Lecture: Section 15.6 Triple Integrals

Example 1: Evaluating over a Box

Evaluate $\iiint_B xyz^2 \,dV$ if $B = [1, 2] \times [-1, 1] \times [0, 3]$.

Solution:

Let's choose the order $dx\,dy\,dz$. This breaks the "all bounds are zero" pattern early!

$$ \int_0^3 \int_{-1}^1 \int_1^2 xyz^2 \,dx \,dy \,dz $$

Inner (with respect to $x$):

$$ \int_1^2 xyz^2 \,dx = yz^2 \left[ \frac{1}{2}x^2 \right]_1^2 = yz^2 \left( \frac{4}{2} - \frac{1}{2} \right) = \frac{3}{2} yz^2 $$

Middle (with respect to $y$):

$$ \int_{-1}^1 \frac{3}{2} yz^2 \,dy = \frac{3}{2}z^2 \left[ \frac{1}{2}y^2 \right]_{-1}^1 = \frac{3}{2}z^2 \left( \frac{1}{2} - \frac{1}{2} \right) = 0 $$

Outer (with respect to $z$):

$$ \int_0^3 0 \,dz = 0 $$

(Note: The integral is zero because the function $f(x,y,z) = xyz^2$ is an odd function with respect to $y$, and we integrated over a symmetric $y$-interval $[-1, 1]$.)

Example 2: Separable Integrals

Evaluate $\iiint_B e^{x+y+z} \,dV$ over the box $B = [0, 1] \times [0, \ln(2)] \times [0, \ln(3)]$.

Solution:

The function can be written as $f(x,y,z) = e^x e^y e^z$. When the function is separable and the limits are constants, we can separate the integral:

$$ \iiint_B e^x e^y e^z \,dV = \left( \int_0^1 e^x \,dx \right) \left( \int_0^{\ln(2)} e^y \,dy \right) \left( \int_0^{\ln(3)} e^z \,dz \right) $$

$$ = \left[ e^x \right]_0^1 \cdot \left[ e^y \right]_0^{\ln(2)} \cdot \left[ e^z \right]_0^{\ln(3)} $$

$$ = (e^1 - e^0) \cdot (e^{\ln(2)} - e^0) \cdot (e^{\ln(3)} - e^0) $$

$$ = (e - 1) \cdot (2 - 1) \cdot (3 - 1) = 2(e - 1) $$

Example 3: Volume of a Tetrahedron

Find the volume of the solid tetrahedron $E$ bounded by the planes $x + 2y + z = 2$, $x = 0$, $y = 0$, and $z = 0$.

Solution:

Step 1: Find the $z$-limits (floor and ceiling).

First, let's visualize the solid.

The solid tetrahedron $E$ in the first octant.

A ray parallel to the $z$-axis enters at the "floor" $z = 0$. It exits at the "ceiling", which is the slanted plane $x + 2y + z = 2$. We solve this for $z$:

$$ z = 2 - x - 2y $$

So, our $z$-limits are $0 \le z \le 2 - x - 2y$.

Step 2: Set up the Inner Integral.

This gives us the innermost integral. The volume $V$ is the integral over the solid $E$, which we can write as a double integral over the shadow $D$ of an inner $z$-integral:

$$ V = \iiint_E 1 \,dV = \iint_D \left[ \int_0^{2-x-2y} 1 \,dz \right] \,dA $$

Step 3: Find the 2D shadow region $D$ in the $xy$-plane.

The shadow $D$ is the 2D region in the $xy$-plane directly "under" the solid. This is the region where the solid meets the $xy$-plane ($z=0$). We set $z=0$ in the ceiling equation: $0 = 2 - x - 2y$, which gives $x + 2y = 2$.

This line, along with $x=0$ and $y=0$, forms our triangular shadow $D$. We can write $D$ as a Type I 2D region:

Solve for $y$: $y = 1 - x/2$. So, $D = \{ (x,y) \mid 0 \le x \le 2, \, 0 \le y \le 1 - x/2 \}$.

