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Integrating over a solid: the six limits of a box, the two surfaces and the shadow that give the limits of a general solid, the six orders and why they differ in effort but not in value, and volume, mass and centre of mass.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to write a triple integral over a box and over a general solid, find the inner limits from the bounding surfaces and the outer four from the projection onto a coordinate plane, choose an order that keeps the solid in one piece, and compute a volume, a mass and a centre of mass. You will also be able to say what Fubini's theorem does and does not promise about the order.
Double integrals over general regions, where the outer limits were constants and the inner limits came off a sketch. A triple integral is the same idea with one more layer: two limits from surfaces, and four more from the flat shadow the solid casts — which is a double integral you already know how to set up.
A solid $E$ is a three-dimensional region. The volume element $dV$ is the small piece of volume, which over a box is $dx\,dy\,dz$. The projection or shadow of a solid onto a coordinate plane is the flat region it covers when light shines along the third axis. Density $\rho(x,y,z)$ is mass per unit volume, and the centre of mass is the point at which the solid would balance — each of its three coordinates is an integral divided by the mass.
Over a box $B = [a_1,a_2] \times [b_1,b_2] \times [c_1,c_2]$,
$$\int\!\!\int\!\!\int_B f\,dV = \int_{a_1}^{a_2}\!\!\int_{b_1}^{b_2}\!\!\int_{c_1}^{c_2} f(x,y,z)\,dz\,dy\,dx,$$
with all six limits constant and all six orders equally easy. This is the last case in which that is true.
A general solid. Suppose $E$ lies between two surfaces in the $z$ direction:
$$E = \{(x,y,z) : (x,y) \in D,\; u_1(x,y) \le z \le u_2(x,y)\},$$
where $D$ is the shadow of $E$ in the $xy$ plane. Then
$$\int\!\!\int\!\!\int_E f\,dV = \int\!\!\int_D\left(\int_{u_1(x,y)}^{u_2(x,y)} f(x,y,z)\,dz\right)dA,$$
and the outer double integral is set up exactly as in the last two lessons. So a triple integral is one new integral wrapped around a problem you have already solved.
Six orders. There are $3! = 6$ orders, and Fubini's theorem says all six give the same number for a continuous integrand. They do not give the same amount of work: one order may describe the solid with a single set of limits while another needs it cut into pieces.
What triple integrals compute.
$$\text{volume}(E) = \int\!\!\int\!\!\int_E 1\,dV, \qquad m = \int\!\!\int\!\!\int_E \rho\,dV,$$
and the centre of mass has coordinates
$$\bar{x} = \frac{1}{m}\int\!\!\int\!\!\int_E x\rho\,dV,$$
with $\bar{y}$ and $\bar{z}$ the same with $y$ or $z$ in place of $x$. A symmetry of both the solid and the density forces the centre of mass onto the axis of that symmetry, which often removes two of the three integrals before any work is done.
Another way: picture
Shine a light straight down on the solid and look at the shadow it casts on the floor. Over each point of that shadow the solid occupies a vertical segment, from the surface underneath to the surface on top. The inner integral walks up that segment; the outer double integral sweeps over the shadow. Choosing a different order is choosing a different wall to shine the light at.
Another way: steps
To evaluate a triple integral over a solid:
It is tempting to think of a triple integral as three times as hard as a single one. In practice the inner integral is almost always the easy one: it runs between two surfaces and is over before the interesting part begins. What is left is a double integral over the shadow, and that is where the effort lives — the shadow may be a disc, a triangle or a region between two curves, and all the technique of the last two lessons applies unchanged.
So the practical question when setting up a triple integral is not which variable shall I integrate first but which projection gives me the shadow I would most like to integrate over. A solid bounded by a paraboloid and a plane projects to a disc; a tetrahedron projects to a triangle; a solid between two cylinders may project to a ring in one direction and to something horrible in another.
Choosing the projection first, and the order afterwards, turns most of these problems from long into short. It also explains why the next lesson matters so much: when the shadow is a disc, polar coordinates in the outer double integral is the obvious next move, and that combination is precisely what cylindrical coordinates are.
Integrating over the bounding box. If the inner limits are constants when the solid is not a box, the integral has quietly included territory that is not in the solid.
Leaving a variable behind. After the $z$ integral no $z$ may survive; after the $y$ integral no $y$. A stray variable means a limit was not substituted, and the final answer will not be a number.
Reading the shadow as a face. The projection of a solid is the whole region it covers, not one of its flat faces. A hemisphere sitting on the plane and a cone standing on the same circle have the same shadow.
