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5.5 Triple integrals in cylindrical and spherical coordinates  (Page 4/12)

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Redo the previous example with the order of integration d θ d z d r .

E = { ( r , θ , z ) | 0 ≤ θ ≤ 2 π , 0 ≤ z ≤ 1 , z ≤ r ≤ 2 − z 2 } and V = ∫ r = 0 r = 1 ∫ z = r z = 2 − r 2 ∫ θ = 0 θ = 2 π r d θ d z d r .

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Finding a volume with triple integrals in two ways

Let E be the region bounded below by the r θ -plane, above by the sphere x 2 + y 2 + z 2 = 4 , and on the sides by the cylinder x 2 + y 2 = 1 ( [link] ). Set up a triple integral in cylindrical coordinates to find the volume of the region using the following orders of integration, and in each case find the volume and check that the answers are the same:

  1. d z d r d θ
  2. d r d z d θ .
    A hemisphere with equation x squared + y squared + z squared = 4 in the upper half plane, and within it, a cylinder with equation x squared + y squared = 1.
    Finding a cylindrical volume with a triple integral in cylindrical coordinates.
  1. Note that the equation for the sphere is
    x 2 + y 2 + z 2 = 4 or r 2 + z 2 = 4

    and the equation for the cylinder is
    x 2 + y 2 = 1 or r 2 = 1 .

    Thus, we have for the region E
    E = { ( r , θ , z ) | 0 ≤ z ≤ 4 − r 2 , 0 ≤ r ≤ 1 , 0 ≤ θ ≤ 2 π }

    Hence the integral for the volume is
    V ( E ) = ∫ θ = 0 θ = 2 π ∫ r = 0 r = 1 ∫ z = 0 z = 4 − r 2 r d z d r d θ = ∫ θ = 0 θ = 2 π ∫ r = 0 r = 1 [ r z | z = 0 z = 4 − r 2 ] d r d θ = ∫ θ = 0 θ = 2 π ∫ r = 0 r = 1 ( r 4 − r 2 ) d r d θ = ∫ 0 2 π ( 8 3 − 3 ) d θ = 2 π ( 8 3 − 3 ) cubic units .
  2. Since the sphere is x 2 + y 2 + z 2 = 4 , which is r 2 + z 2 = 4 , and the cylinder is x 2 + y 2 = 1 , which is r 2 = 1 , we have 1 + z 2 = 4 , that is, z 2 = 3 . Thus we have two regions, since the sphere and the cylinder intersect at ( 1 , 3 ) in the r z -plane
    E 1 = { ( r , θ , z ) | 0 ≤ r ≤ 4 − r 2 , 3 ≤ z ≤ 2 , 0 ≤ θ ≤ 2 π }

    and
    E 2 = { ( r , θ , z ) | 0 ≤ r ≤ 1 , 0 ≤ z ≤ 3 , 0 ≤ θ ≤ 2 π } .

    Hence the integral for the volume is
    V ( E ) = ∫ θ = 0 θ = 2 π ∫ z = 3 z = 2 ∫ r = 0 r = 4 − r 2 r d r d z d θ + ∫ θ = 0 θ = 2 π ∫ z = 0 z = 3 ∫ r = 0 r = 1 r d r d z d θ = 3 π + ( 16 3 − 3 3 ) π = 2 π ( 8 3 − 3 ) cubic units .
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Redo the previous example with the order of integration d θ d z d r .

E 2 = { ( r , θ , z ) | 0 ≤ θ ≤ 2 π , 0 ≤ r ≤ 1 , r ≤ z ≤ 4 − r 2 } and V = ∫ r = 0 r = 1 ∫ z = r z = 4 − r 2 ∫ θ = 0 θ = 2 π r d θ d z d r .

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Review of spherical coordinates

In three-dimensional space ℝ 3 in the spherical coordinate system, we specify a point P by its distance ρ from the origin, the polar angle θ from the positive x -axis (same as in the cylindrical coordinate system), and the angle φ from the positive z -axis and the line O P ( [link] ). Note that ρ ≥ 0 and 0 ≤ φ ≤ π . (Refer to Cylindrical and Spherical Coordinates for a review.) Spherical coordinates are useful for triple integrals over regions that are symmetric with respect to the origin.

A depiction of the spherical coordinate system: a point (x, y, z) is shown, which is equal to (rho, theta, phi) in spherical coordinates. Rho serves as the spherical radius, theta serves as the angle from the x axis in the xy plane, and phi serves as the angle from the z axis.
The spherical coordinate system locates points with two angles and a distance from the origin.

Recall the relationships that connect rectangular coordinates with spherical coordinates.

From spherical coordinates to rectangular coordinates:

x = ρ sin φ cos θ , y = ρ sin φ sin θ , and z = ρ cos φ .

From rectangular coordinates to spherical coordinates:

ρ 2 = x 2 + y 2 + z 2 , tan θ = y x , φ = arccos ( z x 2 + y 2 + z 2 ) .

Other relationships that are important to know for conversions are

• r = ρ sin φ • θ = θ These equations are used to convert from spherical coordinates to cylindrical coordinates • z = ρ cos φ

and

• ρ = r 2 + z 2 • θ = θ These equations are used to convert from cylindrical coordinates to spherical coordinates. • φ = arccos ( z r 2 + z 2 )

The following figure shows a few solid regions that are convenient to express in spherical coordinates.

This figure consists of four figures. In the first, a sphere is shown with the note Sphere rho = c (constant). In the second, a half plane is drawn from the z axis with the note Half plane theta = c (constant). In the last two figures, a half cone is drawn in each with the note Half cone phi = c (constant). In the first of these, the cone opens up and it is marked 0 < c < pi/2. In the second of these, the cone opens down and it is marked pi/2 < c < pi.
Spherical coordinates are especially convenient for working with solids bounded by these types of surfaces. (The letter c indicates a constant.)

Integration in spherical coordinates

We now establish a triple integral in the spherical coordinate system, as we did before in the cylindrical coordinate system. Let the function f ( ρ , θ , φ ) be continuous in a bounded spherical box, B = { ( ρ , θ , φ ) | a ≤ ρ ≤ b , α ≤ θ ≤ β , γ ≤ φ ≤ ψ } . We then divide each interval into l , m and n subdivisions such that Δ ρ = b − a l , Δ θ = β − α m , Δ φ = ψ − γ n .

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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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