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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)
 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
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)
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)
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)
Now we’ll use the trigonometric substitution of $\cos^2 \theta = \frac{1}{2} + \frac{\cos (2 \theta)}{2}$ and so:
(6)
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