Dividing by the volume to get a centre of mass. The denominator is the mass. They coincide only when the density is $1$, and treating them as the same is a mistake that vanishes in every textbook exercise with constant density and reappears immediately in a real one.
Fubini's theorem is often heard as the order does not matter, and that is half right in a way that costs hours. The value does not depend on the order. The description of the solid depends on it entirely, and that description is what you actually have to write down.
So the order is a choice made for convenience, and it should be made deliberately, after looking at the solid, rather than taken as $dz\,dy\,dx$ by habit. The question to ask is which direction a line crosses the solid cleanly in, and which coordinate plane it casts the friendliest shadow on. Getting that choice right is most of the skill in this lesson; getting the arithmetic right afterwards is ordinary one-variable calculus.
Find the volume under $z = 4 - x^2 - y^2$ and above $z = 0$. Integrating in $z$ first runs from $0$ up to $4 - x^2 - y^2$.
The two surfaces give the inner limits.
The shadow is where the paraboloid is above the plane, that is $x^2 + y^2 \le 4$: the disc of radius $2$. So the volume is $\int\int_D (4 - x^2 - y^2)\,dA$.
The shadow is a disc, so polar coordinates follow.
In polar form that is $\int_0^{2\pi}\!\int_0^2 (4 - r^2)r\,dr\,d\theta = 2\pi\left[2r^2 - \tfrac{r^4}{4}\right]_0^2 = 8\pi$ — exactly half the volume of the cylinder of radius $2$ and height $4$ that contains it, which is a property of every paraboloid cap.
A known ratio confirms the answer.
A cone of base radius $1$ and height $1$ stands on the $xy$ plane with uniform density. Where is its centre of mass?
Uniform density, so mass is proportional to volume.
The solid and the density are symmetric about the $z$ axis, so $\bar{x} = \bar{y} = 0$ without any integration at all.
Symmetry removes two of the three integrals.
For $\bar{z}$, the cross-section at height $z$ is a disc of radius $1 - z$, so $\bar{z} = \frac{\int_0^1 z\pi(1-z)^2\,dz}{\int_0^1 \pi(1-z)^2\,dz} = \frac{1/12}{1/3} = \tfrac14$. The balance point sits a quarter of the way up, not half: most of the cone's material is near the wide base.
The answer is lower than intuition suggests, for a reason.
Integrating in $z$ first runs from $0$ up to $y$, so the inner integral gives $y$.
The height of the solid over $(x,y)$ is $y$.
The shadow is the rectangle $0 \le x \le 2$, $0 \le y \le 3$, so the volume is $\int_0^2\!\int_0^3 y\,dy\,dx$.
That is $2 \times \tfrac92 = 9$. The solid is a wedge — half of the box $2 \times 3 \times 3$, whose volume is $18$ — and the factor of a half is the slant, which is a useful way to check any answer of this shape without redoing the integral.
The box $B$ is $0 \le x \le 2$, $0 \le y \le 2$, $0 \le z \le 5$. Give the six limits of $\int\int\int_B f\,dz\,dy\,dx$.
| Value | |
|---|---|
| The innermost lower limit | |
| The innermost upper limit | |
| The middle lower limit | |
| The middle upper limit | |
| The outer lower limit | |
| The outer upper limit |
Put the stages of setting up $\int\int\int_E f\,dV$ over a solid of height $4$ into order.
Number the steps in order (write the number in the box):
Find the volume of the solid in the first octant under the plane $\frac{x}{4} + \frac{y}{3} + \frac{z}{3} = 1$.
Answer:
A block occupies $0 \le x \le 3$, $0 \le y \le 2$, $0 \le z \le 6$ with density $z$. What is its mass?
Answer:
The box $B$ is $0 \le x \le 5$, $0 \le y \le 5$, $0 \le z \le 6$. Give the six limits of $\int\int\int_B f\,dz\,dy\,dx$.
| Value | |
|---|---|
| The innermost lower limit | |
| The innermost upper limit | |
| The middle lower limit | |
| The middle upper limit | |
| The outer lower limit | |
| The outer upper limit |
A solid of height $9$ needs one triple integral in the order $dz\,dy\,dx$ but two in the order $dx\,dy\,dz$. What explains that?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Put the stages of setting up $\int\int\int_E f\,dV$ over a solid of height $2$ into order.
Number the steps in order (write the number in the box):
You can set up a triple integral by finding the bounding surfaces and the shadow, choose an order that suits the solid, and use it for a volume, a mass or a centre of mass. Next: two coordinate systems built for solids with an axis or a centre.
10. Your turn: the volume of the solid bounded by $z = y$, $z = 0$, $0 \le x \le 2$ and $0 \le y \le 3$, step 3