Evaluating Double Integrals over General Domains Examples 4

# Evaluating Double Integrals over General Domains Examples 4

Recall from the Evaluating Double Integrals over General Domains page that we can evaluate double integrals over general domains by breaking them up into their respective iterated integrals.

• For type 1 regions ($y$-simple domains) $D = \{ (x, y) : a ≤ x ≤ b , g_1 (x) ≤ y ≤ g_2(x) \}$ where $g_1$ and $g_2$ are continuous, then:
(1)
\begin{align} \quad \iint_D f(x, y) \: dA = \int_a^b \int_{g_1(x)}^{g_2(x)} f(x, y) \: dy \:dx \end{align}
• For type 2 regions ($x$-simple domains) $D = \{ (x, y) : h_1(y) ≤ x ≤ h_2(y), c ≤ y ≤ d \}$ where $h_1(y)$ and $h_2(y)$ are continuous, then:
(2)
\begin{align} \quad \iint_D f(x, y) \: dA = \int_c^d \int_{h_1(y)}^{h_2(y)} f(x, y) \: dx \: dy \end{align}

We will now look at some more examples of evaluating double integrals over general domains. More examples can be found on the following pages:

## Example 1

Evaluate the double integral $\iint_D h \: dA$ where $h ≥ 0$ and $D$ is the circle centered at the origin with radius $r > 0$.

We note first off that $z = h$ represents a plane that is parallel to the $xy$-plane (i.e, a horizontal plane), and the volume generated by the double integral $\iint_D h \: dA$ over the circle centered at the origin with radius $r$ will be equal to the volume of the cylinder with height $h$ and radius $r$, that is $V = \pi r^2 h$. We will verify this using double integrals.

We first note that $D$ is not a type 1 and not a type 2 domain. We will thus split up $D$ into two subregions $D_1$ and $D_2$. Let $D_1 = \{ (x, y) : -r ≤ x ≤ r, 0 ≤ y ≤ \sqrt{r^2 - x^2} \}$ and let $D_2 = \{ (x, y) : -r ≤ x ≤ r, - \sqrt{r^2 - x^2} ≤ y ≤ 0 \}$. The interiors of these subregions do not intersect, and it's not hard to see that the double integral of $h$ over $D_1$ is equal to the double integral of $h$ over $D_2$, and so:

(3)
\begin{align} \quad \iint_D h \: dA = \iint_{D_1} h \: dA + \iint_{D_2} h \: dA = 2\iint_{D_1} h \: dA \end{align}

Now $D_1$ is a type 1 region, and thus:

(4)
\begin{align} \quad = 2 \int_{-r}^r \int_0^{\sqrt{r^2 - x^2}} h \: dy \: dx = 2 \int_{-r}^r \left [ hy \right ]_{y = 0}^{y = \sqrt{r^2 - x^2}} \: dx = 2h \int_{-r}^{r} \sqrt{r^2 - x^2} \: dx \end{align}

We will now need to use Integration by Trigonometric Substitution to evaluate this integral. Let $x = r\sin \theta$. Then $dx = r \cos \theta \: d \theta$. We will use the trigonometric identities $1 - \sin^2 \theta = \cos^2 \theta$, $\cos^2 \theta = \frac{1 + \cos (2 \theta)}{2}$, and $\sin (2 \theta) = 2 \sin \theta \cos \theta$ and so:

(5)
\begin{align} \quad \int \sqrt{r^2 - x^2} \: dx = \int \sqrt{r^2 - (r \sin \theta)^2} r \cos \theta \: d \theta = \int \sqrt{r^2(1 - \sin^2 \theta)} r \cos \theta \: d \theta = r^2 \int \cos^2 \theta \: d \theta \\ \quad = r^2 \int \frac{1 + \cos (2 \theta)}{2} \: d \theta = r^2 \left [ \frac{\theta}{2} + \frac{\sin (2 \theta)}{4} \right] = \frac{r^2}{2} \left [ \theta + \sin \theta \cos \theta \right ] \end{align}

Now since $x = r \sin \theta$, we have that $\sin \theta = \frac{x}{r}$ and $\cos \theta = \frac{\sqrt{r^2 - x^2}}{r}$. Furthermore, we have that $\theta = \sin^{-1} \left ( \frac{x}{r} \right )$ so:

(6)
\begin{align} \quad = \frac{r^2}{2} \left [ \sin^{-1} \left ( \frac{x}{r} \right ) + \frac{x \sqrt{r^2 - x^2}}{r^2} \right ] \end{align}

Putting this into our integral from earlier and we obtain that:

(7)
\begin{align} \quad 2h \int_{-r}^{r} \sqrt{r^2 - x^2} \: dx = \frac{2hr^2}{2} \left [ \sin^{-1} \left ( \frac{x}{r} \right ) + \frac{x \sqrt{r^2 - x^2}}{r^2} \right ]_{-r}^{r} = r^2 h \left [ \left ( \sin^{-1} (1) + 0 \right ) - \left ( \sin^{-1} (-1) + 0 \right )\right ] = r^2h (\pi) = \pi r^2 h \end{align}

Thus we have shown that the volume of a cylinder is $V = \pi r^2 h$.

Another proof showing that the volume of a cylinder is $V = \pi r^2 h$ was given on the Volumes of Geometric Shapes Examples 1 page and involved only single integrals by rotating a rectangular region in the $xy$-plane about the $y$-axis. Notice that this derivative is MUCH simpler. This is due to the fact that the volume $V$ calculated above could be obtained much more easily using the basic washer method for single integrals.

## Example 2

Evaluate the double integral $\iint_D \frac{x}{1 + y^2} \: dA$ where $D$ is the region in the first quadrant and between the curves $y = \frac{x}{4}$ and $y = \frac{x - 1}{2}$.

We will first sketch our region $D$: This region can be described as $D = \{ (x, y) \in \mathbb{R}^2 : 4y ≤ x ≤ 2y + 1, 0 ≤ y ≤ \frac{1}{2} \}$. Thus we have a type 2 region, and so:

(8)
\begin{align} \quad \iint_D \frac{x}{1 + y^2} \: dA = \int_0^{1/2} \int_{4y}^{2y + 1} \frac{x}{1 + y^2} \: dx \: dy = \int_0^{1/2} \frac{1}{2} \left [ \frac{x^2}{1 + y^2} \right ]_{x = 4y}^{x = 1 + 2y} \: dy = \frac{1}{2} \int_{0}^{1/2} \frac{(2y + 1)^2}{1 + y^2} - \frac{(4y)^2}{1 + y^2} \: dy \\ = \frac{1}{2} \int_0^{1/2} \frac{1 + 4y + 4y^2 - 16y^2}{1 + y^2} = \frac{1}{2} \int_0^{1/2} \frac{1 + 4y - 12y^2}{1 + y^2} \: dy \end{align}

Now we take $\frac{1 + 4y - 12y^2}{1 + y^2}$ and long divide it to break it up into partial fractions. We thus get that:

(9)
\begin{align} \quad \frac{1 + 4y - 12y^2}{1 + y^2} = -12 + \frac{4x + 13}{1 + x^2} = -12 + \frac{4x}{1 + x^2} + \frac{13}{1 + x^2} \end{align}

Therefore we get that:

(10)
\begin{align} \quad = \frac{1}{2} \int_0^{1/2} \frac{1 + 4y - 12y^2}{1 + y^2} \: dy = \frac{1}{2} \left [ \int_0^{1/2} -12 \: dy + \int_{0}^{1/2} \frac{4x}{1 + x^2} \: dy + \int_0^{1/2} \frac{13}{1 + x^2} \: dy\right ] \end{align}

Note the middle integral must be evaluated with substitution. Let $u = 1 + y^2$ so that $du = 2y \: dy$ and $2du = 4y \: dy$, and so $\int \frac{4y}{1 + y^2} \: dy = 2 \ln \mid 1 + y^2 \mid = 2 \ln (1 + y^2)$. The rightmost integral can be evaluated by noting that $\int \frac{1}{1 + y^2} \: dy = \tan^{-1} (y)$.

(11)
\begin{align} \quad = \frac{1}{2} \left [ [-12y]_{y=0}^{y=1/2} + 2 [\ln (1 + y^2)]_{y=0}^{y=1/2} + 13 [\tan^{-1} (y)]_{y=0}^{y=1/2} \right ] = -3 + \ln \left ( \frac{5}{4} \right ) + \frac{13}{2} \tan^{-1} \left ( \frac{1}{2} \right ) \end{align}