This deck covers triple integrals over solid regions: the volume element, finding the six nested-limit orders for a given solid, and projecting onto a coordinate plane, with applications to mass, center of mass, centroids, moments, and moments of inertia. It targets the classic traps of putting an outer variable in the inner limits, forgetting the density factor when computing mass, and projecting onto the wrong plane.
Subject: Calculus III · 108 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Objectives
By the end of this deck you can:
1. Read a triple integral over a solid region and say what it measures.
2. Find the nested limits for a solid, in any of the six orders, by projecting onto a coordinate plane.
3. Compute the volume of a solid as a triple integral of 1.
4. Set up and evaluate mass, center of mass, centroids, moments, and moments of inertia using a density function.
Warm-up
Discussion prompt
Before we open Week 11 - Triple Integrals & Applications: without looking back, what was the main idea of Week 10 - Change of Variables: Polar Coordinates, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
Double integrals in polar coordinates and the general change of variables. Builds the area element from a polar rectangle, explains why the extra factor r must appear, and introduces the Jacobian with polar as the special case.
Concept
A double integral sums a quantity over a flat region in the plane. A triple integral does the same over a solid region in space.
We write the triple integral of a function over a solid region as:
\[ \iiint_E f(x,y,z)\,dV \]
Here the region of integration is a three-dimensional solid, and the tiny piece we sum over is a chunk of volume.
Counterexample
Discussion prompt
A double integral sums a quantity over a flat region in the plane. A triple integral does the same over a solid region in space.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
We write the triple integral of a function over a solid region as:
Intuition
Imagine slicing the solid into a huge number of tiny rectangular boxes, like sugar cubes packed into a shape.
Each little box has a tiny volume. Multiply the function's value there by that tiny volume, then add up over every box.
\[ \Delta V = \Delta x\,\Delta y\,\Delta z \]
As the boxes shrink toward zero size, that sum becomes the triple integral. The integral is a limit of these sums.
Analogy
Discussion prompt
Explain Chop the solid into tiny boxes by analogy to something with no Calculus III in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Imagine slicing the solid into a huge number of tiny rectangular boxes, like sugar cubes packed into a shape.
Concept
Formally, the triple integral is the limit of a Riemann sum over the tiny boxes:
\[ \iiint_E f\,dV = \lim_{\|P\|\to 0} \sum_{i=1}^{n} f(x_i,y_i,z_i)\,\Delta V_i \]
You almost never compute this limit directly. Instead you turn it into three ordinary integrals done one after another.
iterated integral — A triple integral rewritten as three single integrals nested inside one another, each done with respect to one variable while the others are held fixed.
Explain it
Discussion prompt
Explain The formal definition to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
Formally, the triple integral is the limit of a Riemann sum over the tiny boxes:
Concept
The tiny volume, once we pass to the limit, is the volume element. In rectangular coordinates it is the product of three tiny lengths:
\[ dV = dx\,dy\,dz \]
Because multiplication does not care about order, the three differentials can be written in any sequence. That is what lets us choose the order of integration.
Concept
Three variables can be nested in six different sequences. Each is a legal way to write the same triple integral:
\[ dz\,dy\,dx,\quad dz\,dx\,dy,\quad dy\,dz\,dx, \]
\[ dy\,dx\,dz,\quad dx\,dz\,dy,\quad dx\,dy\,dz \]
The value is the same for all six. What changes is how hard the limits and the algebra are. Part of the skill is picking an easy order.
Concept
The simplest solid is a rectangular box, where each variable runs between two fixed numbers.
\[ E = [a,b]\times[c,d]\times[p,q] \]
Then all six limits are constants and the order truly does not matter at all:
\[ \iiint_E f\,dV = \int_a^b\!\int_c^d\!\int_p^q f\,dz\,dy\,dx \]
Ranking
Put in order
Put the moves of Worked example: a triple integral over a box into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Over a box, all limits are constants, so any order works.
Worked example
Evaluate the integral of the product of the three variables over the box shown.
\[ \iiint_E xyz\,dV,\qquad E=[0,1]\times[0,2]\times[0,3] \]
Write it as three nested integrals
Why: Over a box, all limits are constants, so any order works. Take dz first, then dy, then dx.
\[ \int_0^1\!\int_0^2\!\int_0^3 xyz\,dz\,dy\,dx \]
Do the inner integral over z
Why: Hold x and y constant; they factor out. Integrate z from 0 to 3.
\[ \int_0^3 xyz\,dz = xy\cdot\frac{z^2}{2}\Big|_0^3 = \frac{9}{2}xy \]
Do the middle integral over y
Why: Hold x constant. Integrate y from 0 to 2.
\[ \int_0^2 \frac{9}{2}xy\,dy = \frac{9}{2}x\cdot\frac{y^2}{2}\Big|_0^2 = 9x \]
Do the outer integral over x
Why: Integrate x from 0 to 1.
\[ \int_0^1 9x\,dx = 9\cdot\frac{x^2}{2}\Big|_0^1 = \frac{9}{2} \]
Verify by separating the box integral
Why: Over a box with a product integrand, the triple integral is the product of three single integrals. This must match.
\[ \Big(\int_0^1 x\,dx\Big)\Big(\int_0^2 y\,dy\Big)\Big(\int_0^3 z\,dz\Big)=\tfrac12\cdot 2\cdot\tfrac92=\frac{9}{2} \]
Picture it
Animation
Shows: Each line of the worked example "a triple integral over a box", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Over a box with a product integrand, the triple integral is the product of three single integrals. This must match.
