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

  $\begin{aligned} \iiint_\blueE{R} f \, \redE{dV} \end{aligned}$

  where
  • $\blueE{R}$start color #0c7f99, R, end color #0c7f99 is some region in three-dimensional space.
  • $f(x, y, z)$f, left parenthesis, x, comma, y, comma, z, right parenthesis is some scalar-valued function which takes points in three-dimensional space as its input.
  • $\redE{dV}$start color #bc2612, d, V, end color #bc2612 is a tiny unit of volume. In cartesian coordinates, this is expanded as $\redE{dV} = dx\,dy\,dz$start color #bc2612, d, V, end color #bc2612, equals, d, x, d, y, d, z.
• Concretely, these are computed as three embedded integrals:

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

  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 $\blueE{R}$start color #0c7f99, R, end color #0c7f99.
• 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 $dV$d, V.
• 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$3 rests on the positive $x$x-axis.
• One of its edges of length $2$2 rests on the positive $y$y-axis.
• One of its edges of length $5$5 rests on the positive $z$z-axis.

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

• Slice it with planes representing constant values of $\blueE{x}$start color #0c7f99, x, end color #0c7f99.
• Slice it with planes representing constant values of $\redE{y}$start color #bc2612, y, end color #bc2612.
• Slice it with planes representing constant values of $\greenE{z}$start color #0d923f, z, end color #0d923f.

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

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

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

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

• Finally, the outermost integral adds up the masses of these sheets as $z$z ranges from $0$0 to $5$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 $z = x^2 + y^2$z, equals, x, squared, plus, y, squared
• The plane $z = 2(x + y + 1)$z, equals, 2, left parenthesis, x, plus, y, plus, 1, right parenthesis

Choose 1 answer:

Choose 1 answer:

• (Choice A)  

  A circle centered at $(1, 1)$left parenthesis, 1, comma, 1, right parenthesis with radius $2$2

  A

  A circle centered at $(1, 1)$left parenthesis, 1, comma, 1, right parenthesis with radius $2$2
• (Choice B)  

  A circle centered at $(-1, -1)$left parenthesis, minus, 1, comma, minus, 1, right parenthesis with radius $4$4

  B

  A circle centered at $(-1, -1)$left parenthesis, minus, 1, comma, minus, 1, right parenthesis with radius $4$4

Check

[Answer]

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)  

  $\begin{aligned} \int_{0}^{2-\sqrt{x^2+z^2}} \int_{?}^{?} \int_{?}^{?} \,dx \,dz \,dy \end{aligned}$

  A

  $\begin{aligned} \int_{0}^{2-\sqrt{x^2+z^2}} \int_{?}^{?} \int_{?}^{?} \,dx \,dz \,dy \end{aligned}$
• (Choice B)  

  $\begin{aligned} \int_{?}^{?} \int_{0}^{2-\sqrt{x^2+z^2}} \int_{?}^{?} \,dx \,dy \,dz \end{aligned}$

  B

  $\begin{aligned} \int_{?}^{?} \int_{0}^{2-\sqrt{x^2+z^2}} \int_{?}^{?} \,dx \,dy \,dz \end{aligned}$
• (Choice C)  

  $\begin{aligned} \int_{?}^{?} \int_{?}^{?} \int_{0}^{2-\sqrt{x^2+z^2}} \,dy \,dx \,dz \end{aligned}$

  C

  $\begin{aligned} \int_{?}^{?} \int_{?}^{?} \int_{0}^{2-\sqrt{x^2+z^2}} \,dy \,dx \,dz \end{aligned}$

Check

[Answer]

Choose 1 answer:

Choose 1 answer:

• (Choice A)  

  For a given point $(x, y, z)$left parenthesis, x, comma, y, comma, z, right parenthesis in $R$R, the coordinates $x$x and $z$z must satisfy the equation $0 = 2 - \sqrt{x^2 + z^2}$0, equals, 2, minus, square root of, x, squared, plus, z, squared, end square root

  A

  For a given point $(x, y, z)$left parenthesis, x, comma, y, comma, z, right parenthesis in $R$R, the coordinates $x$x and $z$z must satisfy the equation $0 = 2 - \sqrt{x^2 + z^2}$0, equals, 2, minus, square root of, x, squared, plus, z, squared, end square root
• (Choice B)  

