TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL

COORDINATES

PROF. MICHAEL VANVALKENBURGH

1. A Review of Double Integrals in Polar Coordinates

The area of an annulus of inner radius 1 and outer radius 2 is clearly

Area = 4π − π = 3π.

-2 -1 0 1 2

-2

-1

0

1

2

As we learned this semester, we can also calculate areas by setting them up as doubleintegrals. To write the area of the annulus as a double integral, we split it into four parts:

-2 -1 0 1 2

-2

-1

0

1

2

The corresponding double integral is1

TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES 2

∫ −1−2

∫ √4−x2−√4−x2

dy dx+

∫ 1

−1

∫ −√1−x2−√4−x2

dy dx+ · · ·

[There are two more integrals to write down.] Complicated, isn’t it! For something assimple as an annulus. . .

A smarter idea is to use a coordinate system that is better suited to the problem. Insteadof describing points in the annulus in terms of rectangular coordinates (“horizontal andvertical”), we describe the points in terms of polar coordinates (“angle and distance tothe origin”). If r is a variable representing distance to the origin, and θ represents anglecounterclockwise from the positive real axis, the annulus can be expressed simply as theregion

1 < r < 2, 0 ≤ θ < 2π.

The relationship between rectangular and polar coordinates is

x = r cos θ, y = r sin θ.

When integrating with respect to polar coordinates, we split the region of integrationinto a bunch of “polar rectangles”: regions described in polar coordinates (r, θ) as

R = {(r, θ); a < r < b, c < θ < d},

where we write ∆r = b− a and ∆θ = d− c (the change in radius and the change in angle).

As we discussed in class (and you can check for yourself), this polar rectangle R has area

∆A = r∆r∆θ,

where r is the average of the inner and outer radii. This “extra” r is the reason why wehave an “extra” r in the double integral, written in polar coordinates:∫∫

Rf(x, y) dA =

∫ d

c

∫ b

af(r cos θ, r sin θ) r dr dθ.

TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES 3

Notice how easy it is to find the area of an annulus using integration in polar coordinates:

Area =

∫ 2π

0

∫ 2

1r dr dθ

= 2π [12r2]r=2r=1

= 3π.

[We are finding an area, so the function we are integrating is f = 1.]

Problem. Write the integral ∫∫Ax2y2 dA

in polar coordinates, where A is the annulus centered at the origin, with inner radius√

2and outer radius 3.

TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES 4

2. Triple Integrals in Cylindrical Coordinates

It is the same idea with triple integrals: rectangular (x, y, z) coordinates might not bethe best choice. For example, you might be studying an object with cylindrical symmetry:fluid flow in a pipe, heat flow in a metal rod, or light propagated through a cylindricaloptical fiber. In that case, it is best to use a cylindrical coordinate system. This simplymeans to use polar coordinates in the x, y plane, and to leave z alone. The formulasrelating rectangular to cylindrical coordinates are:

x = r cos θ

y = r sin θ

z = z.

Since we are using polar coordinates in the x,y plane, the triple integral in cylindricalcoordinates is: ∫∫∫

Sf(x, y, z) dV =

∫∫∫?f(r cos θ, r sin θ, z) r dr dθ dz.

(The limits of integration depend on the particular problem.)

TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES 5

3. Triple Integrals in Spherical Coordinates

If you are studying an object with spherical symmetry, it makes sense to use coordinatesto reflect that. Spherical coordinates are pictured below:

The volume of the “spherical wedge” pictured is approximately

(∗) ∆V = ρ2 sinφ∆ρ∆θ∆φ.

The ρ2 factor reflects the fact that the wedge has bigger volume when ρ is bigger. Whythe sinφ? Consider the orange slice:

TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES 6

Say the orange slice is determined by 0 ≤ ρ ≤ R and 0 ≤ θ ≤ α. As you can see, theorange slice is further subdivided. A section closer to φ = 0 or φ = π has smaller volume(it is “mostly corner”), and a section closer to φ = π/2 has bigger volume.

