Evaluating Double Integrals in Polar Coordinates Examples 2

# Evaluating Double Integrals in Polar Coordinates Examples 2

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 of the region $D$ that lies between the polar curves $r = 2 + \sin 3 \theta$ and $r = 4 - \cos 3\theta$.

We first note that both of the curves $r = 2 + \sin 3 \theta$ and $r = 4 - \cos 3 \theta$ are fully traced for $0 ≤ \theta ≤ 2 \pi$. We also note that these polar curves do not intersect each other and that $r = 4 - \cos 3 \theta$ surrounds $r = 2 + \sin 3 \theta$. Both of these facts can easily be seen from graphing both curves in the $r\theta$-plane.

The area $D_1 = \{ (r, \theta) : 0 ≤ r ≤ 4 - \cos 3 \theta, 0 ≤ \theta ≤ 2 \pi \}$ which represents the area enclosed by the polar curve $r = 4 - \cos 3 \theta$ can be evaluated with the following double integral:

(3)
\begin{align} \quad \mathrm{Area \: of \: Outer \: Region} = \iint_{D_1} 1 \: dA = \int_0^{2 \pi} \int_0^{4 - \cos 3 \theta} 1 \: dr \: d \theta \end{align}

The area $D_2 = \{ (r, \theta) : 0 ≤ r ≤ 2 + \sin 3 \theta, 0 ≤ \theta ≤ 2 \pi \}$ which represents the area enclosed by the polar curve $r = 2 + \sin 3 \theta$ can be evaluated with the following double integral:

(4)
\begin{align} \quad \mathrm{Area \: of \: Inner \: Region} = \iint_{D_2} 1 \: dA = \int_0^{2\pi} \int_0^{2 + \sin 3 \theta} 1 \: dr \: d \theta \end{align}

Therefore the total area can be computed as follows:

(5)
\begin{align} \quad A =\int_0^{2 \pi} \int_0^{4 - \cos 3 \theta} 1 \: dr \: d \theta - \int_0^{2\pi} \int_0^{2 + \sin 3 \theta} 1 \: dr \: d \theta \\ \quad A = \frac{1}{2} \int_0^{2\pi} (4 - \cos 3 \theta)^2 \: \theta - \frac{1}{2} \int_0^{2 \pi} (2 + \sin 3 \theta)^2 \: d \theta \\ \quad A = \frac{1}{2} \int_0^{2\pi} 16 - 8 \cos 3 \theta + \cos^2 3 \theta - 4 - 4 \sin 3 \theta - \sin^2 3 \theta \: d \theta \\ \quad A = \frac{1}{2} \int_0^{2\pi} 12 + \cos^2 3 \theta - \sin^2 3 \theta - 8 \cos 3 \theta - 4 \sin 3 \theta \: d \theta \end{align}

We will now use the trigonometric identity $\cos 6\theta = \cos^2 3\theta + \sin^2 3\theta$.

(6)
\begin{align} \quad A = \frac{1}{2} \int_0^{2\pi} 12 + \cos 6t - 8 \cos 3 \theta - 4 \sin 3 \theta \: d \theta \\ \quad A = \frac{1}{2} \left [ 12 \theta + \frac{\sin 6\theta}{6} - \frac{8 \sin 3\theta}{3} + \frac{4 \cos 3 \theta}{3} \right ]_{\theta=0}^{\theta = 2\pi} \\ \quad A = \frac{1}{2} \left [ \left ( 24 \pi + \frac{4}{3} \right ) - \left ( \frac{4}{3} \right ) \right ] \\ \quad A = 12 \pi \end{align}

## Example 2

Evaluate the double integral $\iint_D \sin (x^2 + y^2) \: dA$ where $D$ is the region bounded between the the circles $x^2 + y^2 = 1$ and $x^2 + y^2 = 9$ and in the first quadrant.

We first note that the region $D$ can easily be written in polar coordinates as $D = \left \{ (r, \theta) : 1 ≤ r ≤ 3, 0 ≤ \theta ≤ \frac{\pi}{2} \right \}$. Thus by letting $x = r \cos \theta$ and $y = r \sin \theta$ we have that:

(7)
\begin{align} \quad \iint_D \sin (x^2 + y^2) \: dA = \int_0^{\pi /2} \int_1^3 \sin (r^2) r \: dr \: d \theta \\ \end{align}

By using substitution, we can evaluate the inner integral above.

(8)
\begin{align} \quad \iint_D \sin (x^2 + y^2) \: dA = \int_0^{\pi /2} \frac{1}{2} \left [ - \cos (r^2) \right ]_{r=1}^{r=3} \: d \theta \\ \quad \iint_D \sin (x^2 + y^2) \: dA = \int_0^{\pi /2} \frac{cos(1) -cos(9)}{2} \: d \theta \\ \quad \iint_D \sin (x^2 + y^2) \: dA =\frac{cos(1) -cos(9)}{2} [ \theta ]_{\theta=0}^{\theta=\pi/2} \\ \quad \iint_D \sin (x^2 + y^2) \: dA = \frac{[cos(1) -cos(9)]\pi}{4} \end{align}

## Example 3

Evaluate the double integral $\iint_D \sin (x^2 + y^2) \: dA$ where $D$ is the region bounded between the the circles $x^2 + y^2 = a$ and $x^2 + y^2 = b$ where $0 < a ≤ b$ and in the first quadrant.

Like in example 2, we can easily express the region $D$ as $D = \left \{ (r, \theta) : a ≤ r ≤ b, 0 ≤ \theta ≤ \frac{\pi}{2} \right \}$. Hence:

(9)
\begin{align} \quad \iint_D \sin (x^2 + y^2) \: dA = \int_0^{\pi/2} \int_a^b \sin (r^2) r \: dr \: d \theta \\ \quad \iint_D \sin (x^2 + y^2) \: dA = \int_0^{\pi/2} \frac{1}{2} \left [ -\cos (r^2) \right ]_{r=a}^{r=b} \: d \theta \\ \quad \iint_D \sin (x^2 + y^2) \: dA = \frac{1}{2} \int_0^{\pi/2} \cos [(a^2) - \cos(b^2)] \: d \theta \\ \quad \iint_D \sin (X^2 + y^2) \: dA = \frac{[\cos (a^2) - \cos(b^2)] \pi}{4} \end{align}