  Let $R_{xz}$R, start subscript, x, z, end subscript be the set of all points $(x, z)$left parenthesis, x, comma, z, right parenthesis such that $(x, y, z)$left parenthesis, x, comma, y, comma, z, right parenthesis is a point in $R$R for some value of $y$y. $R_{xz}$R, start subscript, x, z, end subscript is bounded by the curve described by the equation $0 = 2-\sqrt{x^2 + z^2}$0, equals, 2, minus, square root of, x, squared, plus, z, squared, end square root

  B

  Let $R_{xz}$R, start subscript, x, z, end subscript be the set of all points $(x, z)$left parenthesis, x, comma, z, right parenthesis such that $(x, y, z)$left parenthesis, x, comma, y, comma, z, right parenthesis is a point in $R$R for some value of $y$y. $R_{xz}$R, start subscript, x, z, end subscript is bounded by the curve described by the equation $0 = 2-\sqrt{x^2 + z^2}$0, equals, 2, minus, square root of, x, squared, plus, z, squared, end square root

Check

[Answer]

Choose 1 answer:

Choose 1 answer:

• (Choice A)  

  A circle of radius $2$2.

  A

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

  A circle of radius $\sqrt{2}$square root of, 2, end square root.

  B

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

  The region inside a circle of radius $2$2.

  C

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

  The region inside a circle of radius $\sqrt{2}$square root of, 2, end square root

  D

  The region inside a circle of radius $\sqrt{2}$square root of, 2, end square root

Check

[Answer]

Choose 1 answer:

Choose 1 answer:

• (Choice A)  

  $\begin{aligned} \int_{-2}^{2} \int_{-\sqrt{4 - z^2}}^{\sqrt{4 - z^2}} \int_{0}^{2-\sqrt{x^2+z^2}} \,dy \,dx \,dz \end{aligned}$

  A

  $\begin{aligned} \int_{-2}^{2} \int_{-\sqrt{4 - z^2}}^{\sqrt{4 - z^2}} \int_{0}^{2-\sqrt{x^2+z^2}} \,dy \,dx \,dz \end{aligned}$
• (Choice B)  

  $\begin{aligned} \int_{-2}^{2} \int_{-\sqrt{2 - z^2}}^{\sqrt{2 - z^2}} \int_{0}^{2-\sqrt{x^2+z^2}} \,dy \,dx \,dz \end{aligned}$

  B

  $\begin{aligned} \int_{-2}^{2} \int_{-\sqrt{2 - z^2}}^{\sqrt{2 - z^2}} \int_{0}^{2-\sqrt{x^2+z^2}} \,dy \,dx \,dz \end{aligned}$
• (Choice C)  

  $\begin{aligned} \int_{-\sqrt{4 - z^2}}^{\sqrt{4 - z^2}} \int_{-2}^{2} \int_{0}^{2-\sqrt{x^2+z^2}} \,dy \,dz \,dx \end{aligned}$

  C

  $\begin{aligned} \int_{-\sqrt{4 - z^2}}^{\sqrt{4 - z^2}} \int_{-2}^{2} \int_{0}^{2-\sqrt{x^2+z^2}} \,dy \,dz \,dx \end{aligned}$
• (Choice D)  

  $\begin{aligned} \int_{-\sqrt{2 - z^2}}^{\sqrt{2 - z^2}} \int_{-2}^{2} \int_{0}^{2-\sqrt{x^2+z^2}} \,dy \,dz \,dx \end{aligned}$

  D

  $\begin{aligned} \int_{-\sqrt{2 - z^2}}^{\sqrt{2 - z^2}} \int_{-2}^{2} \int_{0}^{2-\sqrt{x^2+z^2}} \,dy \,dz \,dx \end{aligned}$

Check

[Answer]

Summary

• Triple integrals are written abstractly as

  $\begin{aligned} \iiint_\blueE{R} f \, \redE{dV} \end{aligned}$

  where
  • $\blueE{R}$start color #0c7f99, R, end color #0c7f99 is some region in three-dimensional space.
  • $f(x, y, z)$f, left parenthesis, x, comma, y, comma, z, right parenthesis is some scalar-valued function which takes points in three-dimensional space as its input.
  • $\redE{dV}$start color #bc2612, d, V, end color #bc2612 is a tiny unit of volume. In cartesian coordinates, this is expanded as $\redE{dV} = dx\,dy\,dz$start color #bc2612, d, V, end color #bc2612, equals, d, x, d, y, d, z.
• Concretely, these are computed as three embedded integrals:

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

  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 $dV$d, V.
• 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.