+\cos \theta dr$$

Like Wise: $$y=r\sin \theta$$

$$dy=r\cos \theta d\theta+\sin \theta dr$$

Now we get:

$$\begin{bmatrix} dx\\ dy \end{bmatrix}=\begin{bmatrix} \cos \theta & -r\sin \theta\\ \sin \theta& r\cos \theta \end{bmatrix} \times \begin{bmatrix} dr\\ d\theta \end{bmatrix}$$

Now we call:

$$J=\begin{vmatrix} \cos \theta &-r\sin \theta \\ \sin \theta& r\cos \theta \end{vmatrix}=r$$

But Now i could not prove:

$$dxdy=Jdr d\theta$$

Answer 1

You can take determinants on both sides. So you'll get, $$ \det(dx dy) = \det(J)*\det(dr d\theta) = r*\det (dr d\theta) $$ Using the property of scalar multiplication for determinants, $ \det(dx dy)=det(r*dr d\theta)$. Now comparing respective elements in both determinants, $dx=r*dr$ and $dy=d\theta$. Hence, $dx*dy=r*dr*d\theta$.