Unit 4, lesson 29

Triple Integrals in Cylindrical Coordinates

Why the volume element in cylindrical coordinates is r dr dθ dz, and how to read bounds for solids built from cylinders, cones and paraboloids, including changing the order and using symmetry.

  • Theory part 57
  • Practice part 58
  • Triple Integrals

A can of soup, a drill bit, a water tower. Many solids are round about a vertical axis, and describing them with xx and yy leads to square roots everywhere. Cylindrical coordinates keep the height zz and use polar coordinates in the horizontal plane, so the round parts become simple bounds on rr and θ\theta.

Key terms

Cylindrical coordinates

Definition A point P=(x,y,z)P = (x, y, z) in a right-handed coordinate system has cylindrical coordinates (r,θ,z)(r, \theta, z), where (r,θ)(r, \theta) are polar coordinates of the projection (x,y,0)(x, y, 0) and zz is unchanged.

x=rcos⁡θ,y=rsin⁡θ,z=z.x = r \cos \theta, \quad y = r \sin \theta, \quad z = z.

We take r≥0r \ge 0, so r=x2+y2r = \sqrt{x^2 + y^2} is the distance from PP to the zz-axis, and θ\theta is measured counterclockwise from the positive xx-axis when viewed from above. The angle is determined only up to multiples of 2π2\pi, and not at all on the zz-axis. The coordinate surfaces are the cylinder r=cr = c for c>0c > 0, the half-plane θ=c\theta = c bounded by the zz-axis, and the horizontal plane z=cz = c.

In plain words Drop the point straight down to the xyxy-plane, describe the shadow in polar coordinates, and keep the height.

Example The point (1,3,4)(1, \sqrt{3}, 4) has r=1+3=2r = \sqrt{1 + 3} = 2, tan⁡θ=3\tan \theta = \sqrt{3} with the shadow in the first quadrant, so θ=π3\theta = \frac{\pi}{3}, and z=4z = 4. It sits where the cylinder r=2r = 2, the half-plane θ=π3\theta = \frac{\pi}{3} and the plane z=4z = 4 meet.

The volume element r dr dθ dz

Definition The small solid with r0≤r≤r0+Δrr_0 \le r \le r_0 + \Delta r, θ0≤θ≤θ0+Δθ\theta_0 \le \theta \le \theta_0 + \Delta \theta and z0≤z≤z0+Δzz_0 \le z \le z_0 + \Delta z, with r0≥0r_0 \ge 0 and 0<Δθ≤2π0 < \Delta \theta \le 2\pi, is a cylindrical wedge. Its base is a piece of an annulus with area 12((r0+Δr)2−r02)Δθ\frac{1}{2}\big((r_0 + \Delta r)^2 - r_0^2\big) \Delta \theta, so its volume is exactly

ΔV=rˉ Δr Δθ Δz,rˉ=r0+Δr2.\begin{gathered} \Delta V = \bar{r} \, \Delta r \, \Delta \theta \, \Delta z, \\[0.4em] \bar{r} = r_0 + \frac{\Delta r}{2}. \end{gathered}

Riemann sums over such wedges lead to the volume element dV=r dz dr dθdV = r \, dz \, dr \, d\theta. The same factor is the Jacobian, since the change of variables has determinant ∂(x,y,z)∂(r,θ,z)=r≥0\frac{\partial(x, y, z)}{\partial(r, \theta, z)} = r \ge 0.

In plain words A step Δθ\Delta \theta in angle covers an arc of length r Δθr \, \Delta \theta, which is longer far from the axis. That is where the extra rr comes from. Forgetting it is the most common mistake in this topic.

Example The wedge 1≤r≤21 \le r \le 2, 0≤θ≤π20 \le \theta \le \frac{\pi}{2}, 0≤z≤30 \le z \le 3 is a quarter of a thick pipe. Its exact volume is

12(4−1)⋅π2⋅3=9π4,\frac{1}{2}(4 - 1) \cdot \frac{\pi}{2} \cdot 3 = \frac{9\pi}{4},

and the formula gives the same, 32⋅1⋅π2⋅3=9π4\frac{3}{2} \cdot 1 \cdot \frac{\pi}{2} \cdot 3 = \frac{9\pi}{4}.