Concept
Always evaluate the innermost integral first, then the middle, then the outer. Each step removes one variable.
While doing an inner integral, every variable that has not yet been integrated is treated as a constant.
By the time you reach the outer integral, only one variable is left and you get a number.
Intuition
Think of the three integrals as layers of an onion. The innermost layer is stripped off first.
The innermost variable is the one allowed to depend on everything outside it. The outermost variable depends on nothing.
This ordering is the single most important idea for getting the limits right.
Concept
If the function is just the constant 1, every tiny box contributes its own volume, and the sum is the total volume of the solid.
\[ \text{Volume}(E) = \iiint_E 1\,dV \]
So volume is the special case where the integrand is 1. Later, replacing that 1 with a density gives mass.
Step zero
Discussion prompt
Worked example: volume of a box — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Set up the integral of 1
Answer:
Worked example
Find the volume of the box with side lengths 2, 3, and 4 as a triple integral, and confirm it matches the length-times-width-times-height rule.
\[ E=[0,2]\times[0,3]\times[0,4] \]
Set up the integral of 1
Why: Volume is the triple integral of the constant 1 over the box.
\[ V=\int_0^2\!\int_0^3\!\int_0^4 1\,dz\,dy\,dx \]
Integrate over z, then y, then x
Why: Each inner integral of 1 just returns the length of that interval.
\[ \int_0^4 dz = 4,\quad \int_0^3 4\,dy = 12,\quad \int_0^2 12\,dx = 24 \]
Verify against length times width times height
Why: The elementary rule gives the same number, confirming the setup.
\[ 2\cdot 3\cdot 4 = 24 \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "volume of a box", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The elementary rule gives the same number, confirming the setup.
Concept
Here is the rule that keeps limits legal: the outermost integral must have constant limits, because after it there is no variable left to depend on.
The middle limits may depend only on the outermost variable. The innermost limits may depend on both variables outside them.
\[ \int_{a}^{b}\!\int_{g_1(x)}^{g_2(x)}\!\int_{h_1(x,y)}^{h_2(x,y)} f\,dz\,dy\,dx \]
Read it as: constants on the outside, then functions of the outer variable, then functions of the two outer variables on the inside.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Tempting mistake: letting the outermost limit depend on a variable that gets integrated away inside.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The outer integral is the last one done.
Fix it: dependence only ever points inward. Constants outside, then functions of the outer variable, then functions of the two outer variables.
Why: The outer integral is the last one done. By then y has already been integrated away, so the limit y refers to nothing. The final answer would still contain y, which cannot be a number.
Trap
Tempting mistake: letting the outermost limit depend on a variable that gets integrated away inside.
\[ \int_{0}^{\,y}\!\int_{0}^{x}\!\int_{0}^{x+y} f\,dz\,dy\,dx \quad(\text{wrong}) \]
See why it is meaningless
Why: The outer integral is the last one done. By then y has already been integrated away, so the limit y refers to nothing. The final answer would still contain y, which cannot be a number.
\[ \text{outer limit } y \text{ is undefined after } y \text{ is gone} \]
Fix it: dependence only ever points inward. Constants outside, then functions of the outer variable, then functions of the two outer variables.
\[ \int_{0}^{1}\!\int_{0}^{x}\!\int_{0}^{x+y} f\,dz\,dy\,dx \quad(\text{right}) \]
Check each level
Why: The outer x limits are constants 0 and 1. The middle y limit depends only on x. The inner z limits depend on x and y. Every dependence points inward, so the result is a pure number.
\[ x:[0,1],\; y:[0,x],\; z:[0,x+y]\;\checkmark \]
Notation
Annotate
From Trap: an outer limit that uses an inner variable — read this one piece at a time. What is each part doing?
On: \( \int_{0}^{\,y}\!\int_{0}^{x}\!\int_{0}^{x+y} f\,dz\,dy\,dx \quad(\text{wrong}) \)
Concept
For a general solid, pick a variable to integrate first, say z. The solid is trapped between a bottom surface and a top surface.
\[ h_1(x,y) \le z \le h_2(x,y) \]
The inner z-integral runs from the bottom surface up to the top surface. Those surfaces are functions of x and y, which is exactly why the inner limits may contain x and y.
Picture it
Figure (svg): A solid casting a shadow straight down onto the xy-plane, with a vertical arrow showing z running from the bottom surface to the top surface.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
After the z-integral, the two remaining integrals cover the solid's shadow on the xy-plane.
Intuition
After the z-integral, the two remaining integrals cover the solid's shadow on the xy-plane.
Figure (svg): A solid casting a shadow straight down onto the xy-plane, with a vertical arrow showing z running from the bottom surface to the top surface.
The shadow is the region you would see if you shone a light straight down. The outer two integrals are just a double integral over that shadow.
Concept
The standard recipe: integrate z first from bottom to top surface, then integrate over the shadow region as an ordinary double integral.
\[ \iiint_E f\,dV = \iint_D\left[\int_{h_1(x,y)}^{h_2(x,y)} f\,dz\right]dA \]
The shadow D is a flat region in the xy-plane, so you already know how to set up its limits: vertical or horizontal strips, just like double integrals.
