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Evaluating Triple Integrals over General Domains Examples 1
Recall from the Triple Integrals over General Domains page that we can evaluate triple integrals depending on what type of region we are integrating over with the following formulas.
 Type 1 Regions: The first type of region we might integrate over is in the form $E = \{ (x, y, z) : (x, y) \in D, u_1(x, y) ≤ z ≤ u_2(x, y) \}$. We are thus integrating $f$ for which $(x, y) \in D$ and $z$ is trapped between the surfaces generated by $u_1$ and $u_2$. Thus we have that for type 1 regions:
(1)
 Type 2 Regions: The second type of region we might integrate over is in the form $E = \{ (x, y, z) : (y, z) \in D, u_1(y, z) ≤ x ≤ u_2(y, z) \}$. We are thus integrating $f$ for which $(y, z) \in D$ and $x$ is trapped between the surfaces generated by $u_1$ and $u_2$. Thus we have that for type 2 regions:
(2)
 Type 3 Regions: The third type of region we might integrate over is in the form $E = \{ (x, y, z) : (x, z) \in D, u_1(x, z) ≤ y ≤ u_2(x, z) \}$. We are thus integrating $f$ for which $(x, z) \in D$ and $y$ is trapped between the surfaces generated by $u_1$ and $u_2$. Thus we have that for type 3 regions:
(3)
Let’s look at some examples.
Example 1
Evaluate the triple integral $\iiint_E \frac{z}{x^2 + z^2} \: dV$ where $E = \{ (x, y, z) \in \mathbb{R}^3 : 0 ≤ x ≤ z, 1 ≤ y ≤ 4, y ≤ z ≤ 4 \}$.
Let’s evaluating this triple integral by treating $E$ as a type two domain. Let $D = \{ (y, z) \in D : 1 ≤ y ≤ 4, y ≤ z ≤ 4 \}$. Then we have that:
(4)
Example 2
Set up double iterated integrals for $\iiint_E (3 + 2xy) \: dV$ where $E$ is the region given by $x^2 + y^2 + z^2 ≤ 4$ above the $xy$plane.
Since the region $E$ is not given explicitly by inequalities, let’s try to determine what this region looks like. We first note that the equation $x^2 + y^2 + z^2 = 4$ represents a sphere centered at the origin with radius $2$. Therefore $x^2 + y^2 + z^2 ≤ 4$ represents a ball of radius $2$ (the inner part of the sphere). We’re also given that $E$ is located above the $xy$plane, that is, $z ≥ 0$, and so $E$ is the upper half ball with radius $2$ as depicted below.
We note that the upper surface of $E$ is given by the equation $z = \sqrt{4 – x^2 – y^2}$ and the lower surface of $E$ is given by $z = 0$. Furthermore, the projection of $E$ onto the $xy$plane is the disk $x^2 + y^2 ≤ 4$ sketched below.
Let’s call this region $D$. $D$ is not $x$simple and not $y$simple, so we split $D$ into two region, $D_1 = \{ (x, y) \in \mathbb{R}^2 : 2 ≤ x ≤ 2, 0 ≤ y ≤ \sqrt{4 – x^2}$ an $D_2 = \{ (x, y) \in \mathbb{R}^2 : 2 ≤ x ≤ 2, \sqrt{4 – x^2} ≤ y ≤ 0 \}$. Therefore:
(5)