Evaluating Double Integrals over General Domains Examples 6

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:

• 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

Set up a double integral to find the volume of the solid that is below the surface $z = x^2 + y^2$ and above the region $D = \{ (x, y) \in \mathbb{R}^2 : 0 ≤ x ≤ 2, x^2 ≤ y ≤ 2x \}$.

We can immediately set up this double integral as follows:

(3)

\begin{align} \quad \: \iint_D x^2 + y^2 \: dA = \int_0^2 \int_{x^2}^{2x} x^2 + y^2 \: dy \: dx = \int_0^2 \left [ yx^2 + \frac{y^3}{3} \right ]_{y = x^2}^{y = 2x} \: dx = \int_0^2 2x^3 + \frac{8x^3}{3} - x^4 - \frac{x^6}{3} \: dx = \int_0^1 \frac{14x^3}{3} - x^4 - \frac{x^6}{3} \: dx \\ = \left [ \frac{7x^4}{6} - \frac{x^5}{5} - \frac{x^7}{21} \right ]_{x=0}^{x=2} = \frac{112}{6} - \frac{32}{5} - \frac{128}{21} = \frac{216}{35} \end{align}

Therefore the volume is $\frac{216}{36}$.

Example 2

Set up a double integral to find the volume of the solid that is above the $xy$ plane and underneath the surface $z = 1 - x^2 - 2y^2$.

The surface $z = 1 - x^2 - 2y^2$ is an elliptical paraboloid:

The elliptical paraboloid intersects the $xy$ plane when $z = 0$, that is all points $(x, y) \in \mathbb{R}^2$ such that $0 = 1 - x^2 - 2y^2$. We will denote this region as $D$. The graph of these region is given below:

This region is not type 1 and is not type 2, however, $D$ can be be split into four regions regions, $D_1$, $D_2$, $D_3$, and $D_4$ whose interiors do not intersect each other if we let each of these regions be a quarter of our ellipse. It's clear that we can take the double integral of $z$ over either region and multiply it by four to get the total volume since this parabolic cylinder is symmetric. Let $D_1 = \{ (x, y) \in \mathbb{R}^2 : 0 ≤ x ≤ 1, 0 ≤ y ≤ \sqrt{\frac{1 - x^2}{2}} \}$. Then our double integral is:

(4)

\begin{align} \quad \iint_D 1 - x^2 - 2y^2 \: dA = 4 \iint_{D_1} 1 - x^2 - 2y^2 \: dA = 4 \int_{0}^{1} \int_{0}^{\sqrt{\frac{1 - x^2}{2}}} 1 - x^2 - 2y^2 \: dy \: dx = 4 \int_0^1 \left [ y - yx^2 - \frac{2}{3} y^3 \right ]_0^{\sqrt{\frac{1 - x^2}{2}}} \: dx \\ = 4 \int_0^1 \sqrt{\frac{1 - x^2}{2}} - x^2 \sqrt{\frac{1 - x^2}{2}} - \frac{2}{3} \left ( \frac{1 - x^2}{2} \right)^{3/2} \: dx = 4 \int_0^1 (1 - x^2) \sqrt{\frac{1 - x^2}{2}} - \frac{2}{3} \left ( \frac{1 - x^2}{2} \right)^{3/2} \: dx \\ = 4 \int_0^1 \frac{1}{\sqrt{2}} (1 - x^2)^{3/2} - \frac{1}{3\sqrt{2}} (1 - x^2)^{3/2} \: dx = 4 \int_0^1 (1 - x^2)^{3/2} \left [ \frac{1}{\sqrt{2}} - \frac{1}{3 \sqrt{2}} \right ] \: dx = \frac{8}{3 \sqrt{2}} \int_0^1 (1 - x^2)^{3/2} \: dx \end{align}

Now we need to use trigonometric substitution to evaluate this integral further. Let $x = \sin \theta$ so that $dx = \cos \: d \theta$. Plugging in $x = 0$ and $x = 1$ and we get that our bounds of integration are $0$ and $\frac{\pi}{2}$ and thus:

(5)

\begin{align} \quad = \frac{8}{3 \sqrt{2}} \int_0^{\pi / 2} (1 - \sin^2 \theta)^{3/2} \cos \theta \: d \theta = \frac{8}{3 \sqrt{2}} \int_0^{\pi / 2} \cos^4 \theta \: d \theta \end{align}

Now we'll use the trigonometric substitution of $\cos^2 \theta = \frac{1}{2} + \frac{\cos (2 \theta)}{2}$ and so:

(6)

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