Cylindrical regions

Definition Let EE be a solid of type 1 whose projection DD onto the xyxy-plane is described in polar coordinates by

α≤θ≤β,h1(θ)≤r≤h2(θ),\alpha \le \theta \le \beta, \qquad h_1(\theta) \le r \le h_2(\theta),

with 0≤β−α≤2π0 \le \beta - \alpha \le 2\pi and h1h_1, h2h_2 continuous with 0≤h1≤h20 \le h_1 \le h_2. Let EE lie between the graphs z=u1(x,y)z = u_1(x, y) and z=u2(x,y)z = u_2(x, y), with u1≤u2u_1 \le u_2 continuous on DD. If ff is continuous on EE, then

∭Ef dV=∫αβ∫h1(θ)h2(θ)∫U1U2f r dz dr dθ,\begin{aligned} &\iiint_E f \, dV \\[0.4em] &= \int_\alpha^\beta \int_{h_1(\theta)}^{h_2(\theta)} \int_{U_1}^{U_2} f \, r \, dz \, dr \, d\theta, \end{aligned}

where Ui=ui(rcos⁡θ,rsin⁡θ)U_i = u_i(r \cos \theta, r \sin \theta) and ff is evaluated at (rcos⁡θ,rsin⁡θ,z)(r \cos \theta, r \sin \theta, z).

In plain words Find the shadow, describe it in polar form, write the bottom and top surfaces in terms of rr and θ\theta, and remember the rr.

Example The solid inside the cylinder x2+y2=4x^2 + y^2 = 4, above z=0z = 0 and below the paraboloid z=9−x2−y2z = 9 - x^2 - y^2 has shadow the disk r≤2r \le 2, and the top is z=9−r2z = 9 - r^2, which stays positive on that disk. So

V=∫02π∫02∫09−r2r dz dr dθ=2π∫02(9r−r3) dr=28π.\begin{aligned} V &= \int_0^{2\pi} \int_0^2 \int_0^{9 - r^2} r \, dz \, dr \, d\theta \\[0.4em] &= 2\pi \int_0^2 (9r - r^3) \, dr = 28\pi. \end{aligned}

Bounds from surfaces and projections

Definition To find the bounds of a solid in cylindrical coordinates, write each bounding surface in terms of rr, θ\theta and zz, using x2+y2=r2x^2 + y^2 = r^2. Decide which surface is the bottom and which is the top for each point of the shadow. The shadow’s outer edge is usually where the bottom and top surfaces meet, found by setting their zz values equal.

In plain words The curve where the lid meets the floor marks the edge of the shadow. Inside it, check one point to see which surface is on top.

Example Take the solid above the cone z=x2+y2z = \sqrt{x^2 + y^2} and below the paraboloid z=2−x2−y2z = 2 - x^2 - y^2. In cylindrical form these are z=rz = r and z=2−r2z = 2 - r^2. They meet where r=2−r2r = 2 - r^2, so r=1r = 1 and z=1z = 1. At r=0r = 0 the paraboloid is higher, so it is the top. The volume is

V=∫02π∫01∫r2−r2r dz dr dθ=2π∫01(2r−r3−r2) dr=2π(1−14−13)=5π6.\begin{aligned} V &= \int_0^{2\pi} \int_0^1 \int_r^{2 - r^2} r \, dz \, dr \, d\theta \\[0.4em] &= 2\pi \int_0^1 (2r - r^3 - r^2) \, dr \\[0.4em] &= 2\pi \left( 1 - \frac{1}{4} - \frac{1}{3} \right) = \frac{5\pi}{6}. \end{aligned}

Integration order

Definition For ff continuous on the solid, Fubini’s theorem, applied to the integrand frf r over the region the solid occupies in (r,θ,z)(r, \theta, z) space, lets the integration in rr, θ\theta and zz be done in any order, provided the bounds of each inner variable are rebuilt from the inequalities that describe the solid. When the outer variable is zz, the slice at height zz is a horizontal region, and if the surface bounding that slice changes at some height, the integral must be split there.

In plain words Slicing by height instead of by radius is a different way of reading the same solid. It is also a good independent check of an answer.

Example Slice the cone and paraboloid solid above by height. For 0≤z≤10 \le z \le 1 the slice is the disk r≤zr \le z, cut by the cone. For 1≤z≤21 \le z \le 2 it is the disk r≤2−zr \le \sqrt{2 - z}, cut by the paraboloid. So

V=∫02π∫01∫0zr dr dz dθ+∫02π∫12∫02−zr dr dz dθ=2π(16+14)=5π6,\begin{aligned} V &= \int_0^{2\pi} \int_0^1 \int_0^z r \, dr \, dz \, d\theta \\[0.4em] &\quad + \int_0^{2\pi} \int_1^2 \int_0^{\sqrt{2 - z}} r \, dr \, dz \, d\theta \\[0.4em] &= 2\pi \left( \frac{1}{6} + \frac{1}{4} \right) = \frac{5\pi}{6}, \end{aligned}

the same as before.

