Evaluating Triple Integrals over General Domains Examples 2

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 $\iiint_E xe^{-y} \: dV$ where $E = \{ (x, y, z) \in \mathbb{R}^3 : 0 ≤ x ≤ 2z, 0 ≤ y ≤ \ln x, 1 ≤ z ≤ 2 \}$.

Note that if we do not want to rewrite $E$, then we will need to integrate with respect to $y$ first, then with respect to $x$ and with respect to $z$ last. We thus get that:

(4)

\begin{align} \quad \iiint_E xe^{-y} \: dV = \int_1^2 \int_0^{2z} \int_0^{\ln x} xe^{-y} \: dy \: dx \: dz \\ \quad \iiint_E xe^{-y} \: dV = \int_1^2 \int_0^{2z} \left [ -xe^{-y} \right ]_{y=0}^{y=\ln x} \: dx \: dz \\ \quad \iiint_E xe^{-y} \: dV = \int_1^2 \int_0^{2z} \left [ -xe^{-\ln y} + x \right ] \: dx \: dz \\ \quad \iiint_E xe^{-y} \: dV = \int_1^2 \int_0^{2z} [-1 + x] \: dx :\ dz \\ \quad \iiint_E xe^{-y} \: dV = \int_1^2 \left [ -x + \frac{x^2}{2} \right ]_{x=0}^{x=2z} \: dz \\ \quad \iiint_E xe^{-y} \: dV = \int_1^2 \left [ -2z + 2z^2 \right ] \: dz \\ \quad \iiint_E xe^{-y} \: dV = \left [ -z^2 + \frac{2z^3}{3} \right ]_{z=1}^{z=2} \\ \quad \iiint_E xe^{-y} \: dV = \left ( -4 + \frac{16}{3} \right ) - \left ( -1 + \frac{2}{3} \right ) \\ \quad \iiint_E xe^{-y} \: dV = 3 + \frac{16}{3} - \frac{2}{3} \\ \quad \iiint_E xe^{-y} \: dV = \frac{5}{3} \end{align}

Example 2

Evaluate the triple integral $\iiint_E x \: dV$ where $E$ is the region bounded by the paraboloid $x = 4y^2 + 4z^2$ and the plane $x = 4$.

The region that we want to integrate over is graphed below:

Now this integral will likely be simplest to evaluate if we regard it as a type 2 region. Since the paraboloid intersects the plane $x = 4$, we have that the intersection is the circle $4 = 4y^2 + x^2$, or more simply, $y^2 + z^2 = 1$. This region can be nicely described in polar coordinates as $y = r \cos \theta$ and $z = r \sin \theta$ and $D = \{ (r, \theta) : 0 ≤ r ≤ 1, 0 ≤ \theta ≤ 2\pi \}$. Furthermore, since we must have that $4y^2 + 4z^2 ≤ z ≤ 4$, and so if $E$ is our region of integration, then:

(5)

\begin{align} \quad \iiint_E x \: dV = \iint_D \left ( \int_{4y^2 + 4z^2}^{4} x \: dx \right ) \: dA \\ \quad \iiint_E x \: dV = \iint_D \left [ \frac{x^2}{2} \right ]_{x=4y^2 + 4z^2}^{x=4} \: dA \\ \quad \iiint_E x \: dV = \iint_D \frac{ 16 - (4y^2 + 4z^2)^2}{2} \: dA \\ \end{align}

Now we can use our polar coordinates:

(6)

\begin{align} \quad \iiint_E x \: dV = \int_0^{2\pi} \int_0^1 \frac{16 - (4 (r \cos \theta)^2 + 4(r \sin \theta)^2)}{2} r \: dr \: d \theta \\ \quad \iiint_E x \: dV = \int_0^{2\pi} \int_0^1 \frac{16 - (4r^2)^2}{2} r \: dr \: d \theta \\ \quad \iiint_E x \: dV = \int_0^{2\pi} \int_0^1 \frac{16 - 16r^4}{2} r \: dr \: d \theta \\ \quad \iiint_E x \: dV = \int_0^{2\pi} \int_0^1 1 8r - 8r^5 \: dr \: d \theta \\ \quad \iiint_E x \: dV = \int_0^{2\pi} \left [ 4^2 - \frac{4r^6}{3} \right ]_{r=0}^{r=1} \: d \theta \\ \quad \iiint_E x \: dV = \int_0^{2\pi} \left [ 4 - \frac{4}{3} \right ] \: d \theta \\ \quad \iiint_E x \: dV = \int_0^{2\pi} \frac{8}{3} \: d \theta \\ \quad \iiint_E x \: dV = \frac{8}{3} [ \theta ]_{\theta = 0}^{\theta = 2\pi} \\ \quad \iiint_E x \: dV = \frac{16 \pi}{3} \end{align}