Hypothesis
Predict first
Worked example: setting up a tetrahedron is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Choose to integrate z first
Why: The solid sits above the xy-plane and below the slanted plane, so z is trapped between two surfaces.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
Set up the volume of the tetrahedron bounded by the three coordinate planes and the plane that cuts the axes at 1.
\[ x+y+z=1,\quad x\ge 0,\; y\ge 0,\; z\ge 0 \]
Choose to integrate z first
Why: The solid sits above the xy-plane and below the slanted plane, so z is trapped between two surfaces.
\[ 0 \le z \le 1-x-y \]
Find the shadow on the xy-plane
Why: The shadow is where the top surface is still above the bottom, that is where 1 minus x minus y is nonnegative: the triangle with x and y nonnegative and x plus y at most 1.
\[ 0 \le y \le 1-x,\qquad 0 \le x \le 1 \]
Assemble the iterated integral
Why: Constants for x, a function of x for y, and a function of x and y for z. Dependence points inward.
\[ V=\int_0^1\!\int_0^{1-x}\!\int_0^{1-x-y} 1\,dz\,dy\,dx \]
Verify the limits are consistent
Why: At x equal to 1 the y-range collapses to a point and the z-range to a point, which matches the tetrahedron narrowing to the vertex. The setup is sound.
\[ x=1\Rightarrow y\in[0,0],\; z\in[0,0]\;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "setting up a tetrahedron", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At x equal to 1 the y-range collapses to a point and the z-range to a point, which matches the tetrahedron narrowing to the vertex. The setup is sound.
Step zero
Discussion prompt
Worked example: evaluating the tetrahedron volume — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Inner integral over z
Answer:
Worked example
Now evaluate the integral we just set up.
\[ V=\int_0^1\!\int_0^{1-x}\!\int_0^{1-x-y} 1\,dz\,dy\,dx \]
Inner integral over z
Why: The integral of 1 in z is just the top minus the bottom limit.
\[ \int_0^{1-x-y} dz = 1-x-y \]
Middle integral over y
Why: Let a stand for 1 minus x. Integrate a minus y from 0 to a.
\[ \int_0^{1-x}(1-x-y)\,dy = \frac{(1-x)^2}{2} \]
Outer integral over x
Why: Integrate one half of the square of (1 minus x) from 0 to 1.
\[ \int_0^1 \frac{(1-x)^2}{2}\,dx = \frac12\cdot\frac{1}{3} = \frac{1}{6} \]
Verify with the pyramid formula
Why: A tetrahedron is a pyramid; its volume is one third the base area times the height. The base is a right triangle of area one half and the height is 1.
\[ \frac13\cdot\frac12\cdot 1 = \frac16 \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "evaluating the tetrahedron volume", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: A tetrahedron is a pyramid; its volume is one third the base area times the height. The base is a right triangle of area one half and the height is 1.
Concept
Once z is integrated out, the shadow is an ordinary plane region. You describe it exactly as you did for double integrals.
A Type I shadow uses vertical strips: x between constants, y between two curves. A Type II shadow uses horizontal strips: y between constants, x between two curves.
\[ \text{Type I: } y\in[g_1(x),g_2(x)]\qquad \text{Type II: } x\in[k_1(y),k_2(y)] \]
Estimation
Predict first
Find the volume of the solid whose shadow is the triangle below the line y equals x in the unit strip, and whose top is the plane z equals x plus y.
Commit before you compute: what does Worked example: a solid with a slanted top come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify the inner dependence points inward
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. The inner z-limit used both x and y, the middle y-limit used only x, and the outer x-limits were constants.
Worked example
Find the volume of the solid whose shadow is the triangle below the line y equals x in the unit strip, and whose top is the plane z equals x plus y.
\[ 0\le x\le 1,\; 0\le y\le x,\; 0\le z\le x+y \]
Inner integral over z
Why: The top surface is z equals x plus y and the bottom is z equals 0, so the inner integral of 1 gives x plus y.
\[ \int_0^{x+y} dz = x+y \]
Middle integral over y
Why: Hold x constant and integrate x plus y as y runs from 0 to x.
\[ \int_0^x (x+y)\,dy = x^2 + \frac{x^2}{2} = \frac{3x^2}{2} \]
Outer integral over x
Why: Integrate three-halves x squared from 0 to 1.
\[ \int_0^1 \frac{3x^2}{2}\,dx = \frac{3}{2}\cdot\frac{1}{3} = \frac12 \]
Verify the inner dependence points inward
Why: The inner z-limit used both x and y, the middle y-limit used only x, and the outer x-limits were constants. A legal nesting producing a pure number, so one half is trustworthy.
\[ V=\tfrac12 \]
Picture it
Animation
Shows: Each line of the worked example "a solid with a slanted top", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The inner z-limit used both x and y, the middle y-limit used only x, and the outer x-limits were constants. A legal nesting producing a pure number, so one half is trustworthy.
Concept
Projecting onto the xy-plane and integrating z first is only one choice. You could integrate x first and project onto the yz-plane, or integrate y first and project onto the xz-plane.
Each of the three projection planes gives two nesting orders, which is where all six orders come from.
\[ \text{project on } xy \to \text{integrate } z\text{ first};\;\; xz \to y;\;\; yz \to x \]
Anomaly
Predict first
A student writes this, and it looks reasonable:
Suppose the solid is a wedge whose top and bottom are given as functions of x and y. Someone integrates x first anyway, projecting onto the yz-plane.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: To integrate x first you need x expressed between two surfaces as functions of y and z.
