Triple integrals

A solid ball whose density varies from point to point has a definite mass, and a double integral, which sums over a flat region, has no way to compute it.

The previous lesson fitted coordinates to regions of the plane, with polar coordinates and their area element $d A = r d r d θ$ for discs, but all of it happened in the plane. This lesson repeats the construction one dimension up: sums over small boxes, evaluated as three single integrals one inside another, then carried into the two coordinate systems that fit round solids. It assumes the double integral as an iterated integral (integrate in one variable with the other held fixed, then integrate the result) and polar coordinates.

Sums over small boxes

Take a function $f (x, y, z)$ defined on the box $B = [a, b] \times [c, d] \times [p, q]$ . Cut each of the three edges into equal pieces, of widths $Δ x$ , $Δ y$ and $Δ z$ . The cutting planes divide $B$ into small boxes, each of volume $Δ V = Δ x Δ y Δ z$ . Pick a sample point in each small box, multiply the value of $f$ there by $Δ V$ , and add. The triple integral of $f$ over $B$ is the limit of these sums as all three widths shrink to zero:

\underset{B}{∭} f d V = lim \sum_{i} \sum_{j} \sum_{k} f (x_{i j k}^{*}, y_{i j k}^{*}, z_{i j k}^{*}) Δ V

A continuous function on a closed box always has one, by the same argument as in two variables.

The graph of a function of three variables lives in four dimensions, so the picture to keep is density instead of volume under a graph: if $f$ is a density, each term $f Δ V$ is the mass of a small box and the integral is the total mass. If $f = 1$ , the integral $\underset{E}{∭} 1 d V$ is the volume of the solid $E$ .

Fubini's theorem carries over, with the same proof: for continuous $f$ on a box, the triple integral equals each of the six iterated integrals, one per order of $x$ , $y$ and $z$ . So

\underset{B}{∭} x y z^{2} d V = \int_{0}^{1} \int_{0}^{2} \int_{0}^{3} x y z^{2} d z d y d x

over $B = [0, 1] \times [0, 2] \times [0, 3]$ . The inner integral gives $x y \cdot 9$ , the middle one $9 x \cdot 2$ , and the outer one $18 \cdot \frac{1}{2} = 9$ . Because the integrand is a product of one function of each variable, the answer is also the product $\frac{1}{2} \cdot 2 \cdot 9$ of three separate integrals. A midpoint sum over a million small boxes, $100$ along each edge, gives $8.99978$ .

Solids described by bounds

A solid that is not a box is handled as the region between two curves was in the plane. Suppose the solid $E$ lies over a region $D$ of the $x y$ plane, between a floor $z = u_{1} (x, y)$ and a roof $z = u_{2} (x, y)$ . Then the innermost integral runs up the vertical segment through each point of $D$ , and what remains is a double integral over $D$ :

\underset{E}{∭} f d V = \iint_{D} (\int_{u_{1} (x, y)}^{u_{2} (x, y)} f d z) d A

The rule for the limits is the one from the plane, extended by one level: the outermost limits are constants, the middle ones may depend on the outer variable, and the innermost ones may depend on both others.

The standard test case is the tetrahedron cut from the first octant ( $x, y, z \geq 0$ ) by the plane $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ , which meets the axes at $a$ , $b$ and $c$ . The floor is $z = 0$ and the roof is the plane, $z = c (1 - \frac{x}{a} - \frac{y}{b})$ . The shadow $D$ on the floor is where the roof is above zero: the triangle $0 \leq x \leq a$ , $0 \leq y \leq b (1 - \frac{x}{a})$ . Write $s = 1 - \frac{x}{a}$ for short. The inner integral of $1$ gives the height $c (s - \frac{y}{b})$ , and integrating that from $y = 0$ to $y = b s$ gives $c (b s^{2} - \frac{b s^{2}}{2}) = \frac{b c s^{2}}{2}$ . Finally $\int_{0}^{a} \frac{b c}{2} {(1 - \frac{x}{a})}^{2} d x = \frac{b c}{2} \cdot \frac{a}{3}$ , so

volume = \frac{a b c}{6}

The tetrahedron fills exactly one sixth of the box $[0, a] \times [0, b] \times [0, c]$ that contains it. That agrees with the school formula, one third of base area times height, since the base triangle has area $\frac{a b}{2}$ and the height is $c$ .

