Triple integrals

Background

• Double integrals beyond volume

What we're building to

• At the risk of sounding obvious, triple integrals are just like double integrals, but in three dimensions. They are written abstractly as

  $\begin{array}{r}{\iiint }_{R}fdV\end{array}$‍

  where
  • $R$‍ is some region in three-dimensional space.
  • $f(x,y,z)$‍ is some scalar-valued function which takes points in three-dimensional space as its input.
  • $dV$‍ is a tiny unit of volume. In cartesian coordinates, this is expanded as $dV=dxdydz$‍.
• Concretely, these are computed as three embedded integrals:

  $\begin{array}{r}{\int }_{z_1}^{z_2}\underset{\text{This is a function purely of }z}{\underbrace{{\int }_{y_1(z)}^{y_2(z)}\stackrel{\text{This is a function purely of }y\text{ and }z}{\overbrace{{\int }_{x_1(y,z)}^{x_2(y,z)}f(x,y,z)dx}}dy}}dz\end{array}$‍

  As with double integrals, the bounds of inner integrals might be functions of the outer variables. These bound functions are what encodes the shape of $R$‍.
• Use a three-dimensional integral anytime you get that sensation of wanting to chop up a three-dimensional region into infinitely many pieces, associate each piece with a value, then add them all up. One place where this is surprisingly useful is just finding the volume of three-dimensional regions by adding up all the tiny volumes $dV$‍.
• As with double integrals, the hard part is finding the right bounds which encode your region. This just takes some practice, and a willingness to roll up your sleeves and dive into the muck of a problem.

Example 1: Rectangular prism with variable density

• One corner is sitting at the origin.
• One of its edges of length $3$‍ rests on the positive $x$‍-axis.
• One of its edges of length $2$‍ rests on the positive $y$‍-axis.
• One of its edges of length $5$‍ rests on the positive $z$‍-axis.

Key question: What is the mass of the entire block?

• Slice it with planes representing constant values of $x$‍.
• Slice it with planes representing constant values of $y$‍.
• Slice it with planes representing constant values of $z$‍.

• You can think of the inner most integral as adding up little bits of mass along lines parallel to the $x$‍-axis. It returns some expression of $y$‍ and $z$‍, which is a way of saying

  "Depending on the choice for the $y$‍ and $z$‍ coordinates of your line, which is parallel to the $x$‍-axis, this is what the sum of the infinitesimal masses along that line will be."

• The next integral, with respect to $y$‍, adds up the infinitesimal masses of those lines in the $y$‍-direction, giving the infinitesimal mass of a sheet parallel to the $xy$‍-plane. It will return an expression purely in terms of $z$‍, which says

  "Depending on the height of your sheet above the $xy$‍-plane, this is what its infinitesimal mass will be".

• Finally, the outermost integral adds up the masses of these sheets as $z$‍ ranges from $0$‍ to $5$‍. It returns a constant, which is the (no-longer-infinitesimal) mass of the block of metal as a whole.

Example 2: Using a triple integral to compute volume.

• The paraboloid $z=x^2+y^2$‍
• The plane $z=2(x+y+1)$‍

Choose 1 answer:

Choose 1 answer:

• (Choice A)  

  A circle centered at $(1,1)$‍ with radius $2$‍

  A

  A circle centered at $(1,1)$‍ with radius $2$‍
• (Choice B)  

  A circle centered at $(-1,-1)$‍ with radius $4$‍

  B

  A circle centered at $(-1,-1)$‍ with radius $4$‍

Check

Answer

The equation

$(x-1{)}^2+(y-1{)}^2=4$‍

represents a circle centered at $(1,1)$‍ with radius $\sqrt{4}=2$‍.

Example 3: Volume of a conical region

"But wait,"

"I already know how to compute the volume of a cone!"