The triangular shadow $D$ in the $xy$-plane, bounded by $x=0$, $y=0$, and $x+2y=2$.

Step 4: Set up and evaluate the full integral.

$$ V = \int_0^2 \int_0^{1-x/2} \left( \int_0^{2-x-2y} 1 \,dz \right) \,dy \,dx $$

Inner ($dz$): $\int_0^{2-x-2y} 1 \,dz = [z]_0^{2-x-2y} = 2 - x - 2y$.

Middle ($dy$): $\int_0^{1-x/2} (2 - x - 2y) \,dy = [2y - xy - y^2]_0^{1-x/2}$

$$ = \left( 2(1-\frac{x}{2}) - x(1-\frac{x}{2}) - (1-\frac{x}{2})^2 \right) - 0 $$

$$ = (2 - x) - (x - \frac{x^2}{2}) - (1 - x + \frac{x^2}{4}) $$

$$ = 2 - x - x + \frac{x^2}{2} - 1 + x - \frac{x^2}{4} = 1 - x + \frac{x^2}{4} $$

Outer ($dx$): $\int_0^2 (1 - x + \frac{x^2}{4}) \,dx = [x - \frac{x^2}{2} + \frac{x^3}{12}]_0^2$

$$ = (2 - \frac{4}{2} + \frac{8}{12}) - 0 = (2 - 2 + \frac{2}{3}) = \frac{2}{3} $$

Example 4: Volume Between Two Surfaces

Find the volume of the solid $E$ bounded *between* the two paraboloids $z_1 = x^2+y^2$ (the "floor") and $z_2 = 8 - x^2 - y^2$ (the "ceiling").

Solution:

Step 1 ($z$-limits): The problem gives them to us. The floor is $z_1 = x^2+y^2$. The ceiling is $z_2 = 8 - x^2 - y^2$.

Step 2 (Shadow $D$): To find the $xy$-shadow, we find where the floor and ceiling intersect:

$$ x^2+y^2 = 8 - x^2 - y^2 \implies 2x^2 + 2y^2 = 8 \implies x^2+y^2 = 4 $$

The shadow is a circle in the $xy$-plane of radius 2, centered at the origin.

click off f(x,y) and g(x,y) to see the volume to be computed and see the domain D in the xy plane

Step 3 (Integral): We set up the $...dz \,dy \,dx$ integral. We describe the circular shadow $D$ as a 2D region:

• $-2 \le x \le 2$
• $-\sqrt{4-x^2} \le y \le \sqrt{4-x^2}$

Inner ($dz$): $\int_{x^2+y^2}^{8-x^2-y^2} 1 \,dz = [z]_{x^2+y^2}^{8-x^2-y^2} = (8 - x^2 - y^2) - (x^2+y^2) = 8 - 2x^2 - 2y^2$.

Outer Double Integral:

$$ V = \int_{-2}^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} (8 - 2(x^2+y^2)) \,dy \,dx $$

Converting to Polar Coordinates

This integral is very difficult in rectangular coordinates. We can solve it by converting the *double integral* to polar coordinates.

• The Region $D$: The shadow is a circle $x^2+y^2 \le 4$. In polar, this is simply $0 \le r \le 2$ and $0 \le \theta \le 2\pi$.
• The Integrand: $8 - 2(x^2+y^2)$ becomes $8 - 2r^2$.
• The Differential: $dy \,dx$ becomes $r \,dr \,d\theta$.

