How to Solve Double Integrals by Using Polar Coordinates – Example 2 – Fusion of Engineering, Control, Coding, Machine Learning, and Science

In this mathematics and calculus tutorial, we explain how to solve double integrals by using polar coordinates. The YouTube tutorial is given below.

The problem of Solving Double Integrals

Calculate the value of the double integral

(1) $\begin{align*}\iint_{D} (1-x^{2}-y^{2})dA\end{align*}$

where the domain of integration $D$ is the circle with the radius of 1.

Solve Double Integrals by Using Polar Coordinates

To solve this integral, we introduce the polar coordinates

(2) $\begin{align*}x & =r\cos (\theta) \\y & =r\sin (\theta)\end{align*}$

The integration domain $D$ is a circle with the radius of $1$ . The equation of the circle boundary in the Cartesian coordinates is

(3) $\begin{align*}x^{2}+y^{2}=1\end{align*}$

The inside of the circle is described by the following inequality

(4) $\begin{align*}x^{2}+y^{2}<1\end{align*}$

Consequently, in the Cartesian coordinates, the integration domain $D$ is

(5) $\begin{align*}D=\{(x,y)| x^{2}+y^{2}\le 1 }\end{align*}$

By substituting the polar coordinates (6) in

(6) $\begin{align*}x^{2}+y^{2}\le 1 \end{align*}$

We obtain

(7) $\begin{align*}& r^{2}\cos^{2}(\theta) +r^{2}\sin^{2}(\theta)\le 1 \\& r^{2}\big( \cos^{2}(\theta) +\sin^{2}(\theta)\big)\le 1 \\& r^{2} \le 1 \\& r \le 1\end{align*}$

The angle $\theta$ is in the interval from $0$ to $2\pi$ since we need to describe the complete circle. Consequently, the integration domain in the polar coordinates takes the following form

(8) $\begin{align*}D=\{(r,\theta)| 0\le r \le 1,\, 0\le \theta \le 2\pi \}\end{align*}$

On the other hand, the term $dA$ in the original integral (1) can be expressed as follows in the polar coordinates

(9) $\begin{align*}dA=r\cdot dr \cdot d\theta\end{align*}$

By substituting (6) and (9) in the original integral (1) and by taking into account the integration domain in the polar form (8), the original integral becomes:

(10) $\begin{align*}\iint_{D} (1-x^{2}-y^{2})dA & =\int_{\theta=0}^{\theta=2\pi} \int_{r=0}^{r=1} (1-r^{2})r\cdot dr \cdot d\theta \\& =\int_{\theta=0}^{\theta=2\pi} \int_{r=0}^{r=1} (r-r^{3})dr \cdot d\theta = \\& = \int_{\theta=0}^{\theta=2\pi}(\frac{r^{2}}{2}-\frac{r^{4}}{4})\Big|_{0}^{1}d\theta=\frac{1}{4} \int_{\theta=0}^{\theta=2\pi}d\theta=\frac{\pi}{2}\end{align*}$

The problem of Solving Double Integrals

Solve Double Integrals by Using Polar Coordinates

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