15.7 Triple Integrals in Cylindrical and SphericalCoordinates

Ulrich Hoensch

Friday, November 9, 2012

Cylindrical CoordinatesIn R3, we can represent each point (x , y , z) as follows.

z

x

y

Hx,y,0L

Hx,y,zL

Θ

Hx,y,zL

r

h

This means:

x = r cos θ r =√

x2 + y2

y = r sin θ or tan θ = y/x

z = h h = z .

Cylindrical CoordinatesThe coordinates (r , θ, h) are the cylindrical coordinates of thepoint P. The Cartesian coordinates of this point are obtained bythe transformation

(x , y , z) = T (r , h, θ) = (r cos θ, r sin θ, h).

This transformation is one-to-one if r > 0 and 0 ≤ θ < 2π. ItsJacobian determinant is

∂(x , y , z)

∂(r , θ, h)=

∣∣∣∣∣∣cos θ −r sin θ 0sin θ r cos θ 0

0 0 1

∣∣∣∣∣∣ = r cos2 θ + r sin2 θ = r .

Applying Theorem 15.6 gives the following integration formulawhen using cylindrical coordinates:∫∫∫

T (S)f (x , y , z) dV (x , y , z) =

∫∫∫S

f (r cos θ, r sin θ, h)r dV (r , θ, h).

Example

Find the value of the integral∫ 1

−1

∫ √1−x2

−√

1−x2

∫ 2−x2−y2

x2+y2

(x2 + y2)3/2 dz dy dx

by expressing the region of integration in cylindrical coordinates.

We have:

I −1 ≤ x ≤ 1;

I −√

1− x2 ≤ y ≤√

1− x2, so x2 + y2 ≤ 1;

I x2 + y2 ≤ z ≤ 2− x2 − y2.

When using cylindrical coordinates, we get:

I 0 < r ≤ 1;

I 0 ≤ θ < 2π;

I r2 ≤ h ≤ 2− r2.

Example

This means the region Q over which we integrate is the solid thatlies between the two paraboloids z = r2 and z = 2− r2, andQ = T (S), where S = {0 < r ≤ 1, 0 ≤ θ < 2π, r2 ≤ h ≤ 2− r2}.

Example

Using that x2 + y2 = r2 and switching to cylindrical coordinatesgives ∫ 1

−1

∫ √1−x2

−√

1−x2

∫ 2−x2−y2

x2+y2

(x2 + y2)3/2 dz dy dx

=

∫ 1

0

∫ 2π

0

∫ 2−r2

r2

(r2)3/2r dh dθ dr

= . . .

=8π

35.

Example

We want to find the volume of the interior of the torus with innerradius A and tube radius R. The figure below shows a torus withA = 1 and R = 1.

Example

The torus can be obtained by rotating the circle with center atr = A + R and radius R about the vertical axis:

A A+R A+2Rr

-R

R

h

In cylindrical coordinates, this gives us:

I A ≤ r ≤ A + 2R;

I 0 ≤ θ < 2π;

I −√

R2 − (r − (A + R))2 ≤ h ≤√

R2 − (r − (A + R))2.

To see the last inequality, note that h2 = R2 − (r − (A + R))2.

ExampleThe volume of the interior Q of the torus is∫∫∫

QdV (x , y , z) =

∫ A+2R

A

∫ 2π

0

∫ √R2−(r−(A+R))2

−√

R2−(r−(A+R))2r dh dθ dr

= 2π

∫ A+2R

A2r√

R2 − (r − (A + R))2 dr

Using the substitution r − (A + R) = R cos u, −π ≤ u ≤ 0 givesr = (A + R) + R cos u, dr = −R sin u du, and√

R2 − (r − (A + R))2 =√

R2 − R2 cos2 u = −R sin u.

Thus, the integral becomes:

4π

∫ 0

−π((A + R) + R cos u)(−R sin u)(−R sin u) du

= 4πR2

∫ 0

−π(A + R) sin2 u + R cos u sin2 u du.

Example

The first term can be integrated by using the power-reducingformula sin2 u = (1− cos(2u))/2 and the second term via thesubstitution v = sin u. We obtain∫ 0

−πsin2 u du = π/2,∫ 0

−πcos u sin2 u du = 0.

Thus,

vol(Q) = 4πR2

∫ 0

−π(A + R) sin2 u + R cos u sin2 u du

= 4πR2(A + R)(π/2) = 2π2R2(A + R).

Spherical CoordinatesWe can also represent a point (x , y , z) ∈ R3 like this:

z

x

y

Hx,y,0L

Hx,y,zL

Θ

Hx,y,zL

r sinΦ

rΦ

h

Note that cosφ = z/r , so z = r cosφ. Since x2 + y2 + z2 = r2, wehave x2 + y2 = r2 − r2 cos2 φ = r2 sin2 φ. This gives the sphericalcoordinates

x = r sinφ cos θ, y = r sinφ sin θ, z = r cosφ.

Spherical CoordinatesThe Cartesian coordinates of this point are obtained by thetransformation

(x , y , z) = T (r , h, θ) = (r sinφ cos θ, r sinφ sin θ, r cosφ),

where r > 0, 0 ≤ θ ≤ 2π, 0 ≤ φ < π. The Jacobian determinant is

∂(x , y , z)

∂(r , θ, φ)=

∣∣∣∣∣∣sinφ cos θ −r sinφ sin θ r cosφ cos θsinφ sin θ r sinφ cos θ r cosφ sin θ

cosφ 0 −r sinφ

∣∣∣∣∣∣ = −r2 sinφ.

Applying Theorem 15.6 gives the following integration formulawhen using spherical coordinates:∫∫∫

T (S)f (x , y , z) dV (x , y , z) =

∫∫∫S

f (r sinφ cos θ, r sinφ sin θ, r cosφ)r2 sinφ dV (r , θ, φ).

ExampleFind the volume of the spherical sector Q shown here. The radiusof the sphere is

√2, and the interior angle of the sector is π/2.

In spherical coordinates, Q is given by

0 ≤ r ≤√

2, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π/4.

Example

So,

vol(Q) =

∫∫∫Q

dV (x , y , z)

=

∫ √2

0

∫ 2π

0

∫ π/4

0r2 sinφ dφ dθ dr

= . . .

=4π(√

2− 1)

3.

Example

Find the value of the integral∫ 2

−2

∫ √4−x2

−√

4−x2

∫ √4−x2−y2

0x2 + y2 + z2 dz dy dx

by expressing the region of integration in spherical coordinates.

We have:

I −2 ≤ x ≤ 2;

I −√

4− x2 ≤ y ≤√

4− x2, so x2 + y2 ≤ 4;

I 0 ≤ z ≤√

4− x2 − y2, so x2 + y2 + z2 ≤ 4.

The region of integration Q is the part of the ball with radius 2that lies above the xy -plane. Q can be expressed in sphericalcoordinates as:

0 ≤ r ≤ 2, 0 ≤ θ < 2π, 0 ≤ φ ≤ π/2.

Example

Note that x2 + y2 + z2 = r2 when using spherical coordinates.The integral becomes∫∫∫

Qx2 + y2 + z2 dV =

∫ 2

0

∫ 2π

0

∫ π/2

0(r2)(r2 sinφ)dφ dθ dr

= . . .

=64π

5.

Practice Problems for Section 15.7

p.901-903: 1, 3, 5, 7, 11, 15, 21, 23, 27, 43, 45.