Example. Let $E$ be the tetrahedron in the first octant under the plane $2 x + 3 y + z = 6$ . Evaluate $\underset{E}{∭} x d V$ .

The plane meets the axes at $x = 3$ , $y = 2$ and $z = 6$ , so the volume is $\frac{3 \cdot 2 \cdot 6}{6} = 6$ . The roof is $z = 6 - 2 x - 3 y$ , so the inner integral is $\int_{0}^{6 - 2 x - 3 y} x d z = x (6 - 2 x - 3 y)$ . The shadow is $0 \leq x \leq 3$ , $0 \leq y \leq \frac{6 - 2 x}{3}$ , and

\int_{0}^{(6 - 2 x) / 3} x (6 - 2 x - 3 y) d y = x {[(6 - 2 x) y - \frac{3 y^{2}}{2}]}_{0}^{(6 - 2 x) / 3} = \frac{x (6 - 2 x)^{2}}{6}

which is $\frac{2}{3} x (3 - x)^{2}$ . Then $\int_{0}^{3} \frac{2}{3} (9 x - 6 x^{2} + x^{3}) d x = \frac{2}{3} (\frac{81}{2} - 54 + \frac{81}{4}) = \frac{2}{3} \cdot \frac{27}{4} = \frac{9}{2}$ . A midpoint sum on a $400 \times 400 \times 400$ grid over the enclosing box, counting only the points inside the tetrahedron, gives $4.49997$ .

Now you. Let $E$ be the tetrahedron in the first octant under the plane $x + y + 2 z = 4$ . Find its volume and evaluate $\underset{E}{∭} y d V$ .

Answer

The intercepts are $4$ , $4$ and $2$ , so the volume is $\frac{4 \cdot 4 \cdot 2}{6} = \frac{16}{3}$ . Integrating in $x$ first, from $0$ to $4 - y - 2 z$ , gives $y (4 - y - 2 z)$ ; then in $z$ from $0$ to $\frac{4 - y}{2}$ gives $\frac{y (4 - y)^{2}}{4}$ ; and $\int_{0}^{4} \frac{y (4 - y)^{2}}{4} d y = \frac{1}{4} \cdot \frac{64}{3} = \frac{16}{3}$ , about $5.333$ .

Cylindrical coordinates

A solid with an axis of symmetry (a cylinder, a cone, a bowl) has round cross sections, which want polar coordinates. Cylindrical coordinates are polar coordinates in the $x y$ plane with the height $z$ kept as it is: a point is located by $(r, θ, z)$ , with

x = r \cos θ, y = r \sin θ, z = z

so $r$ is the distance from the $z$ axis and $x^{2} + y^{2} = r^{2}$ .

The volume element follows from the area element. Cutting space along cylinders of constant $r$ , half planes of constant $θ$ and horizontal planes of constant $z$ produces small pieces whose base is a polar rectangle, of area close to $r Δ r Δ θ$ , and whose height is $Δ z$ . Their volume is close to $r Δ r Δ θ Δ z$ , and in the limit

$d V = r d r d θ d z$

The factor $r$ is there because a step $Δ θ$ sweeps an arc of length $r Δ θ$ , long far from the axis and short near it.

For a first check, take the solid cone with its vertex at the origin, opening upwards, of height $h$ and base radius $a$ . Its side is the surface $z = \frac{h}{a} r$ , so the cone is the solid between that surface and the plane $z = h$ , over the disc $r \leq a$ . Then

V = \int_{0}^{2 π} \int_{0}^{a} \int_{h r / a}^{h} r d z d r d θ = 2 π \int_{0}^{a} (h - \frac{h r}{a}) r d r = 2 π h (\frac{a^{2}}{2} - \frac{a^{2}}{3}) = \frac{π a^{2} h}{3}

which is one third of the base area times the height, as Euclid proved in the Elements by a much longer route.