Choose 1 answer:

Choose 1 answer:

• (Choice A)  

  $\begin{array}{r}{\int }_0^{2-\sqrt{x^2+z^2}}{\int }_{?}^{?}{\int }_{?}^{?}dxdzdy\end{array}$‍

  A

  $\begin{array}{r}{\int }_0^{2-\sqrt{x^2+z^2}}{\int }_{?}^{?}{\int }_{?}^{?}dxdzdy\end{array}$‍
• (Choice B)  

  $\begin{array}{r}{\int }_{?}^{?}{\int }_0^{2-\sqrt{x^2+z^2}}{\int }_{?}^{?}dxdydz\end{array}$‍

  B

  $\begin{array}{r}{\int }_{?}^{?}{\int }_0^{2-\sqrt{x^2+z^2}}{\int }_{?}^{?}dxdydz\end{array}$‍
• (Choice C)  

  $\begin{array}{r}{\int }_{?}^{?}{\int }_{?}^{?}{\int }_0^{2-\sqrt{x^2+z^2}}dydxdz\end{array}$‍

  C

  $\begin{array}{r}{\int }_{?}^{?}{\int }_{?}^{?}{\int }_0^{2-\sqrt{x^2+z^2}}dydxdz\end{array}$‍

Check

Answer

The third choice is correct:

$\begin{array}{r}{\int }_{?}^{?}{\int }_{?}^{?}{\int }_0^{2-\sqrt{x^2+z^2}}dydxdz\end{array}$‍

Since the upper bound of $y$‍ is given as a function of $x$‍ and $z$‍, the integral handling $y$‍ must be inside the integrals handling $x$‍ and $z$‍. This way, the variables $x$‍ and $z$‍ can be integrated out as we work through the middle and outer integrals.

Choose 1 answer:

Choose 1 answer:

• (Choice A)  

  For a given point $(x,y,z)$‍ in $R$‍, the coordinates $x$‍ and $z$‍ must satisfy the equation $0=2-\sqrt{x^2+z^2}$‍

  A

  For a given point $(x,y,z)$‍ in $R$‍, the coordinates $x$‍ and $z$‍ must satisfy the equation $0=2-\sqrt{x^2+z^2}$‍
• (Choice B)  

  Let $R_{xz}$‍ be the set of all points $(x,z)$‍ such that $(x,y,z)$‍ is a point in $R$‍ for some value of $y$‍. $R_{xz}$‍ is bounded by the curve described by the equation $0=2-\sqrt{x^2+z^2}$‍

  B

  Let $R_{xz}$‍ be the set of all points $(x,z)$‍ such that $(x,y,z)$‍ is a point in $R$‍ for some value of $y$‍. $R_{xz}$‍ is bounded by the curve described by the equation $0=2-\sqrt{x^2+z^2}$‍

Check

Answer

The second choice is correct. The hard part of this question is just parsing what is being said (as happens so often in math).

To be clear about what the question is asking, imagine projecting the region $R$‍ onto the $xz$‍-plane.

Khan Academy video wrapper

See video transcript

This will give you all of the values $(x,z)$‍ such that there was a point $(x,y,z)$‍ inside $R$‍ for some value of $y$‍. For example, the value $(x,z)=(0,1)$‍ is inside this projection, because the point $(0,0.75,1)$‍, to take one of many examples, is inside $R$‍. However, the value $(3,4)$‍ is not insdide this projection, because there is no value of $y$‍ such that $(3,y,4)$‍ is inside $R$‍.

The boundary of this two-dimensional region on the $xz$‍-plane is determined by where the condition $y=0$‍ meets with the condition $y=2-\sqrt{x^2+z^2}$‍. In other words, the boundary is given by the equation

$0=2-\sqrt{x^2+z^2}$‍

Choose 1 answer:

Choose 1 answer:

• (Choice A)  

  A circle of radius $2$‍.

  A

  A circle of radius $2$‍.
• (Choice B)  

  A circle of radius $\sqrt{2}$‍.

  B

  A circle of radius $\sqrt{2}$‍.
• (Choice C)  

  The region inside a circle of radius $2$‍.

  C

  The region inside a circle of radius $2$‍.
• (Choice D)  

  The region inside a circle of radius $\sqrt{2}$‍

  D

  The region inside a circle of radius $\sqrt{2}$‍

Check

Answer

The third answer choice is correct: The region inside a circle of radius $2$‍.

