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5.5 Triple integrals in cylindrical and spherical coordinates  (Page 3/12)

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Evaluate the triple integral ∫ θ = 0 θ = π ∫ r = 0 r = 1 ∫ z = 0 z = 4 r z sin θ r d z d r d θ .

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If the cylindrical region over which we have to integrate is a general solid, we look at the projections onto the coordinate planes. Hence the triple integral of a continuous function f ( r , θ , z ) over a general solid region E = { ( r , θ , z ) | ( r , θ ) ∈ D , u 1 ( r , θ ) ≤ z ≤ u 2 ( r , θ ) } in ℝ 3 , where D is the projection of E onto the r θ -plane, is

∭ E f ( r , θ , z ) r d r d θ d z = ∬ D [ ∫ u 1 ( r , θ ) u 2 ( r , θ ) f ( r , θ , z ) d z ] r d r d θ .

In particular, if D = { ( r , θ ) | g 1 ( θ ) ≤ r ≤ g 2 ( θ ) , α ≤ θ ≤ β } , then we have

∭ E f ( r , θ , z ) r d r d θ = ∫ θ = α θ = β ∫ r = g 1 ( θ ) r = g 2 ( θ ) ∫ z = u 1 ( r , θ ) z = u 2 ( r , θ ) f ( r , θ , z ) r d z d r d θ .

Similar formulas exist for projections onto the other coordinate planes. We can use polar coordinates in those planes if necessary.

Setting up a triple integral in cylindrical coordinates over a general region

Consider the region E inside the right circular cylinder with equation r = 2 sin θ , bounded below by the r θ -plane and bounded above by the sphere with radius 4 centered at the origin ( [link] ). Set up a triple integral over this region with a function f ( r , θ , z ) in cylindrical coordinates.

In polar coordinate space, a sphere of radius 4 is shown with equation r squared + z squared = 16 and center being the origin. There is also a cylinder described by r = 2 sin theta inside the sphere.
Setting up a triple integral in cylindrical coordinates over a cylindrical region.

First, identify that the equation for the sphere is r 2 + z 2 = 16 . We can see that the limits for z are from 0 to z = 16 − r 2 . Then the limits for r are from 0 to r = 2 sin θ . Finally, the limits for θ are from 0 to π . Hence the region is

E = { ( r , θ , z ) | 0 ≤ θ ≤ π , 0 ≤ r ≤ 2 sin θ , 0 ≤ z ≤ 16 − r 2 } .

Therefore, the triple integral is

∭ E f ( r , θ , z ) r d z d r d θ = ∫ θ = 0 θ = π ∫ r = 0 r = 2 sin θ ∫ z = 0 z = 16 − r 2 f ( r , θ , z ) r d z d r d θ .
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Consider the region E inside the right circular cylinder with equation r = 2 sin θ , bounded below by the r θ -plane and bounded above by z = 4 − y . Set up a triple integral with a function f ( r , θ , z ) in cylindrical coordinates.

∭ E f ( r , θ , z ) r d z d r d θ = ∫ θ = 0 θ = π ∫ r = 0 r = 2 sin θ ∫ z = 0 z = 4 − r sin θ f ( r , θ , z ) r d z d r d θ .

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Setting up a triple integral in two ways

Let E be the region bounded below by the cone z = x 2 + y 2 and above by the paraboloid z = 2 − x 2 − y 2 . ( [link] ). Set up a triple integral in cylindrical coordinates to find the volume of the region, using the following orders of integration:

  1. d z d r d θ
  2. d r d z d θ .
    A paraboloid with equation z = 2 minus x squared minus y squared opening down, and within it, a cone with equation z = the square root of (x squared + y squared) pointing down.
    Setting up a triple integral in cylindrical coordinates over a conical region.
  1. The cone is of radius 1 where it meets the paraboloid. Since z = 2 − x 2 − y 2 = 2 − r 2 and z = x 2 + y 2 = r (assuming r is nonnegative), we have 2 − r 2 = r . Solving, we have r 2 + r − 2 = ( r + 2 ) ( r − 1 ) = 0 . Since r ≥ 0 , we have r = 1 . Therefore z = 1 . So the intersection of these two surfaces is a circle of radius 1 in the plane z = 1 . The cone is the lower bound for z and the paraboloid is the upper bound. The projection of the region onto the x y -plane is the circle of radius 1 centered at the origin.
    Thus, we can describe the region as
    E = { ( r , θ , z ) | 0 ≤ θ ≤ 2 π , 0 ≤ r ≤ 1 , r ≤ z ≤ 2 − r 2 } .

    Hence the integral for the volume is
    V = ∫ θ = 0 θ = 2 π ∫ r = 0 r = 1 ∫ z = r z = 2 − r 2 r d z d r d θ .
  2. We can also write the cone surface as r = z and the paraboloid as r 2 = 2 − z . The lower bound for r is zero, but the upper bound is sometimes the cone and the other times it is the paraboloid. The plane z = 1 divides the region into two regions. Then the region can be described as
    E = { ( r , θ , z ) | 0 ≤ θ ≤ 2 π , 0 ≤ z ≤ 1 , 0 ≤ r ≤ z } ∪ { ( r , θ , z ) | 0 ≤ θ ≤ 2 π , 1 ≤ z ≤ 2 , 0 ≤ r ≤ 2 − z } .

    Now the integral for the volume becomes
    V = ∫ θ = 0 θ = 2 π ∫ z = 0 z = 1 ∫ r = 0 r = z r d r d z d θ + ∫ θ = 0 θ = 2 π ∫ z = 1 z = 2 ∫ r = 0 r = 2 − z r d r d z d θ .
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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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