Here ρ is distance to the origin, 0 ≤ θ < 2π, and 0 ≤ φ ≤ π. Using trigonometry, wefind the relationship between rectangular and spherical coordinates:

x = ρ sinφ cos θ

y = ρ sinφ sin θ

z = ρ cosφ.

The equation ρ = constant describes a sphere, the equation θ = constant describes ahalf-plane, and the equation φ = constant describes a cone. See the figure below and thinkabout it. Where is the point P in the figure?

Another way to put it: first consider a sphere of radius ρ centered at the origin. Thenintersect it with the half-plane with one edge on the z-axis, determined by the angle θ(0 ≤ θ < 2π) in the x, y-plane. The intersection is a half-circle. Then φ (0 ≤ φ ≤ π)determines where the point P is on the half-circle.

TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES 7

A triple integral in spherical coordinates is∫∫∫Sf(x, y, z) dV =

∫∫∫?f(ρ sinφ cos θ, ρ sinφ sin θ, ρ cosφ) ρ2 sinφdρ dθ dφ.

The “distortion factor” ρ2 sinφ comes from the volume of a spherical wedge as in (∗).Again, the limits of integration depend on the particular problem.

Problem. Let B = {(x, y, z); x2 + y2 + z2 ≤ 1} and

f(x, y, z) =1√

2 + x2 + y2 + z2.

Evaluate ∫∫∫Bf(x, y, z) dV.

TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES 8

Solution. Writing the integral in rectangular coordinates, we get∫ 1

−1

∫ √1−z2−√1−z2

∫ √1−y2−z2

−√

1−y2−z2

1√2 + x2 + y2 + z2

dx dy dz.

Right? We can probably do this integral using the methods of Calculus II, but it is simplerin spherical coordinates. After all, in spherical coordinates f is:

f(ρ sinφ cos θ, ρ sinφ sin θ, ρ cosφ) =1√

2 + ρ2,

and B can be described simply as 0 ≤ ρ ≤ 1, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π. So the triple integralis ∫ π

0

∫ 2π

0

∫ 1

0

1√2 + ρ2

ρ2 sinφdρ dθ dφ = 4π

∫ 1

0

ρ2√2 + ρ2

dρ

= 2π√

3− 4π ln(1 +√

3) + 2π ln 2.

In case your Calc II is rusty, here is one way to do that last integral:

Problem. Let S be the intersection of a sphere of radius R centered at the origin and acone of angle 2α with tip at the origin. Find the volume of S. (Sorry for the confusingdescription—the meaning should be clear from the solution.)

TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES 9

Setting up the volume as a triple integral in spherical coordinates, we have:∫∫∫SdV =

∫ α

0

∫ 2π

0

∫ R

0ρ2 sinφdρ dθ dφ

=

∫ α

0

∫ 2π

0[13ρ

3]ρ=Rρ=0 sinφdθ dφ

= 13R

3(2π)[− cosφ]φ=αφ=0

= 23πR

3(1− cosα).

In the special case α = π, we recover the well-known formula that a sphere of radius R hasvolume 4

3πR3.

TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES 10

Problems taken from Marsden and Weinstein’s Calculus, III. The text is availablefree online: http://authors.library.caltech.edu/25043/

(1) Integrate z ex2+y2 over the cylinder x2 + y2 ≤ 4, 2 ≤ z ≤ 3.

(2) Integrate x2 + y2 + z2 over the cylinder given by x2 + z2 ≤ 2, −2 ≤ y ≤ 3.

(3) Evaluate the integral ∫∫∫B

(x2 + y2 + z2)5/2 dx dy dz,

where B is the ball x2 + y2 + z2 ≤ 1.

(4) Evaluate the integral∫∫∫S

(x2 + y2 + z2)−3/2 dx dy dz,

where S is the solid bounded by the spheres x2+y2+z2 = a2 and x2+y2+z2 = b2,where a > b > 0.

(5) Integrate√x2 + y2 + z2 e−(x

2+y2+z2) over the region in the previous problem.

(6) Find the volume of the region enclosed by the surfaces x2 + y2 + z2 = 1 andx2 + y2 = 1

4 .

(7) Find the volume of the region enclosed by the cones z =√x2 + y2 and

z = 1− 2√x2 + y2.