Earlier in this chapter we showed how to convert a double integral in rectangular coordinates into a double integral in polar coordinates in order to deal more conveniently with problems involving circular symmetry. A similar situation occurs with triple integrals, but here we need to distinguish between cylindrical symmetry and spherical symmetry. In this section we convert triple integrals in rectangular coordinates into a triple integral in either cylindrical or spherical coordinates.
Also recall the chapter opener, which showed the opera house l’Hemisphèric in Valencia, Spain. It has four sections with one of the sections being a theater in a five-story-high sphere (ball) under an oval roof as long as a football field. Inside is an IMAX screen that changes the sphere into a planetarium with a sky full of 9000 9000 twinkling stars. Using triple integrals in spherical coordinates, we can find the volumes of different geometric shapes like these.
As we have seen earlier, in two-dimensional space ℝ 2 , ℝ 2 , a point with rectangular coordinates ( x , y ) ( x , y ) can be identified with ( r , θ ) ( r , θ ) in polar coordinates and vice versa, where x = r cos θ , x = r cos θ , y = r sin θ , y = r sin θ , r 2 = x 2 + y 2 r 2 = x 2 + y 2 and tan θ = ( y x ) tan θ = ( y x ) are the relationships between the variables.
In three-dimensional space ℝ 3 , ℝ 3 , a point with rectangular coordinates ( x , y , z ) ( x , y , z ) can be identified with cylindrical coordinates ( r , θ , z ) ( r , θ , z ) and vice versa. We can use these same conversion relationships, adding z z as the vertical distance to the point from the x y x y -plane as shown in the following figure.
Figure 5.50 Cylindrical coordinates are similar to polar coordinates with a vertical z z coordinate added.
To convert from rectangular to cylindrical coordinates, we use the conversion x = r cos θ x = r cos θ and y = r sin θ . y = r sin θ . To convert from cylindrical to rectangular coordinates, we use r 2 = x 2 + y 2 r 2 = x 2 + y 2 and tan θ = y x . tan θ = y x . The z z -coordinate remains the same in both cases.
In the two-dimensional plane with a rectangular coordinate system, when we say x = k x = k (constant) we mean an unbounded vertical line parallel to the y y -axis and when y = l y = l (constant) we mean an unbounded horizontal line parallel to the x x -axis. With the polar coordinate system, when we say r = c r = c (constant), we mean a circle of radius c c units and when θ = α θ = α (constant) we mean an infinite ray making an angle α α with the positive x x -axis.
Similarly, in three-dimensional space with rectangular coordinates ( x , y , z ) , ( x , y , z ) , the equations x = k , y = l , x = k , y = l , and z = m , z = m , where k , l , k , l , and m m are constants, represent unbounded planes parallel to the y z y z -plane, x z x z -plane and x y x y -plane, respectively. With cylindrical coordinates ( r , θ , z ) , ( r , θ , z ) , by r = c , θ = α , r = c , θ = α , and z = m , z = m , where c , α , c , α , and m m are constants, we mean an unbounded vertical cylinder with the z z -axis as its radial axis; a plane making a constant angle α α with the x y x y -plane; and an unbounded horizontal plane parallel to the x z x z -plane, respectively. This means that the circular cylinder x 2 + y 2 = c 2 x 2 + y 2 = c 2 in rectangular coordinates can be represented simply as r = c r = c in cylindrical coordinates. (Refer to Cylindrical and Spherical Coordinates for more review.)
Triple integrals can often be more readily evaluated by using cylindrical coordinates instead of rectangular coordinates. Some common equations of surfaces in rectangular coordinates along with corresponding equations in cylindrical coordinates are listed in Table 5.1. These equations will become handy as we proceed with solving problems using triple integrals.
| Circular cylinder | Circular cone | Sphere | Paraboloid | |
|---|---|---|---|---|
| Rectangular | x 2 + y 2 = c 2 x 2 + y 2 = c 2 | z 2 = c 2 ( x 2 + y 2 ) z 2 = c 2 ( x 2 + y 2 ) | x 2 + y 2 + z 2 = c 2 x 2 + y 2 + z 2 = c 2 | z = c ( x 2 + y 2 ) z = c ( x 2 + y 2 ) |
| Cylindrical | r = c r = c | z = c r z = c r | r 2 + z 2 = c 2 r 2 + z 2 = c 2 | z = c r 2 z = c r 2 |
As before, we start with the simplest bounded region B B in ℝ 3 , ℝ 3 , to describe in cylindrical coordinates, in the form of a cylindrical box, B = < ( r , θ , z ) | a ≤ r ≤ b , α ≤ θ ≤ β , c ≤ z ≤ d >B = < ( r , θ , z ) | a ≤ r ≤ b , α ≤ θ ≤ β , c ≤ z ≤ d >(Figure 5.51). Suppose we divide each interval into l , m and n l , m and n subdivisions such that Δ r = b − a l , Δ θ = β − α m , Δ r = b − a l , Δ θ = β − α m , and Δ z = d − c n . Δ z = d − c n . Then we can state the following definition for a triple integral in cylindrical coordinates.
