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16.4 Triple Integrals

\(\iint_R f(x,y) dA\) is the accumulation of \(f(x,y)\) over the region \(R\).

If \(f(x,y) = 1\), then \(\iint_R dA\) is the area of \(R\).

A 2D region R in the xy-plane with a small differential area element dA labeled as dx by dy.
\[dA = dx dy = dy dx\]

Two possible orders

\(\iiint_D f(x,y,z) dV\) is the accumulation of \(f(x,y,z)\) over the volume \(D\).

If \(f(x,y,z) = 1\), then \(\iiint_D dV\) is the volume of \(D\).

A 3D volume D in xyz-space with a small differential volume element dV shown as a cube with sides dx, dy, dz.
\[dV = dx dy dz\]\[= dx dz dy\]\[= dy dx dz\]\[= dy dz dx\]\[= dz dx dy\]\[= dz dy dx\]

Six possible orders

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Example

Use a triple integral to calculate the volume of the solid under \(x + 2y + 3z = 6\) in the first octant.

Plane equation: \(x + 2y + 3z = 6\)

  • x-int: 6
  • y-int: 3
  • z-int: 2

First octant: 3D equivalent of quadrant, where \(x \ge 0, y \ge 0, z \ge 0\).

A 3D graph showing a plane intersecting the axes at x=6, y=3, and z=2, forming a tetrahedron in the first octant.
\[\text{volume} = \iiint_D dV\]

Let's look at the "floor" of the shape using the xy-plane as a Type I region.

A 2D graph of the triangular region in the xy-plane bounded by x=6 and the line y = 3 - 1/2x.

\(0 \le x \le 6\)

\(0 \le y \le 3 - \frac{1}{2}x\)

Then we will go up as high as the plane allows (from plane equation):

\[0 \le z \le 2 - \frac{1}{3}x - \frac{2}{3}y\]
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Triple Integral Bounds and Setup

bounds:

\[ \begin{aligned} 0 &\le x \le 6 \\ 0 &\le y \le 3 - \frac{1}{2}x \\ 0 &\le z \le 2 - \frac{1}{3}x - \frac{2}{3}y \end{aligned} \]

Basic rule:

  • integrate constant-bounded LAST (outside)
  • integrate the one with most variables FIRST (inside)

So, here

\[ \int_{0}^{6} \int_{0}^{3-\frac{1}{2}x} \int_{0}^{2-\frac{1}{3}x-\frac{2}{3}y} dV \]

Where \( dV = dz \, dy \, dx \). Note the order of integration corresponds to the variables \( z \), then \( y \), then \( x \).

\[ = \int_{0}^{6} \int_{0}^{3-\frac{1}{2}x} \int_{0}^{2-\frac{1}{3}x-\frac{2}{3}y} dz \, dy \, dx \]
\[ = \int_{0}^{6} \int_{0}^{3-\frac{1}{2}x} (2 - \frac{1}{3}x - \frac{2}{3}y) \, dy \, dx = \dots = \boxed{6} \]
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Alternative Integration Order: Different "Floor"

let's try a different "floor"

yz-plane is the "floor" now

3D graph of a plane in the first octant with intercepts at x=6, y=3, and z=2.
2D graph of the yz-plane projection showing a line from z=2 to y=3.

Type I:

\[ \begin{aligned} 0 &\le y \le 3 \\ 0 &\le z \le 2 - \frac{2}{3}y \end{aligned} \]

Type II:

\[ \begin{aligned} 0 &\le z \le 2 \\ 0 &\le y \le 3 - \frac{3}{2}z \end{aligned} \]

we can go in x-direction as far as the plane \( x + 2y + 3z = 6 \) allows us

  • can't go below yz-plane: \( x = 0 \)
  • plane allows max: \( x = 6 - 2y - 3z \)
\[ 0 \le x \le 6 - 2y - 3z \]

So, volume is

\[ \int_{0}^{2} \int_{0}^{3-\frac{3}{2}z} \int_{0}^{6-2y-3z} dx \, dy \, dz = \dots = \boxed{6} \]

where \( dV = dx \, dy \, dz \).

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Example: Finding Volume Between Two Cones

Find volume above \( z = \sqrt{x^2 + y^2} \) and below \( z = 2 - \sqrt{x^2 + y^2} \).

\( z = \sqrt{x^2 + y^2} \): cone opening up

\( z = 2 - \sqrt{x^2 + y^2} \): cone opening down, vertex at \( z = 2 \)

the volume of

3D plot of two cones intersecting to form a diamond-like solid volume.

Let's use the projection of this thing on the \( xy \)-plane as the "floor".

A unit circle in the xy-plane representing the projection of the intersection.

Setting the equations equal to find the intersection:

\[ \sqrt{x^2 + y^2} = 2 - \sqrt{x^2 + y^2} \]\[ 2\sqrt{x^2 + y^2} = 2 \]\[ \sqrt{x^2 + y^2} = 1 \]\[ x^2 + y^2 = 1 \]
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Unit circle with y limits defined by square root functions of x.

As Type I:

  • \( -1 \le x \le 1 \)
  • \( -\sqrt{1 - x^2} \le y \le \sqrt{1 - x^2} \)

Inside this region, the lowest possible \( z \) is \( z = \sqrt{x^2 + y^2} \)

Inside this region, the highest possible \( z \) is \( z = 2 - \sqrt{x^2 + y^2} \)

\[ \sqrt{x^2 + y^2} \le z \le 2 - \sqrt{x^2 + y^2} \]

Volume Integral

\[ \text{Volume} = \int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{\sqrt{x^2+y^2}}^{2-\sqrt{x^2+y^2}} dz \, dy \, dx \]

Terrible in Cartesian, let's go to polar.

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Triple Integration in Cylindrical Coordinates

The region of integration is defined by the following bounds in polar coordinates:

  • \( 0 \leq r \leq 1 \)
  • \( 0 \leq \theta \leq 2\pi \)

The "floor" was the \( xy \)-plane, which implies the differential area element is:

\[ dy \, dx = r \, dr \, d\theta \]
A unit circle centered at the origin on a Cartesian coordinate system with x and y axes.

Setting up the Triple Integral

The bounds for \( z \) are given by the surfaces:

\[ \underbrace{\sqrt{x^2+y^2}}_{r} \leq z \leq \underbrace{2 - \sqrt{x^2+y^2}}_{2-r} \]

The volume integral is set up as follows:

\[ \int_{0}^{2\pi} \int_{0}^{1} \int_{r}^{2-r} r \, dz \, dr \, d\theta \]

Evaluating the inner integral with respect to \( z \):

\[ = \int_{0}^{2\pi} \int_{0}^{1} rz \bigg|_{z=r}^{z=2-r} dr \, d\theta = \int_{0}^{2\pi} \int_{0}^{1} r(2 - r - r) \, dr \, d\theta \]

Continuing the calculation:

\[ = \dots = \boxed{\frac{2\pi}{3}} \]