Evaluating Double Integrals in Polar Coordinates Examples 1

# Evaluating Double Integrals in Polar Coordinates Examples 1

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

Evaluate the double integral $\iint_D x^2 y \: dA$ where $D$ is the top semicircle centered at $(0, 0)$ and with radius $5$.

Our region of integration is $D = \{ (r, \theta) : 0 ≤ r ≤ 5, 0 ≤ \theta ≤ \pi \}$. The region $D$ is sketched below:

Furthermore, evaluating $f$ at $x = r \cos \theta$ and $y = r \sin \theta$ and applying the trigonometric substitution of $\cos ^2 \theta = 1 - \sin ^2 \theta$ and we have that:

(3)
\begin{align} \quad f(r \cos \theta, r \sin \theta) = (r \cos \theta)^2 (r \sin \theta) = r^3 \cos^2 \theta \sin \theta \end{align}

Thus we have that:

(4)
\begin{align} \quad \iint_D x^2 y \: dA = \int_{0}^{\pi} \int_0^5 r^4 \cos^2 \theta \sin \theta \: dr \: d \theta \end{align}

Notice that we can split these integrals into the product of iterated integrals:

(5)
\begin{align} \quad = \int_0^5 r^4 \: dr \cdot \int_0^{\pi} \cos^2 \theta \sin \theta \: d \theta = \left [ \frac{r^5}{5} \right ]_{r=0}^{r=5} \cdot \left [ - \frac{\cos^3 \theta}{3} \right ]_0^{\pi} = 625 \cdot \frac{2}{3} = \frac{1250}{3} \end{align}

## Example 2

Evaluate the double integral $\iint_D \sqrt{a^2 -x^2 - y^2} \: dA$ where $D$ is the circle of radius $a ≥ 0$ centered at the origin, and prove that the volume of a sphere is $V = \frac{4}{3} \pi a^3$.

We first note that $z = \sqrt{a^2 -x^2 - y^2}$ represents the top hemisphere of the sphere centered at the origin with radius $r$. The region $D$ can be described in polar coordinates as $D = \{ (r, \theta) : 0 ≤ r ≤ a, 0 ≤ \theta ≤ 2\pi \}$. Now:

(6)
\begin{align} \quad f(r \cos \theta, r \sin \theta) = \sqrt{ a^2 - (r \cos \theta)^2 - (r \sin \theta)^2} = \sqrt{a^2 - r^2 \cos^2 \theta - r^2 \sin^2 \theta} = \sqrt{a^2 - r^2(\cos^2 \theta + \sin^2 \theta)} = \sqrt{a^2 - r^2} \end{align}

Now we can evaluate our double integral as follows:

(7)
\begin{align} \quad \iint_D \sqrt{a^2 - x^2 - y^2} \: dA = \int_0^{2\pi} \int_0^a r \sqrt{a^2 - r^2} \: dr \: d \theta \end{align}

We now need to use substitution to evaluate the inner integral. Let $u = a^2 - r^2$. Then $du = -2r \: dr$ and so $- \frac{1}{2} \: du = r \: dr$ and thus:

(8)
\begin{align} \quad \int r \sqrt{a^2 - r^2} \: dr = -\frac{1}{2} \int u^{1/2} \: du = - \frac{1}{3} u^{3/2} = - \frac{1}{3} (a^2 - r^2)^{3/2} \end{align}

Putting this into our iterated integrals from earlier and we have that:

(9)
\begin{align} \quad \int_0^{2\pi} \int_0^a r \sqrt{a^2 - r^2} \: dr \: d \theta = \int_0^{2\pi} -\frac{1}{3} \left [ (a^2 - r^2)^{3/2} \right ]_{r=0}^{r=a} \: d \theta = \int_0^{2\pi} = \int_0^{2\pi} \frac{1}{3}a^3 \: d \theta = \left [ \frac{1}{3} a^3 \theta \right ]_{\theta = 0}^{\theta = 2\pi} = \frac{2}{3} \pi a^3 \end{align}

Since this double integral represents the volume of half of a sphere, then the volume of a sphere is twice the double integral above and so $V = \frac{4}{3} \pi a^3$.