Set up the Polar Integral:

$$ V = \int_0^{2\pi} \int_0^2 (8 - 2r^2) \cdot r \,dr \,d\theta $$

$$ V = \int_0^{2\pi} \int_0^2 (8r - 2r^3) \,dr \,d\theta $$

Inner ($dr$): $\int_0^2 (8r - 2r^3) \,dr = \left[ 4r^2 - \frac{2r^4}{4} \right]_0^2 = \left[ 4r^2 - \frac{r^4}{2} \right]_0^2$

$$ = \left( 4(2)^2 - \frac{(2)^4}{2} \right) - 0 = \left( 16 - \frac{16}{2} \right) = 16 - 8 = 8 $$

Outer ($d\theta$):

$$ V = \int_0^{2\pi} 8 \,d\theta = [8\theta]_0^{2\pi} = 8(2\pi) - 0 = 16\pi $$

Example 5: A New Solid - Order 1 ($dz \,dy \,dx$)

Find the volume of the solid $E$ bounded by the surfaces $z=y+1$, $z=x^2+1$, and $y=1$.

Solution (a): Order $dz \,dy \,dx$

Let's start by visualizing the solid. The boundaries are two "troughs" ($z=y+1$ and $z=x^2+1$) cut by the vertical plane $y=1$.

The solid $E$ bounded by $z=y+1$ (red), $z=x^2+1$ (blue), and $y=1$ (green).

We want to set up $V = \iiint_E 1 \,dz \,dy \,dx$. This means our shadow $D$ is in the $xy$-plane. Let's use this view to find the bounds:

Visualizing the order dzdydx, we will go from the bottom surface z = x^2+1 to the top surface z = y+1. the middle integral is y = x^2 to y = 1 and the outer integral is -1 to 1

1. Inner ($z$) bounds: A ray parallel to the $z$-axis (the "Ray") enters the solid at the "floor" $z = x^2+1$ and exits at the "ceiling" $z=y+1$. So, $x^2+1 \le z \le y+1$.

2. Shadow ($D_{xy}$) bounds: The shadow in the $xy$-plane is the 2D region where the solid exists. It's bounded by the plane $y=1$ and the intersection of the two surfaces. Let's find the intersection: $y+1 = x^2+1 \implies y = x^2$. So the shadow is bounded by the line $y=1$ and the parabola $y=x^2$.

For $...dy \,dx$, we have outer $x$-bounds from $-1$ to $1$, and inner $y$-bounds from $y=x^2$ to $y=1$.

3. Set up and Integrate:

$$ V = \int_{-1}^1 \int_{x^2}^1 \int_{x^2+1}^{y+1} 1 \,dz \,dy \,dx $$

Inner ($dz$): $\int_{x^2+1}^{y+1} 1 \,dz = [z]_{x^2+1}^{y+1} = (y+1) - (x^2+1) = y - x^2$.

Middle ($dy$): $\int_{x^2}^1 (y - x^2) \,dy = \left[ \frac{1}{2}y^2 - x^2y \right]_{x^2}^1$

$$ = \left( \frac{1}{2}(1)^2 - x^2(1) \right) - \left( \frac{1}{2}(x^2)^2 - x^2(x^2) \right) = \left(\frac{1}{2} - x^2\right) - \left(\frac{1}{2}x^4 - x^4\right) = \frac{1}{2} - x^2 + \frac{1}{2}x^4 $$

Outer ($dx$): $\int_{-1}^1 \left(\frac{1}{2} - x^2 + \frac{1}{2}x^4\right) \,dx$. This is an even function, so we can simplify:

$$ = 2 \int_0^1 \left(\frac{1}{2} - x^2 + \frac{1}{2}x^4\right) \,dx = 2 \left[ \frac{1}{2}x - \frac{1}{3}x^3 + \frac{1}{10}x^5 \right]_0^1 $$

$$ = 2 \left( \left(\frac{1}{2} - \frac{1}{3} + \frac{1}{10}\right) - 0 \right) = 2 \left( \frac{15 - 10 + 3}{30} \right) = 2 \left( \frac{8}{30} \right) = \frac{16}{30} = \frac{8}{15} $$

Solution (b): Same Solid, New Order ($dx \,dz \,dy$)

Now, what if we want to change the order to $dx \,dz \,dy$? This means our shadow $D$ is on the $yz$-plane.

1. Inner ($x$) bounds: First, solve any surface involving $x$ for $x$. We have $z = x^2+1 \implies x^2 = z-1 \implies x = \pm\sqrt{z-1}$. This also tells us we must have $z \ge 1$.