Match the projection to how the solid is described. Surfaces given as z equals a function of x and y call for integrating z first and projecting onto the xy-plane.
Why: To integrate x first you need x expressed between two surfaces as functions of y and z. But the bounding surfaces are given in terms of x and y, so the inner x-limits cannot be written and the limits come out inconsistent.
Trap
Suppose the solid is a wedge whose top and bottom are given as functions of x and y. Someone integrates x first anyway, projecting onto the yz-plane.
\[ z=h_1(x,y),\; z=h_2(x,y)\;\text{ but integrate } x \text{ first (wrong)} \]
See the breakdown
Why: To integrate x first you need x expressed between two surfaces as functions of y and z. But the bounding surfaces are given in terms of x and y, so the inner x-limits cannot be written and the limits come out inconsistent.
\[ x \in [\,?\,,\,?\,]\;\text{ cannot be read off } z=h(x,y) \]
Match the projection to how the solid is described. Surfaces given as z equals a function of x and y call for integrating z first and projecting onto the xy-plane.
\[ \iint_D\!\left[\int_{h_1(x,y)}^{h_2(x,y)} f\,dz\right]dA \;\text{ (right)} \]
Check that the inner limits are readable
Why: With z integrated first, the inner limits are exactly the two given surfaces, and the shadow D on the xy-plane is easy to describe. Everything is expressible, so the projection is the correct one.
\[ z:[h_1,h_2],\; (x,y)\in D\;\checkmark \]
Break the constraint
Discussion prompt
The rule this trap just fixed:
With z integrated first, the inner limits are exactly the two given surfaces, and the shadow D on the xy-plane is easy to describe. Everything is expressible, so the projection is the correct one.
Now break it on purpose. Build a case that violates it and follow the consequences until something visibly fails. Where does the failure first show up — and would you have noticed it if you had not been looking?
Hint: The dangerous rules are the ones whose violation still produces an answer. If yours fails loudly, try to find one that fails quietly.
Answer:
To integrate x first you need x expressed between two surfaces as functions of y and z. But the bounding surfaces are given in terms of x and y, so the inner x-limits cannot be written and the limits come out inconsistent.
Concept
Ask which variable is cleanly trapped between two single surfaces across the whole solid. Integrate that one first.
Then look at the shadow on the opposite plane. If that shadow is awkward in one strip direction, switch the remaining two to the other order.
A good order can turn an impossible-looking integral into a routine one. The value never changes, only the effort.
Ranking
Put in order
Put the moves of Worked example: rewriting the order of integration into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Integrating x first means x is trapped from the plane x equals 0 up to x equals 1 minus y minus z.
Worked example
Rewrite the tetrahedron volume with x integrated first instead of z, projecting onto the yz-plane, and confirm the same answer.
\[ x+y+z=1,\; x,y,z\ge 0 \]
Solve the plane for x
Why: Integrating x first means x is trapped from the plane x equals 0 up to x equals 1 minus y minus z.
\[ 0 \le x \le 1-y-z \]
Describe the shadow on the yz-plane
Why: The shadow is where 1 minus y minus z is nonnegative: the triangle with y and z nonnegative and their sum at most 1.
\[ 0 \le z \le 1-y,\qquad 0 \le y \le 1 \]
Evaluate the new integral
Why: The structure is identical to the earlier calculation with the letters relabeled, so it produces the same middle and outer integrals.
\[ \int_0^1\!\int_0^{1-y}\!\int_0^{1-y-z} dx\,dz\,dy = \frac16 \]
Verify against the first order
Why: The dz dy dx order gave one sixth, and this dx dz dy order gives one sixth. The value is independent of order, as it must be.
\[ \frac16 = \frac16 \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "rewriting the order of integration", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The dz dy dx order gave one sixth, and this dx dz dy order gives one sixth. The value is independent of order, as it must be.
Concept
Triple integrals are linear: a sum splits into two integrals, and a constant factor pulls out front.
\[ \iiint_E (a f + b g)\,dV = a\iiint_E f\,dV + b\iiint_E g\,dV \]
They are also additive over regions: if a solid splits into two non-overlapping pieces, integrate over each and add.
\[ \iiint_{E_1\cup E_2} f\,dV = \iiint_{E_1} f\,dV + \iiint_{E_2} f\,dV \]
Concept
Another way to see the inner integral: it computes the area of a cross-section, and the outer integrals stack those slices.
This is the same idea as finding a volume by integrating cross-sectional area in single-variable calculus, now built automatically into the nesting.
\[ V=\int \big(\text{area of a slice}\big)\,d(\text{stacking variable}) \]
Concept
The average value of a function over a solid is its integral divided by the volume of the solid.
\[ \bar f = \frac{1}{V}\iiint_E f\,dV,\qquad V=\iiint_E 1\,dV \]
This is the natural three-dimensional version of adding values and dividing by how many there are.
Estimation
Predict first
Find the average value of the function equal to x over the unit cube.
Commit before you compute: what does Worked example: average value on a cube come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify by symmetry
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. On the interval from 0 to 1 the values of x are symmetric about one half, so the average must be one half.
