Evaluating Double Integrals in Polar Coordinates Examples 3

# Evaluating Double Integrals in Polar Coordinates Examples 3

Recall from the Evaluating Double Integrals in Polar Coordinates page that sometimes evaluating a double integral over a region may be difficult due to the nature of the region, and the double integral may be more easily expressible in terms of polar coordinates.

For regions in the form $D = \{ (r, \theta) : a ≤ r ≤ b, \alpha ≤ \theta ≤ \beta \}$ we have that:

(1)
\begin{align} \quad \iint_D f(x, y) \: dA = \int_{\alpha}^{\beta} \int_a^b f(r \cos \theta, r \sin \theta) r \: dr \: d \theta \end{align}

For regions in the form $D = \{ (r, \theta) : h_1(\theta) ≤ r ≤ h_2(\theta), \alpha ≤ \theta ≤ \beta \}$ we have that:

(2)
\begin{align} \quad \iint_D f(x, y) \: dA = \int_{\alpha}^{\beta} \int_{h_1(\theta)}^{h_2(\theta)} f(r \cos \theta, r \sin \theta) r \: dr \: d \theta \end{align}

We will now look at some examples of evaluating double integrals on regions defined by polar coordinates.

## Example 1

Find the area trapped between the spiral $r = \theta$ for $0 ≤ \theta ≤ 2\pi$.

We can solve this problem using double integrals as $A = \iint_D 1 \: dA$ where $D$ is the region we're calculating the area of. We can describe this region as $D = \{ (r, \theta) : 0 ≤ r ≤ \theta, 0 ≤ \theta ≤ 2\pi \}$ and so:

(3)
\begin{align} \quad A = \iint_D 1 \: dA \\ \quad A = \int_0^{2\pi} \int_0^{\theta} r \: dr \: d \theta \\ \quad A = \int_0^{2\pi} \left [ \frac{r^2}{2} \right ]_{r =0}^{r=\theta} \: d \theta \\ \quad A = \int_0^{2\pi} \frac{\theta^2}{2} \: d \theta \\ \quad A = \left [ \frac{\theta^3}{3} \right ]_{\theta=0}^{\theta=2\pi} \\ \quad A = \frac{4 \pi^3}{3} \end{align} ## Example 2

Find the volume above the disk $x^2 + y^2 ≤ 4$ and under the cone $z = \sqrt{x^2 + y^2}$.

We can express the region of integration nicely as $D = \{ (r, \theta) : 0 ≤ r ≤ 2, 0 ≤ \theta ≤ 2\pi \}$ and thus using polar coordinates we have that:

(4)
\begin{align} \quad \iint_D \sqrt{x^2 + y^2} \: dA = \int_0^{2\pi} \int_0^2 \sqrt{r^2} r \: dr \: d \theta \\ \quad \iint_D \sqrt{x^2 + y^2} \: dA = \int_0^{2\pi} \int_0^2 \mid r \mid r \: dr \: d \theta \end{align}

We note that $0 ≤ r ≤ 2$ and so $\mid r \mid = r$:

(5)
\begin{align} \quad \iint_D \sqrt{x^2 + y^2} \: dA = \int_0^{2\pi} \int_0^2 r^2 \: dr \: d \theta \\ \quad \iint_D \sqrt{x^2 + y^2} \: dA = \int_0^{2\pi} \left [ \frac{r^3}{3} \right ]_{r=0}^{r=2} \: d \theta \\ \quad \iint_D \sqrt{x^2 + y^2} \: dA = \int_0^{2\pi} \frac{8}{3} \: d \theta \\ \quad \iint_D \sqrt{X^2 + y^2} \: dA = \frac{8}{3} \left [ \theta \right ]_{\theta = 0}^{\theta = 2\pi} \\ \quad \iint_D \sqrt{x^2 + y^2} \: dA = \frac{16 \pi}{3} \end{align}

## Example 3

Evaluate the double integral $\iint_D \tan^{-1} \left ( \frac{y}{x} \right ) \: dA$ where $D = \{ (x, y) : 1 ≤ x^2 + y^2 ≤ 4, 0 ≤ y ≤ x \}$.

We first note that we can rewrite $D$ in polar coordinates. Note that $1 ≤ r ≤ 4$. Furthermore, it's not hard to see that if $0 ≤ y ≤ x$, then $0 ≤ \frac{y}{x} = \frac{r \sin \theta}{r \cos \theta} = \tan \theta ≤ 1$. Therefore $0 ≤ \theta ≤ \frac{\pi}{4}$, so $D = \left \{ 1 ≤ r ≤ 4, 0 ≤ \theta ≤ \frac{\pi}{4} \right \}$. Thus we have that:

(6)
\begin{align} \quad \iint_D \tan^{-1} \left ( \frac{y}{x} \right ) \: dA = \int_0^{\pi/4} \int_1^4 \tan^{-1} (\tan (\theta)) r \: dr \: d \theta \\ \quad \iint_D \tan^{-1} \left ( \frac{y}{x} \right ) \: dA = \int_0^{\pi/4} \int_1^4 \theta r \: dr \: d \theta \\ \quad \iint_D \tan^{-1} \left ( \frac{y}{x} \right ) \: dA = \int_0^{\pi/4} \theta \left [ \frac{r^2}{2} \right ]_{r=1}^{r=4} \: d \theta \\ \quad \iint_D \tan^{-1} \left ( \frac{y}{x} \right ) \: dA = \int_0^{\pi/4} \frac{3 \theta}{2} \: d \theta \\ \quad \iint_D \tan^{-1} \left ( \frac{y}{x} \right ) \: dA = \left [ \frac{3 \theta^2}{4} \right ]_{\theta =0}^{\theta = \pi/4} \\ \quad \iint_D \tan^{-1} \left ( \frac{y}{x} \right ) \: dA =\frac{3 \pi^2}{64} \end{align}