A ray parallel to the $x$-axis ("piercing the solid") enters at the "floor" $x = -\sqrt{z-1}$ and exits at the "ceiling" $x = \sqrt{z-1}$. So, $-\sqrt{z-1} \le x \le \sqrt{z-1}$.

2. Shadow ($D_{yz}$) bounds: Now we find the shadow in the $yz$-plane. We are bounded by $y=1$ (a line) and $z=y+1$ (a line). What about $z=x^2+1$? Since $x$ is "crushed" onto this plane, its smallest value is $x=0$. This occurs on the $yz$-plane itself. The surface $z=x^2+1$ starts at $z=1$ (when $x=0$).

So, our shadow is bounded by the three lines: $y=1$, $z=y+1$, and $z=1$.

The $yz$-shadow (in purple) is bounded by $y=1$, $z=y+1$, and $z=1$.

For $...dz \,dy$, our outer $y$-bounds are $0 \le y \le 1$. Our inner $z$-bounds are from the floor $z=1$ up to the ceiling $z=y+1$.

3. Set up and Integrate:

$$ V = \int_0^1 \int_1^{y+1} \int_{-\sqrt{z-1}}^{\sqrt{z-1}} 1 \,dx \,dz \,dy $$

Inner ($dx$): $\int_{-\sqrt{z-1}}^{\sqrt{z-1}} 1 \,dx = [x]_{-\sqrt{z-1}}^{\sqrt{z-1}} = \sqrt{z-1} - (-\sqrt{z-1}) = 2\sqrt{z-1}$.

Middle ($dz$): $\int_1^{y+1} 2(z-1)^{1/2} \,dz = 2 \left[ \frac{2}{3}(z-1)^{3/2} \right]_1^{y+1}$

$$ = \frac{4}{3} \left[ ((y+1)-1)^{3/2} - (1-1)^{3/2} \right] = \frac{4}{3} [y^{3/2} - 0] = \frac{4}{3}y^{3/2} $$

Outer ($dy$): $\int_0^1 \frac{4}{3}y^{3/2} \,dy = \frac{4}{3} \left[ \frac{2}{5}y^{5/2} \right]_0^1 = \frac{8}{15}(1 - 0) = \frac{8}{15}$.

Both methods give the same volume, $V = 8/15$, which confirms our bounds are correct!

Example 6: Evaluating an "Impossible" Integral

Evaluate $\int_0^2 \int_0^{4-x^2} \int_0^x \frac{\sin(2z)}{4-z} \,dy \,dz \,dx$.

Solution:

Let's try to integrate as written.

Inner ($dy$): $\int_0^x \frac{\sin(2z)}{4-z} \,dy = x \frac{\sin(2z)}{4-z}$.

Middle ($dz$): $\int_0^{4-x^2} x \frac{\sin(2z)}{4-z} \,dz$. This requires us to find the antiderivative of $\frac{\sin(2z)}{4-z}$, which is not possible with standard functions . We must change the order.

Our goal is to make $dz$ the *outer* integral, hoping the inner two integrals simplify the $z$-term.

The original bounds are: $0 \le x \le 2$, $0 \le z \le 4-x^2$, and $0 \le y \le x$. Let's visualize this solid.

Visualize the surface bounded by all constraints.

We want the new order $dy\,dx\,dz$. This means our shadow is on the $xz$-plane. Let's look at that shadow $D_{xz}$ and its new $dx\,dz$ order.

The shadow $D_{xz}$ (purple) is bounded by $x=0, z=0$, and $z=4-x^2$.

To change to a $dx \,dz$ order, we solve for $x$: $z=4-x^2 \implies x^2=4-z \implies x = \sqrt{4-z}$ (since $x \ge 0$).

Our new outer $z$-bounds are $0 \le z \le 4$. Our new middle $x$-bounds are $0 \le x \le \sqrt{4-z}$.