Worked example
Find the average value of the function equal to x over the unit cube.
\[ E=[0,1]\times[0,1]\times[0,1],\quad f=x \]
Compute the volume
Why: The unit cube has volume 1.
\[ V=1\cdot 1\cdot 1 = 1 \]
Integrate x over the cube
Why: Over a box with a separable integrand, the triple integral is a product of single integrals; the y and z integrals each give 1.
\[ \iiint_E x\,dV = \Big(\int_0^1 x\,dx\Big)(1)(1) = \frac12 \]
Divide by the volume
Why: Average value is the integral over the volume.
\[ \bar f = \frac{1/2}{1} = \frac12 \]
Verify by symmetry
Why: On the interval from 0 to 1 the values of x are symmetric about one half, so the average must be one half. It matches.
\[ \bar f = \tfrac12 \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "average value on a cube", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: On the interval from 0 to 1 the values of x are symmetric about one half, so the average must be one half. It matches.
Concept
Applications of triple integrals answer two kinds of questions about a solid object: how much of something there is, and where it is concentrated.
How much leads to mass and moments. Where leads to the center of mass and the centroid. All of them start from a density.
Concept
density — The mass packed into each unit of volume at a point. It may vary from point to point, written as a function of position.
We usually write density with the Greek letter rho, as a function of the three coordinates.
\[ \rho = \rho(x,y,z) \]
If the density is the same everywhere the object is called homogeneous, and rho is just a constant.
Intuition
Take one of the tiny boxes. Its volume is small, and the density there tells you how much mass sits in each unit of volume.
\[ \Delta m \approx \rho\,\Delta V \]
Add up the little masses over the whole solid and you get the total mass. That sum is a triple integral.
Concept
The total mass of a solid is the triple integral of its density.
\[ m = \iiint_E \rho(x,y,z)\,dV \]
Notice that volume is just the special case where the density is 1. Volume counts space; mass weights that space by density.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Asked for the mass of a solid with density equal to z on the unit cube, a common slip is to integrate 1 and report the volume.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Integrating 1 answers how much space the solid takes up, not how much mass it holds.
Keep the density inside the integral. Integrate rho, not 1.
Why: Integrating 1 answers how much space the solid takes up, not how much mass it holds. The density z was dropped entirely.
Trap
Asked for the mass of a solid with density equal to z on the unit cube, a common slip is to integrate 1 and report the volume.
\[ \iiint_E 1\,dV = 1 \quad(\text{this is volume, not mass}) \]
Spot the error
Why: Integrating 1 answers how much space the solid takes up, not how much mass it holds. The density z was dropped entirely.
\[ \text{answer } 1 \text{ ignores } \rho=z \]
Keep the density inside the integral. Integrate rho, not 1.
\[ m=\iiint_E z\,dV = (1)(1)\int_0^1 z\,dz = \frac12 \]
Check the two answers differ
Why: The volume is 1 but the mass is one half, because the lower half of the cube where z is small holds less mass. The density genuinely changes the answer.
\[ V=1,\quad m=\tfrac12 \]
Translation
\( \iiint_E 1\,dV = 1 \quad(\text{this is volume, not mass}) \)
Draw it
Translate both ways. First write the expression above as a sentence with no symbols in it at all. Then cover it, and write your sentence back as notation. If the two versions disagree, the disagreement is the thing to fix.
Step zero
Discussion prompt
Worked example: mass with variable density — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Write the mass integral
Answer:
Worked example
Find the mass of the unit cube whose density at a point is the sum of the coordinates.
\[ E=[0,1]^3,\qquad \rho = x+y+z \]
Write the mass integral
Why: Mass is the triple integral of the density over the solid.
\[ m=\iiint_E (x+y+z)\,dV \]
Split using linearity and symmetry
Why: The three terms are symmetric over the cube, so each integrates to the same amount; compute one and triple it.
\[ m = 3\iiint_E x\,dV \]
Evaluate the single piece
Why: Over the box the x-integral separates; the y and z integrals each give 1.
\[ \iiint_E x\,dV = \Big(\int_0^1 x\,dx\Big)(1)(1) = \frac12 \]
Combine
Why: Multiply the single piece by three.
\[ m = 3\cdot\frac12 = \frac32 \]
Verify the average density is reasonable
Why: Average density is mass over volume, which is three halves. Since each coordinate averages one half, the density x plus y plus z should average three halves. It matches.
\[ \frac{m}{V} = \frac{3/2}{1} = \frac32 \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "mass with variable density", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Average density is mass over volume, which is three halves. Since each coordinate averages one half, the density x plus y plus z should average three halves. It matches.
Concept
A first moment measures how the mass is distributed relative to a plane. Each tiny mass is weighted by its distance to that plane.
The distance to the yz-plane is x, so the moment about the yz-plane weights the density by x.
\[ M_{yz} = \iiint_E x\,\rho\,dV \]
Larger moment means the mass sits farther from that plane on average, in the positive direction.
Concept
There is one first moment for each coordinate plane, each weighting the density by the matching coordinate.
\[ M_{yz}=\iiint_E x\rho\,dV,\quad M_{xz}=\iiint_E y\rho\,dV,\quad M_{xy}=\iiint_E z\rho\,dV \]
Read the subscript as the plane you are measuring distance to: the moment about the xy-plane uses the z-coordinate, and so on.
Concept
The center of mass is the balance point of the solid. Each coordinate is the matching first moment divided by the total mass.
\[ \bar x = \frac{M_{yz}}{m},\qquad \bar y = \frac{M_{xz}}{m},\qquad \bar z = \frac{M_{xy}}{m} \]
Dividing by the mass turns a mass-weighted total into an average position.