Example. Find the volume of the solid above the cone $z = \sqrt{x^{2} + y^{2}}$ and below the paraboloid $z = 2 - x^{2} - y^{2}$ .

In cylindrical coordinates the floor is $z = r$ and the roof is $z = 2 - r^{2}$ . They meet where $r = 2 - r^{2}$ , that is $r^{2} + r - 2 = 0$ , whose positive root is $r = 1$ . Inside that circle the roof is higher (at $r = 0$ it is at $2$ and the floor at $0$ ), so the solid lies over the disc $r \leq 1$ , and

V = \int_{0}^{2 π} \int_{0}^{1} \int_{r}^{2 - r^{2}} r d z d r d θ = 2 π \int_{0}^{1} (2 r - r^{3} - r^{2}) d r = 2 π (1 - \frac{1}{4} - \frac{1}{3}) = \frac{5 π}{6}

about $2.618$ . The integrand does not depend on $θ$ , so the outer integral only multiplies by $2 π$ . A Cartesian midpoint sum on a $400 \times 400 \times 400$ grid gives $2.617$ , the missing digit being lost at the curved edges, where a grid of cubes fits badly.

Now you. Find the volume of the solid above the cone $z = \sqrt{x^{2} + y^{2}}$ and below the paraboloid $z = 6 - x^{2} - y^{2}$ , to two decimal places.

Answer

The surfaces meet where $r = 6 - r^{2}$ , at $r = 2$ . Then $V = 2 π \int_{0}^{2} (6 r - r^{3} - r^{2}) d r = 2 π (12 - 4 - \frac{8}{3}) = \frac{32 π}{3} \approx 33.51$ .

Spherical coordinates

A ball has no preferred axis at all, and cylindrical coordinates describe it clumsily: the ball of radius $a$ is $- \sqrt{a^{2} - r^{2}} \leq z \leq \sqrt{a^{2} - r^{2}}$ , square roots again. What fits it is the distance from the origin. Spherical coordinates locate a point by three numbers: $ρ \geq 0$ , its distance from the origin; $φ$ , the angle between the positive $z$ axis and the line from the origin to the point, running from $0$ at the north pole to $π$ at the south pole; and $θ$ , the same angle around the $z$ axis as in cylindrical coordinates. On the Earth, $θ$ is longitude and $φ$ is ninety degrees minus latitude.

This convention, $φ$ from the axis and $θ$ around it, is that of most mathematics texts. Physics texts and the standard ISO 80000-2 swap the two letters; the geometry is identical, and in either notation the sine in the volume element belongs to the angle measured from the axis.

The point at distance $ρ$ and angle $φ$ from the axis is at height $z = ρ \cos φ$ and at distance $r = ρ \sin φ$ from the axis, by right angled trigonometry in the vertical half plane through the point. Feeding that $r$ into the cylindrical formulas gives

x = ρ \sin φ \cos θ, y = ρ \sin φ \sin θ, z = ρ \cos φ

and $x^{2} + y^{2} + z^{2} = ρ^{2}$ . The ball of radius $a$ is now simply $0 \leq ρ \leq a$ , $0 \leq φ \leq π$ , $0 \leq θ \leq 2 π$ : a box in the new coordinates.

For the volume element, look at the small piece of space cut out by increasing each coordinate a little from $(ρ, φ, θ)$ . It is nearly a rectangular box, because the three coordinate directions meet at right angles, and its three edges can be measured one at a time. Increasing $ρ$ by $Δ ρ$ moves the point straight outwards, a distance $Δ ρ$ . Increasing $φ$ by $Δ φ$ moves it along a circle of radius $ρ$ centred at the origin (a meridian), so the edge is an arc of length $ρ Δ φ$ . Increasing $θ$ by $Δ θ$ moves it along a horizontal circle around the $z$ axis, a line of latitude, whose radius is the distance $r = ρ \sin φ$ from the axis, so that edge has length $ρ \sin φ Δ θ$ . Multiplying the three edges,

Δ V \approx Δ ρ \cdot ρ Δ φ \cdot ρ \sin φ Δ θ, d V = ρ^{2} \sin φ d ρ d φ d θ

The $ρ^{2}$ says that pieces of the same angular size are bigger further out; the $\sin φ$ says that lines of latitude shrink towards the poles. The Jacobian determinant of the previous lesson, written out for $(ρ, φ, θ)$ , gives the same $ρ^{2} \sin φ$ after a page of algebra.

