Double Integrals with Polar Coordinates

Polar coordinates are $(r,\theta )$ such that $x=r\cos (\theta )$ and $y=r\sin (\theta )$.

How do we write an integral $\int {\int }_{R}f(x,y)dA$ in terms of $r$ and $\theta$?

Two options:

• Partition a polar sector by concentric circles with radii $a=r_0<r_1<\dots r_n=b$ and radial lines $\alpha <{\theta }_0<{\theta }_1<\cdots <{\theta }_m=\beta$ and then use the definition of the integral to find what $dA$ changes to.

• Linear algebra: Recall that the determinate tells us the area of a parallelogram given by the vectors that form its rows. Hence the change in area caused by changing coordinate systems can be found computing the Jacobian?

Theorem 1

If $f(x,y)$ is some function in Cartesian coordinates, $R$ is some region in space, $x=g(u,v)$ and $y=h(u,v)$ and $S$ is the region $R$ written in $(u,v)$ coordinates then:

$$\int {\int }_{R}f(x,y)dA=\int {\int }_{S}f(g(u,v),h(u,v))\left|\frac{\partial (x,y)}{\partial (u,v)}\right|d{A}^{\prime }$$

where $d{A}^{\prime }$ is the area with the respect to $u$ and $v$. We take the absolute values of the Jacobian matrix not just the determinate.

Theorem 2

In polar coordinates, $dA=rdrd\theta$.

Proof

In polar coordinates $x=r\cos (\theta )$ and $y=r\sin (\theta )$. Hence,

\begin{align}

\frac{ \partial (x,y) }{ \partial (r,\theta) } & = \begin{vmatrix}

\frac{ \partial x }{ \partial r } & \frac{ \partial x }{ \partial \theta } \ \frac{ \partial y }{ \partial r } & \frac{ \partial x }{ \partial \theta } \end{vmatrix} \[3ex] & = \begin{vmatrix} \cos \theta & -r\sin \theta \ \sin \theta & r \cos \theta \end{vmatrix} \[3ex] &=r \end{align}

Hence $dA = | r |dr d\theta=r dr d\theta$. This means that

\int \int _{R} f(x,y) , dA = \int \int _{s} f(r\cos(\theta),r\sin \theta)r , dr , d\theta