Intuition
Picture balancing the solid on a fingertip. The single point where it balances in every direction is the center of mass.
Heavy regions pull the balance point toward themselves. That is exactly the mass-weighting in the moment integrals.
Concept
When the density is constant, it cancels between each moment and the mass. The balance point then depends only on the shape.
centroid — The center of mass of a solid with constant (homogeneous) density. It is the purely geometric center of the region, found by averaging each coordinate over the volume.
\[ \bar x = \frac1V\iiint_E x\,dV,\quad \bar y = \frac1V\iiint_E y\,dV,\quad \bar z = \frac1V\iiint_E z\,dV \]
Matching
Match the pairs
Match each term to the definition this lesson gave it — not the one you would guess from the word.
Why: These are the working definitions of iterated integral, density, centroid as Week 11 - Triple Integrals & Applications uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Estimation
Predict first
Find the z-coordinate of the centroid of the tetrahedron bounded by the coordinate planes and the plane x plus y plus z equals 1.
Commit before you compute: what does Worked example: a centroid coordinate come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify by symmetry of the tetrahedron
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. The solid treats x, y, and z identically, so all three centroid coordinates must be equal.
Worked example
Find the z-coordinate of the centroid of the tetrahedron bounded by the coordinate planes and the plane x plus y plus z equals 1.
\[ \bar z = \frac{1}{V}\iiint_E z\,dV,\qquad V=\frac16 \]
Inner integral over z
Why: Integrate z from the bottom plane 0 up to the top surface 1 minus x minus y.
\[ \int_0^{1-x-y} z\,dz = \frac{(1-x-y)^2}{2} \]
Middle integral over y
Why: Let a be 1 minus x. Integrating one half of the square of (a minus y) from 0 to a gives a cubed over six.
\[ \int_0^{1-x}\frac{(1-x-y)^2}{2}\,dy = \frac{(1-x)^3}{6} \]
Outer integral over x
Why: Integrate one sixth of (1 minus x) cubed from 0 to 1 to get the moment about the xy-plane.
\[ M_{xy}=\int_0^1 \frac{(1-x)^3}{6}\,dx = \frac16\cdot\frac14 = \frac{1}{24} \]
Divide by the volume
Why: The centroid coordinate is the moment over the volume.
\[ \bar z = \frac{1/24}{1/6} = \frac14 \]
Verify by symmetry of the tetrahedron
Why: The solid treats x, y, and z identically, so all three centroid coordinates must be equal. Each equals one quarter, consistent with the standard result.
\[ \bar x=\bar y=\bar z=\tfrac14 \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "a centroid coordinate", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The solid treats x, y, and z identically, so all three centroid coordinates must be equal. Each equals one quarter, consistent with the standard result.
Anomaly
Predict first
A student writes this, and it looks reasonable:
With a varying density, someone finds the geometric centroid and calls it the center of mass, dropping rho from the moments.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Leaving out the density assumes every part of the solid is equally heavy.
Keep the density in both the moment and the mass. Only when rho is constant does it cancel and reduce to the centroid.
Why: Leaving out the density assumes every part of the solid is equally heavy. If denser material sits high up, the true balance point is higher than the geometric center.
Trap
With a varying density, someone finds the geometric centroid and calls it the center of mass, dropping rho from the moments.
\[ \bar z \stackrel{?}{=} \frac1V\iiint_E z\,dV \quad(\text{wrong when } \rho \text{ varies}) \]
See the error
Why: Leaving out the density assumes every part of the solid is equally heavy. If denser material sits high up, the true balance point is higher than the geometric center.
\[ \text{missing } \rho \text{ in both } M_{xy} \text{ and } m \]
Keep the density in both the moment and the mass. Only when rho is constant does it cancel and reduce to the centroid.
\[ \bar z = \frac{\iiint_E z\,\rho\,dV}{\iiint_E \rho\,dV} \]
Check the reduction
Why: Set rho to a constant and it factors out of the top and bottom, cancelling to leave the pure shape average. So the centroid is the special case, not the general rule.
\[ \rho=\text{const}\Rightarrow \bar z = \frac1V\iiint_E z\,dV\;\checkmark \]
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Concept
If the solid and its density are symmetric across a plane, the center of mass lies on that plane. You get that coordinate for free, without integrating.
For example, a solid symmetric about the plane where x equals 0, with matching density, has its x-coordinate of the center of mass equal to 0.
Always check for symmetry before grinding through an integral; it can save an entire computation.
Concept
A moment of inertia measures how mass is spread around an axis, weighting each tiny mass by the square of its distance to that axis.
Because it uses distance squared, mass far from the axis counts much more than mass close to it. These are called second moments.
Intuition
The moment of inertia is how hard it is to start or stop something spinning about an axis. Mass placed far out makes it harder to spin.
That is why a figure skater spins faster when pulling their arms in: less mass far from the axis means a smaller moment of inertia.