The volume of a ball

With the ball described as a box in spherical coordinates, its volume is an integral whose limits are all constants and whose integrand is a product, so it splits into three single integrals:

V = \int_{0}^{2 π} \int_{0}^{π} \int_{0}^{a} ρ^{2} \sin φ d ρ d φ d θ = \frac{a^{3}}{3} \cdot {[- \cos φ]}_{0}^{π} \cdot 2 π = \frac{a^{3}}{3} \cdot 2 \cdot 2 π = \frac{4}{3} π a^{3}

Archimedes reached the same result around 225 BC, in On the Sphere and Cylinder, in the form that a ball is two thirds of the cylinder that just contains it: that cylinder has radius $a$ and height $2 a$ , so volume $2 π a^{3}$ , and two thirds of it is $\frac{4}{3} π a^{3}$ .

Cutting the $φ$ integral short gives the solid shaped like an ice cream cone: the part of the ball inside the cone $φ \leq α$ around the positive $z$ axis. The only change is the middle factor, $\int_{0}^{α} \sin φ d φ = 1 - \cos α$ , so its volume is $\frac{2}{3} π a^{3} (1 - \cos α)$ . For $a = 1$ and $α = \frac{π}{4}$ that is $\frac{2 π}{3} (1 - \frac{\sqrt{2}}{2}) \approx 0.6134$ , and a Cartesian midpoint sum with $300$ points along each edge gives $0.6135$ . At $α = \frac{π}{2}$ the cone opens flat and the formula gives the hemisphere, $\frac{2}{3} π a^{3}$ .

Mass and centre of mass

Since $ρ$ now names a coordinate, write $δ (x, y, z)$ for the density of a solid, in kilograms per cubic metre. Summing the masses $δ Δ V$ of small boxes gives the mass $m = \underset{E}{∭} δ d V$ . The centre of mass is the average position, each point weighted by its mass:

\overline{x} = \frac{1}{m} \underset{E}{∭} x δ d V, \overline{y} = \frac{1}{m} \underset{E}{∭} y δ d V, \overline{z} = \frac{1}{m} \underset{E}{∭} z δ d V

This is $\overline{x} = \frac{\sum m_{i} x_{i}}{\sum m_{i}}$ from mechanics, with the sum over particles become an integral. A constant density cancels, and the centre of mass then depends only on the shape.

Example. Find the centre of mass of a uniform solid hemisphere of radius $a$ resting on its flat face.

Put the flat face on the $x y$ plane, centred at the origin, so the solid is $0 \leq ρ \leq a$ , $0 \leq φ \leq \frac{π}{2}$ , $0 \leq θ \leq 2 π$ . It is unchanged by rotation about the $z$ axis, so $\overline{x} = \overline{y} = 0$ and only the height is in question. The density cancels, the volume is $\frac{2}{3} π a^{3}$ , and with $z = ρ \cos φ$ ,

\underset{E}{∭} z d V = \int_{0}^{2 π} \int_{0}^{π / 2} \int_{0}^{a} ρ^{3} \cos φ \sin φ d ρ d φ d θ = \frac{a^{4}}{4} \cdot \frac{1}{2} \cdot 2 π = \frac{π a^{4}}{4}

using $\int_{0}^{π / 2} \sin φ \cos φ d φ = {[\frac{\sin^{2} φ}{2}]}_{0}^{π / 2} = \frac{1}{2}$ . So $\overline{z} = \frac{π a^{4}}{4} \div \frac{2 π a^{3}}{3} = \frac{3 a}{8}$ . A glass paperweight shaped as a hemisphere of radius $4$ cm balances at a point $1.5$ cm above its base. A Cartesian midpoint sum for $a = 1$ gives $0.37499$ .

