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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)
 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)
We will now look at some more examples of evaluating double integrals over general domains. More examples can be found on the following pages:
 Evaluating Double Integrals over General Domains Examples 1
 Evaluating Double Integrals over General Domains Examples 2
 Evaluating Double Integrals over General Domains Examples 3
 Evaluating Double Integrals over General Domains Examples 4
 Evaluating Double Integrals over General Domains Examples 5
 Evaluating Double Integrals over General Domains Examples 6
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)
Now $D_1$ is a type 1 region, and thus:
(4)
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)
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)
Putting this into our integral from earlier and we obtain that:
(7)
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)
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)
Therefore we get that:
(10)
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)
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