Concept
For each axis, the squared distance uses the other two coordinates. Distance to the z-axis, for instance, involves x and y.
\[ I_x=\iiint_E (y^2+z^2)\rho\,dV,\quad I_y=\iiint_E (x^2+z^2)\rho\,dV \]
\[ I_z=\iiint_E (x^2+y^2)\rho\,dV \]
Concept
There is also a second moment about the origin, using the squared distance to the origin, which involves all three coordinates.
\[ I_0 = \iiint_E (x^2+y^2+z^2)\rho\,dV \]
It is the sum of the three axis moments cut in half, and a useful check: it must not equal any single axis moment. Do not confuse it with the axis moments.
\[ I_0 = \tfrac12\,(I_x+I_y+I_z) \]
Step zero
Discussion prompt
Worked example: moment of inertia of a cube — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Split the integrand
Answer:
Worked example
Find the moment of inertia about the z-axis for the unit cube with density 1.
\[ E=[0,1]^3,\;\rho=1,\qquad I_z=\iiint_E (x^2+y^2)\,dV \]
Split the integrand
Why: By linearity, integrate the x-squared term and the y-squared term separately.
\[ I_z = \iiint_E x^2\,dV + \iiint_E y^2\,dV \]
Evaluate one term
Why: Over the box the x-squared integral separates; the y and z integrals each give 1, and the integral of x squared from 0 to 1 is one third.
\[ \iiint_E x^2\,dV = \Big(\int_0^1 x^2\,dx\Big)(1)(1) = \frac13 \]
Add the two equal terms
Why: By symmetry the y-squared term is also one third, so the sum is two thirds.
\[ I_z = \frac13 + \frac13 = \frac23 \]
Verify against the box formula
Why: For a homogeneous box the z-axis moment is the mass times the sum of the two in-plane side squares, over three. Here the mass is 1 and each side is 1.
\[ \frac{m\,(a^2+b^2)}{3} = \frac{1\,(1+1)}{3} = \frac23 \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "moment of inertia of a cube", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: For a homogeneous box the z-axis moment is the mass times the sum of the two in-plane side squares, over three. Here the mass is 1 and each side is 1.
Concept
The radius of gyration is the single distance at which you could place all the mass, as a thin ring, to get the same moment of inertia.
\[ r = \sqrt{\frac{I}{m}} \]
It repackages a moment of inertia as an effective distance, which is often easier to interpret.
Concept
Often a solid is handed to you as a set of inequalities rather than a picture. Each inequality is a wall of the region.
\[ E=\{(x,y,z): 0\le x\le 1,\; 0\le y\le x,\; 0\le z\le x+y\} \]
The inequalities that bound one variable between the other two become that variable's integration limits directly. The description is already the limits in disguise.
Intuition
Scan the inequalities for the one variable trapped by expressions in the others. That is your innermost variable.
Then find the variable trapped only by constants and a single other variable; that is the middle. The one bounded purely by numbers is outermost.
This is just the inward-pointing rule read backward, from a written region straight to an iterated integral.
Concept
When a solid is caught between a lower graph and an upper graph over a shadow region, integrating z first automatically produces the height as top minus bottom.
\[ V=\iint_D \big(z_{\text{top}}-z_{\text{bottom}}\big)\,dA = \iint_D\!\int_{z_{\text{bottom}}}^{z_{\text{top}}} 1\,dz\,dA \]
So the double-integral height formula you already know is exactly the inner z-integral of the triple integral, done for you.
Explain it
Discussion prompt
Explain Volume between two surfaces to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
When a solid is caught between a lower graph and an upper graph over a shadow region, integrating z first automatically produces the height as top minus bottom.
Ranking
Put in order
Put the moves of Worked example: center of mass with varying density into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Mass is the density x integrated over the cube; the y and z integrals each give 1.
Worked example
Find the x-coordinate of the center of mass of the unit cube whose density at a point equals x.
\[ E=[0,1]^3,\qquad \rho = x \]
Find the mass
Why: Mass is the density x integrated over the cube; the y and z integrals each give 1.
\[ m=\iiint_E x\,dV = \int_0^1 x\,dx = \frac12 \]
Find the moment about the yz-plane
Why: Weight the density by x, giving x times x, so integrate x squared over the cube.
\[ M_{yz}=\iiint_E x\cdot x\,dV = \int_0^1 x^2\,dx = \frac13 \]
Divide moment by mass
Why: The x-coordinate of the center of mass is the moment over the mass.
\[ \bar x = \frac{1/3}{1/2} = \frac23 \]
Verify the direction of the shift
Why: The density grows with x, so more mass sits near x equal to 1 and the balance point should lie beyond the geometric center one half. Two thirds is greater than one half, as expected.
\[ \bar x = \tfrac23 > \tfrac12 \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "center of mass with varying density", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The density grows with x, so more mass sits near x equal to 1 and the balance point should lie beyond the geometric center one half. Two thirds is greater than one half, as expected.
Concept
A few quick checks catch most errors. A volume and a mass must both come out positive.
The center of mass must lie inside, or on, the solid. If a coordinate lands outside the region, a limit or a sign is wrong.
With a density that increases in some direction, the center of mass should shift that way compared with the plain centroid.
Analogy
Discussion prompt
Explain Sanity checks on your answer by analogy to something with no Calculus III in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
A few quick checks catch most errors. A volume and a mass must both come out positive.
Concept
Some solids fight rectangular coordinates: cylinders, cones, spheres, and anything with round symmetry give ugly square-root limits.
The next deck switches to cylindrical and spherical coordinates, where the volume element picks up an extra factor but the limits become simple constants.
\[ dV = r\,dz\,dr\,d\theta \quad\text{or}\quad dV=\rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta \]
The choice of coordinate system is guided by the symmetry of the solid; match the coordinates to the shape and the work collapses.