Now you. Find the centre of mass of the uniform ice cream cone $0 \leq ρ \leq a$ , $0 \leq φ \leq \frac{π}{3}$ , as a multiple of $a$ .

Answer

The volume is $\frac{2}{3} π a^{3} (1 - \cos \frac{π}{3}) = \frac{π a^{3}}{3}$ . The moment is $∭ z d V = \frac{a^{4}}{4} \cdot {[\frac{\sin^{2} φ}{2}]}_{0}^{π / 3} \cdot 2 π = \frac{a^{4}}{4} \cdot \frac{3}{8} \cdot 2 π = \frac{3 π a^{4}}{16}$ . So $\overline{z} = \frac{3 π a^{4}}{16} \div \frac{π a^{3}}{3} = \frac{9 a}{16} = 0.5625 a$ , higher than the hemisphere's because the narrow cone puts less material near the base.

Moment of inertia

A body turning about an axis resists changes of spin in proportion to its moment of inertia, $I = \sum m_{i} d_{i}^{2}$ , where $d_{i}$ is the distance of the particle $m_{i}$ from the axis. For a solid the sum becomes $I = \underset{E}{∭} d^{2} δ d V$ .

For a uniform ball of radius $a$ about a diameter, take the diameter as the $z$ axis. The distance from it is $r = ρ \sin φ$ , so $d^{2} = ρ^{2} \sin^{2} φ$ and

I = δ \int_{0}^{2 π} \int_{0}^{π} \int_{0}^{a} ρ^{4} \sin^{3} φ d ρ d φ d θ = δ \cdot \frac{a^{5}}{5} \cdot \frac{4}{3} \cdot 2 π = \frac{8 π δ a^{5}}{15}

where $\int_{0}^{π} \sin^{3} φ d φ = \int_{- 1}^{1} (1 - u^{2}) d u = \frac{4}{3}$ by the substitution $u = \cos φ$ . The mass is $M = \frac{4}{3} π a^{3} δ$ , and dividing,

I = \frac{8 π δ a^{5}}{15} \cdot \frac{3}{4 π a^{3} δ} M = \frac{2}{5} M a^{2}

The lesson on rotation in Classical Mechanics quotes this value, together with $\frac{1}{2} M a^{2}$ for a solid cylinder about its axis; the cylinder's comes the same way in cylindrical coordinates, where $d = r$ and $δ \int_{0}^{h} \int_{0}^{2 π} \int_{0}^{a} r^{3} d r d θ d z = \frac{1}{2} π δ a^{4} h$ , which is $\frac{1}{2} M a^{2}$ since $M = π a^{2} h δ$ . The number in front measures how far out the mass sits. If a ball's density instead falls steadily from the centre to zero at the surface, $δ = δ_{0} (1 - \frac{ρ}{a})$ , the same integrals give $\frac{4}{15} M a^{2}$ , about $0.267 M a^{2}$ . The Earth's measured value, $0.3307 M a^{2}$ , lies between the two, one of the main pieces of evidence that it has a dense core.

What has been built

This course has taken the two operations of calculus, differentiation and integration, from functions of one variable to functions of several. On the differential side: partial derivatives, the tangent plane and linear approximation, the chain rule, the gradient, and the search for maxima and minima with and without constraints. On the integral side: sums over small rectangles and boxes, evaluated as iterated single integrals, in whatever coordinates fit the region, with the Jacobian factor paying for the change. Every integral so far has been over a flat region or a solid, a piece of the plane or of space of full dimension.

Vector Calculus, which follows, integrates over things of lower dimension. An integral along a curve adds up the work a force does on a particle moving along a path; an integral over a surface adds up the flow of a fluid or an electric field through it. Then come the theorems of Green, Stokes and Gauss, which say that an integral over a region equals an integral over its boundary: over a plane region and its edge, a surface and the curve around it, a solid and the surface enclosing it. They are the Fundamental Theorem of Calculus in higher dimensions, and they are the language in which Maxwell's equations and the laws of fluid flow are written.