As I described in the answer to the last question, the relevant region of the $xz$‍ plane has a boundary described by the equation

$\begin{array}{rl}0& =2-\sqrt{x^2+z^2}\\ \sqrt{x^2+z^2}& =2\\ x^2+z^2& =4\end{array}$‍

This describes a circle of radius $2$‍. But remember, this is just the boundary of the region in the $xz$‍-plane that we care about; most of the points $(x,z)$‍ that our triple integrals needs to cover lie within this region.

Choose 1 answer:

Choose 1 answer:

• (Choice A)  

  $\begin{array}{r}{\int }_{-2}^2{\int }_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}{\int }_0^{2-\sqrt{x^2+z^2}}dydxdz\end{array}$‍

  A

  $\begin{array}{r}{\int }_{-2}^2{\int }_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}{\int }_0^{2-\sqrt{x^2+z^2}}dydxdz\end{array}$‍
• (Choice B)  

  $\begin{array}{r}{\int }_{-2}^2{\int }_{-\sqrt{2-z^2}}^{\sqrt{2-z^2}}{\int }_0^{2-\sqrt{x^2+z^2}}dydxdz\end{array}$‍

  B

  $\begin{array}{r}{\int }_{-2}^2{\int }_{-\sqrt{2-z^2}}^{\sqrt{2-z^2}}{\int }_0^{2-\sqrt{x^2+z^2}}dydxdz\end{array}$‍
• (Choice C)  

  $\begin{array}{r}{\int }_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}{\int }_{-2}^2{\int }_0^{2-\sqrt{x^2+z^2}}dydzdx\end{array}$‍

  C

  $\begin{array}{r}{\int }_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}{\int }_{-2}^2{\int }_0^{2-\sqrt{x^2+z^2}}dydzdx\end{array}$‍
• (Choice D)  

  $\begin{array}{r}{\int }_{-\sqrt{2-z^2}}^{\sqrt{2-z^2}}{\int }_{-2}^2{\int }_0^{2-\sqrt{x^2+z^2}}dydzdx\end{array}$‍

  D

  $\begin{array}{r}{\int }_{-\sqrt{2-z^2}}^{\sqrt{2-z^2}}{\int }_{-2}^2{\int }_0^{2-\sqrt{x^2+z^2}}dydzdx\end{array}$‍

Check

Answer

The first choice is correct. This corresponds to breaking up the circle in the $xz$‍-plane into horizontal stripes:

Within each stripe, $x$‍ ranges from $-\sqrt{4-z^2}$‍ to $\sqrt{4-z^2}$‍, and the height of the stripes themselves ranges from $z=-2$‍ to $z=2$‍. Therefore, the bounds on the outer integrals look like this:

$\begin{array}{r}{\int }_{-2}^2{\int }_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}{\int }_0^{2-\sqrt{x^2+z^2}}dydxdz\end{array}$‍

Summary

• Triple integrals are written abstractly as

  $\begin{array}{r}{\iiint }_{R}fdV\end{array}$‍

  where
  • $R$‍ is some region in three-dimensional space.
  • $f(x,y,z)$‍ is some scalar-valued function which takes points in three-dimensional space as its input.
  • $dV$‍ is a tiny unit of volume. In cartesian coordinates, this is expanded as $dV=dxdydz$‍.
• Concretely, these are computed as three embedded integrals:

  $\begin{array}{r}{\int }_{z_1}^{z_2}\underset{\text{This is a function purely of }z}{\underbrace{{\int }_{y_1(z)}^{y_2(z)}\stackrel{\text{This is a function purely of }y\text{ and }z}{\overbrace{{\int }_{x_1(y,z)}^{x_2(y,z)}f(x,y,z)dx}}dy}}dz\end{array}$‍

  As with double integrals, the bounds of inner integrals might be functions of the outer variables.
• Use a three-dimensional integral anytime you get that sensation of wanting to chop up a three-dimensional region into infinitely many pieces, associate each piece with a value, then add them all up. One place where this is surprisingly useful is just finding the volume of three-dimensional regions by adding up all the tiny volumes $dV$‍.
• As with double integrals, the hard part is finding the right bounds which encode your region. This just takes some practice, and a willingness to roll up your sleeves and dive into the muck of a problem.