Counterexample
Discussion prompt
Some solids fight rectangular coordinates: cylinders, cones, spheres, and anything with round symmetry give ugly square-root limits.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
The next deck switches to cylindrical and spherical coordinates, where the volume element picks up an extra factor but the limits become simple constants.
Pattern
1. Sketch the solid and pick a first variable
Why: Choose the variable that is cleanly trapped between two single surfaces across the whole solid; that becomes the innermost integral.
2. Write the inner limits as those two surfaces
Why: Bottom surface up to top surface. These limits may depend on the two remaining variables.
3. Project onto the matching coordinate plane
Why: The shadow is a flat region; describe it with Type I or Type II strips to get the middle and outer limits.
4. Confirm dependence points inward
Why: Outer limits constant, middle depend only on the outer variable, inner may depend on both. Then integrate from the inside out.
Fill the middle
Fill in the blanks
From Recipe: mass, center of mass, inertia — finish the line. Write what belongs on the right of the equals sign before you look.
M_\iiint_E x\rho\,dV,\;\ldots = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Integrate rho over the solid. If rho is 1 you are only finding volume.
Pattern
1. Mass is the density integrated
Why: Integrate rho over the solid. If rho is 1 you are only finding volume.
\[ m=\iiint_E \rho\,dV \]
2. First moments weight by one coordinate
Why: Multiply the density by x, y, or z inside the integral to get the moment about the opposite plane.
\[ M_{yz}=\iiint_E x\rho\,dV,\;\ldots \]
3. Center of mass divides moment by mass
Why: Each coordinate of the balance point is its first moment over the total mass.
\[ \bar x=\frac{M_{yz}}{m},\;\ldots \]
4. Moments of inertia weight by squared distance
Why: For an axis, weight the density by the square of the distance to that axis, using the other two coordinates.
\[ I_z=\iiint_E (x^2+y^2)\rho\,dV \]
Notation
Annotate
From Recipe: mass, center of mass, inertia — read this one piece at a time. What is each part doing?
On: \( I_z=\iiint_E (x^2+y^2)\rho\,dV \)
Elimination
Eliminate the wrong options
Which iterated integral is set up correctly, with dependence pointing inward?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: The outer x-limits are constants 0 and 1, the middle y-limit depends only on the outer variable x, and the inner z-limit depends on both x and y. Every dependence points inward, so the result is a pure number.
Check
A solid has z running from 0 to x plus y, and its shadow on the xy-plane is the triangle where y runs from 0 to x and x runs from 0 to 1.
Check your understanding
Which iterated integral is set up correctly, with dependence pointing inward?
Answer: A
Why: The outer x-limits are constants 0 and 1, the middle y-limit depends only on the outer variable x, and the inner z-limit depends on both x and y. Every dependence points inward, so the result is a pure number.
Check
Consider the solid where x runs from 0 to 1, y runs from 0 to x, and z runs from 0 to x plus y.
Check your understanding
What is the volume of this solid?
Answer: A
Why: Integrating 1 gives inner x+y, then the middle integral over y from 0 to x is 3x-squared over 2, and the outer integral from 0 to 1 gives 1/2. This matches the worked example.
Check
The unit cube has density equal to z at each point.
Check your understanding
What is the mass of the cube?
Answer: A
Why: Mass is the integral of the density z over the cube. The x and y integrals each give 1 and the integral of z from 0 to 1 is 1/2, so the mass is 1/2.
Prediction
Predict first
What is the z-coordinate of its center of mass?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: 1/4
Why: The z-coordinate is the moment about the xy-plane divided by the volume, that is 1/24 divided by 1/6, which equals 1/4.
Check
The tetrahedron bounded by the coordinate planes and x plus y plus z equals 1 has constant density. Its volume is 1/6 and its moment about the xy-plane is 1/24.
Check your understanding
What is the z-coordinate of its center of mass?
Answer: A
Why: The z-coordinate is the moment about the xy-plane divided by the volume, that is 1/24 divided by 1/6, which equals 1/4.
Elimination
Eliminate the wrong options
What is the moment of inertia about the z-axis?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: The z-axis moment integrates x-squared plus y-squared. Each term is 1/3 over the unit cube, so the total is 2/3.
Check
The unit cube has density 1. Recall the integral of a squared coordinate over the cube is 1/3.
Check your understanding
What is the moment of inertia about the z-axis?
Answer: A
Why: The z-axis moment integrates x-squared plus y-squared. Each term is 1/3 over the unit cube, so the total is 2/3.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Recipe: set up any triple integral · Recipe: mass, center of mass, inertia · From double integrals to triple integrals · Chop the solid into tiny boxes · The formal definition. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
A triple integral sums a quantity over a solid, built from tiny boxes of volume.
\[ \iiint_E f\,dV \]
Set the limits by choosing a first variable trapped between two surfaces, then projecting onto a coordinate plane for the outer two. Outer limits are constant, and dependence always points inward.
Volume is the integral of 1; replace that 1 with a density to get mass.
\[ V=\iiint_E 1\,dV,\qquad m=\iiint_E \rho\,dV \]
Weight the density by a coordinate for a first moment and center of mass, or by a squared distance for a moment of inertia.
\[ \bar x=\frac{M_{yz}}{m},\qquad I_z=\iiint_E (x^2+y^2)\rho\,dV \]
| Quantity | Integrand |
|---|---|
| Volume | 1 |
| Mass | density |
| First moment | coordinate times density |
| Moment of inertia | squared distance times density |
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