The inner $y$-bounds are given by $y=x$ (ceiling) and $y=0$ (floor), so $0 \le y \le x$.

Set up and Integrate:

$$ \int_0^4 \int_0^{\sqrt{4-z}} \int_0^x \frac{\sin(2z)}{4-z} \,dy \,dx \,dz $$

Inner ($dy$): $\int_0^x \frac{\sin(2z)}{4-z} \,dy = x \frac{\sin(2z)}{4-z}$.

Middle ($dx$): $\int_0^{\sqrt{4-z}} x \frac{\sin(2z)}{4-z} \,dx = \frac{\sin(2z)}{4-z} \left[ \frac{1}{2}x^2 \right]_0^{\sqrt{4-z}}$

$$ = \frac{\sin(2z)}{4-z} \left( \frac{(\sqrt{4-z})^2}{2} - 0 \right) = \frac{\sin(2z)}{4-z} \left( \frac{4-z}{2} \right) = \frac{1}{2}\sin(2z) $$

Outer ($dz$): $\int_0^4 \frac{1}{2}\sin(2z) \,dz = \left[ -\frac{1}{4}\cos(2z) \right]_0^4 $

$$ = \left(-\frac{1}{4}\cos(8)\right) - \left(-\frac{1}{4}\cos(0)\right) = \frac{1 - \cos(8)}{4} $$

Example 7: Average Value

Find the average value of $f(x,y,z) = xyz$ over the box $B = [0, 1] \times [0, 2] \times [0, 3]$.

Solution:

Step 1: Find the Volume $V(B)$.

$V = (1-0) \cdot (2-0) \cdot (3-0) = 6$.

Step 2: Find the integral $\iiint_B f \,dV$.

$$ \iiint_B xyz \,dV = \left( \int_0^1 x \,dx \right) \left( \int_0^2 y \,dy \right) \left( \int_0^3 z \,dz \right) $$

$$ = \left[ \frac{x^2}{2} \right]_0^1 \cdot \left[ \frac{y^2}{2} \right]_0^2 \cdot \left[ \frac{z^2}{2} \right]_0^3 $$

$$ = \left( \frac{1}{2} \right) \cdot \left( \frac{4}{2} \right) \cdot \left( \frac{9}{2} \right) = \frac{1}{2} \cdot 2 \cdot \frac{9}{2} = \frac{9}{2} $$

Step 3: Calculate the average.

$$ f_{\text{avg}} = \frac{1}{V(B)} \iiint_B f \,dV = \frac{1}{6} \left( \frac{9}{2} \right) = \frac{9}{12} = \frac{3}{4} $$

Example 8: Setting up Center of Mass

Set up the integrals needed to find the center of mass of the tetrahedron from Example 3 ($E$ bounded by $x+2y+z=2$ and the coordinate planes) assuming constant density $\rho(x,y,z) = k$.

Solution:

The limits of integration are the same as in Example 3: $0 \le x \le 2$, $0 \le y \le 1-x/2$, $0 \le z \le 2-x-2y$.

Mass $m = \iiint_E k \,dV = k \iiint_E 1 \,dV = k \cdot V(E)$. From Example 3, $V(E) = 2/3$, so $m = 2k/3$.

We just need to set up the moments:

$$ M_{yz} = \iiint_E x \rho \,dV = \int_0^2 \int_0^{1-x/2} \int_0^{2-x-2y} (x \cdot k) \,dz \,dy \,dx $$

$$ M_{xz} = \iiint_E y \rho \,dV = \int_0^2 \int_0^{1-x/2} \int_0^{2-x-2y} (y \cdot k) \,dz \,dy \,dx $$

$$ M_{xy} = \iiint_E z \rho \,dV = \int_0^2 \int_0^{1-x/2} \int_0^{2-x-2y} (z \cdot k) \,dz \,dy \,dx $$

Then $\bar{x} = M_{yz} / m$, $\bar{y} = M_{xz} / m$, $\bar{z} = M_{xy} / m$. Notice that the constant $k$ will cancel from the numerator and denominator.