Symmetry

Definition Suppose ff is continuous on a solid EE. If EE is unchanged by the reflection (x,y,z)↦(−x,y,z)(x, y, z) \mapsto (-x, y, z) and f(−x,y,z)=−f(x,y,z)f(-x, y, z) = -f(x, y, z) on EE, then ∭Ef dV=0\iiint_E f \, dV = 0. The same holds with the roles of xx and yy exchanged, for the reflection (x,y,z)↦(x,−y,z)(x, y, z) \mapsto (x, -y, z) and ff odd in yy. If EE is unchanged by every rotation about the zz-axis, so that its bounds do not involve θ\theta and θ\theta runs over a full turn, and ff depends only on rr and zz, then the θ\theta integral gives a factor 2π2\pi. For such a solid with a density that depends only on rr and zz, the center of mass lies on the zz-axis.

In plain words Before integrating, look for halves that cancel and for a full turn that contributes just 2π2\pi.

Example The cone and paraboloid solid with constant density is symmetric about the zz-axis, so xˉ=yˉ=0\bar{x} = \bar{y} = 0 with no computation. For the height,

Mxy=∫02π∫01∫r2−r2zr dz dr dθ=π∫01(4r−5r3+r5) dr=11π12,\begin{aligned} &M_{xy} \\[0.4em] &= \int_0^{2\pi} \int_0^1 \int_r^{2 - r^2} z r \, dz \, dr \, d\theta \\[0.4em] &= \pi \int_0^1 (4r - 5r^3 + r^5) \, dr = \frac{11\pi}{12}, \end{aligned}

so dividing by V=5π6V = \frac{5\pi}{6} gives zˉ=1110\bar{z} = \frac{11}{10}. The centroid is (0,0,1110)\left( 0, 0, \frac{11}{10} \right).

Mass and moment of inertia

Definition For a solid EE with continuous density δ≥0\delta \ge 0, the mass and the moment of inertia about the zz-axis are

m=∭Eδ dV,Iz=∭Er2 δ dV,\begin{gathered} m = \iiint_E \delta \, dV, \\[0.4em] I_z = \iiint_E r^2 \, \delta \, dV, \end{gathered}

because x2+y2=r2x^2 + y^2 = r^2 is the squared distance to the zz-axis.

In plain words The moment of inertia measures how hard the solid is to spin about the axis. Mass far from the axis counts more, by the square of its distance.

Example A cylinder of radius 22 and height 33 has density δ=r\delta = r, so it is denser near the rim. Then

m=∫02π∫02∫03r⋅r dz dr dθ=16π,Iz=∫02π∫02∫03r2⋅r⋅r dz dr dθ=192π5.\begin{aligned} m &= \int_0^{2\pi} \int_0^2 \int_0^3 r \cdot r \, dz \, dr \, d\theta \\[0.4em] &= 16\pi, \\[0.4em] I_z &= \int_0^{2\pi} \int_0^2 \int_0^3 r^2 \cdot r \cdot r \, dz \, dr \, d\theta \\[0.4em] &= \frac{192\pi}{5}. \end{aligned}

The ratio Iz/(m⋅22)I_z / (m \cdot 2^2) is 35\frac{3}{5}, larger than the value 12\frac{1}{2} for a uniform cylinder, because more of the mass sits near the rim.

Check yourself

  1. Find cylindrical coordinates of the point (−2,2,1)(-2, 2, 1).
Show the answer

The distance to the axis is r=4+4=22r = \sqrt{4 + 4} = 2\sqrt{2}. The shadow (−2,2)(-2, 2) lies in the second quadrant with tan⁡θ=−1\tan \theta = -1, so θ=3π4\theta = \frac{3\pi}{4}. The point is (22,3π4,1)\left( 2\sqrt{2}, \frac{3\pi}{4}, 1 \right).

  1. Evaluate the integral of x+zx + z over the solid cylinder x2+y2≤1x^2 + y^2 \le 1, 0≤z≤20 \le z \le 2.
Show the answer

The cylinder is symmetric under x↦−xx \mapsto -x and xx is odd, so the xx part is 00. For the zz part, every horizontal slice has area π\pi, so the integral is π∫02z dz=2π\pi \int_0^2 z \, dz = 2\pi.

  1. Find the mass of the solid between the paraboloid z=x2+y2z = x^2 + y^2 and the plane z=4z = 4 with density δ=z\delta = z, and check it by slicing.
Show the answer

The surfaces meet at r=2r = 2, and the plane is on top. So

m=∫02π∫02∫r24z r dz dr dθ=π∫02(16r−r5) dr=64π3.\begin{aligned} m &= \int_0^{2\pi} \int_0^2 \int_{r^2}^4 z \, r \, dz \, dr \, d\theta \\[0.4em] &= \pi \int_0^2 (16r - r^5) \, dr = \frac{64\pi}{3}. \end{aligned}

As a check, the slice at height zz is a disk of radius z\sqrt{z} and area πz\pi z, so the mass is ∫04z⋅πz dz=64π3\int_0^4 z \cdot \pi z \, dz = \frac{64\pi}{3}.

How this lesson connects

It also uses Polar coordinates and triple integrals and Cylinders, cones and paraboloids.