Evaluating Triple Integrals over General Domains Examples 1

# 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)
\begin{align} \quad \iiint_E f(x, y, z) \: dV = \iint_D \left [ \int_{u_1(x, y)}^{u_2(x, u)} f(x, y, z) \: dz \right ] \: dA \end{align}
• 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)
\begin{align} \quad \iiint_E f(x, y, z) \: dV = \iint_D \left [ \int_{u_1(y, z)}^{u_2(y, z)} f(x, y, z) \: dx \right ] \: dA \end{align}
• 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)
\begin{align} \quad \iiint_E f(x, y, z) \: dV = \iint_D \left [ \int_{u_1(x, z)}^{u_2(x, z)} f(x, y, z) \: dy \right ] \: dA \end{align}

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)
\begin{align} \quad \iiint_D \frac{z}{x^2 + z^2} \: dV = \iint_D \left [ \int_0^z \frac{z}{x^2 + z^2} \: dx \right ] \: dz \: dy = \iint_D z \left [ \int_0^z \frac{1}{1 + \left ( \frac{x}{z}\right )^2} \: dx \right ] \: dy \: dz = \iint_D z \: \cdot \frac{1}{z} \left [ \tan^{-1} \left ( \frac{x}{z} \right ) \: dz\right]_{x=0}^{x=z} \: dz \: dy \\ = \iint_D \tan^{-1} (1) - \tan^{-1} (0) \: dz \: dy = \iint_D \frac{\pi}{4} \: dz \: dy = \int_1^4 \int_y^4 \frac{\pi}{4} \: dz \: dy = \int_1^4 \frac{\pi}{4} \left [ z \right ]_{z=y}^{z=4} \: dy = \int_1^4 \frac{\pi}{4} [4 - y] \: dy \\ = \frac{\pi}{4} \left [4y - \frac{y^2}{2} \right ]_1^4 = \frac{\pi}{4} \left [(16 - 8) - \left ( 4 - \frac{1}{2} \right ) \right ] = \frac{\pi}{4} \left [8 - \frac{7}{2} \right ] = \frac{\pi}{4} \cdot \frac{9}{2}= \frac{9 \pi}{8} \end{align}

## 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)
\begin{align} \quad \iiint_E (3 + 2xy) \: dV = \iint_D \left [ \int_0^{\sqrt{4 - x^2 - y^2}} (3 + 2xy) \: dz \right ] \: dy \: dx = \iint_{D_1} \left [ \int_0^{\sqrt{4 - x^2 - y^2}} (3 + 2xy) \: dz \right ] \: dy \: dx + \iint_{D_2} \left [ \int_0^{\sqrt{4 - x^2 - y^2}} (3 + 2xy) \: dz \right ] \: dy \: dx \\ \quad \int_{-2}^{2} \int_{0}^{\sqrt{4 - x^2}} \int_{0}^{\sqrt{4 - x^2 - y^2}} (3 + 2xy) \: dz \: dy \: dx + \int_{-2}^{2} \int_{-\sqrt{4 - x^2}}^{0} \int_{0}^{\sqrt{4 - x^2 - y^2}} (3 + 2xy) \: dz \: dy \: dx \\ \quad \int_{-2}^{2} \int_0^{\sqrt{4 - x^2}} \sqrt{4 - x^2 - y^2} (3 + 2xy) \: dy \: dx + \int_{-2}^{2} \int_{-\sqrt{4 - x^2}}^{0} \sqrt{4 - x^2 - y^2} (3 + 2xy) \: dy \: dx \\ \end{align}