Consider the cylindrical box (expressed in cylindrical coordinates)
If the function f ( r , θ , z ) f ( r , θ , z ) is continuous on B B and if ( r i j k * , θ i j k * , z i j k * ) ( r i j k * , θ i j k * , z i j k * ) is any sample point in the cylindrical subbox B i j k = [ r i − 1 , r i ] × [ θ j − 1 , θ j ] × [ z k − 1 , z k ] B i j k = [ r i − 1 , r i ] × [ θ j − 1 , θ j ] × [ z k − 1 , z k ] (Figure 5.51), then we can define the triple integral in cylindrical coordinates as the limit of a triple Riemann sum, provided the following limit exists:
lim l , m , n → ∞ ∑ i = 1 l ∑ j = 1 m ∑ k = 1 n f ( r i j k * , θ i j k * , z i j k * ) r i j k * Δ r Δ θ Δ z . lim l , m , n → ∞ ∑ i = 1 l ∑ j = 1 m ∑ k = 1 n f ( r i j k * , θ i j k * , z i j k * ) r i j k * Δ r Δ θ Δ z .
Note that if g ( x , y , z ) g ( x , y , z ) is the function in rectangular coordinates and the box B B is expressed in rectangular coordinates, then the triple integral ∭ B g ( x , y , z ) d V ∭ B g ( x , y , z ) d V is equal to the triple integral ∭ B g ( r cos θ , r sin θ , z ) r d r d θ d z ∭ B g ( r cos θ , r sin θ , z ) r d r d θ d z and we have
∭ B g ( x , y , z ) d V = ∭ B g ( r cos θ , r sin θ , z ) r d r d θ d z = ∭ B f ( r , θ , z ) r d r d θ d z . ∭ B g ( x , y , z ) d V = ∭ B g ( r cos θ , r sin θ , z ) r d r d θ d z = ∭ B f ( r , θ , z ) r d r d θ d z .
As mentioned in the preceding section, all the properties of a double integral work well in triple integrals, whether in rectangular coordinates or cylindrical coordinates. They also hold for iterated integrals. To reiterate, in cylindrical coordinates, Fubini’s theorem takes the following form:
Suppose that g ( x , y , z ) g ( x , y , z ) is continuous on a portion of a circular cylinder B , B , which when described in cylindrical coordinates looks like B = < ( r , θ , z ) | a ≤ r ≤ b , α ≤ θ ≤ β , c ≤ z ≤ d >. B = < ( r , θ , z ) | a ≤ r ≤ b , α ≤ θ ≤ β , c ≤ z ≤ d >.
Then g ( x , y , z ) = g ( r cos θ , r sin θ , z ) = f ( r , θ , z ) g ( x , y , z ) = g ( r cos θ , r sin θ , z ) = f ( r , θ , z ) and
∭ B g ( x , y , z ) d V = ∫ c d ∫ α β ∫ a b f ( r , θ , z ) r d r d θ d z . ∭ B g ( x , y , z ) d V = ∫ c d ∫ α β ∫ a b f ( r , θ , z ) r d r d θ d z .
The iterated integral may be replaced equivalently by any one of the other five iterated integrals obtained by integrating with respect to the three variables in other orders.
Cylindrical coordinate systems work well for solids that are symmetric around an axis, such as cylinders and cones. Let us look at some examples before we define the triple integral in cylindrical coordinates on general cylindrical regions.
Evaluate the triple integral ∭ B ( z r sin θ ) r d r d θ d z ∭ B ( z r sin θ ) r d r d θ d z where the cylindrical box B B is B = < ( r , θ , z ) | 0 ≤ r ≤ 2 , 0 ≤ θ ≤ π / 2 , 0 ≤ z ≤ 4 >. B = < ( r , θ , z ) | 0 ≤ r ≤ 2 , 0 ≤ θ ≤ π / 2 , 0 ≤ z ≤ 4 >.
As stated in Fubini’s theorem, we can write the triple integral as the iterated integral
∭ B ( z r sin θ ) r d r d θ d z = ∫ θ = 0 θ = π / 2 ∫ r = 0 r = 2 ∫ z = 0 z = 4 ( z r sin θ ) r d z d r d θ . ∭ B ( z r sin θ ) r d r d θ d z = ∫ θ = 0 θ = π / 2 ∫ r = 0 r = 2 ∫ z = 0 z = 4 ( z r sin θ ) r d z d r d θ .
The evaluation of the iterated integral is straightforward. Each variable in the integral is independent of the others, so we can integrate each variable separately and multiply the results together. This makes the computation much easier:
∫ θ = 0 θ = π / 2 ∫ r = 0 r = 2 ∫ z = 0 z = 4 ( z r sin θ ) r d z d r d θ = ( ∫ 0 π / 2 sin θ d θ ) ( ∫ 0 2 r 2 d r ) ( ∫ 0 4 z d z ) = ( − cos θ | 0 π / 2 ) ( r 3 3 | 0 2 ) ( z 2 2 | 0 4 ) = 64 3 . ∫ θ = 0 θ = π / 2 ∫ r = 0 r = 2 ∫ z = 0 z = 4 ( z r sin θ ) r d z d r d θ = ( ∫ 0 π / 2 sin θ d θ ) ( ∫ 0 2 r 2 d r ) ( ∫ 0 4 z d z ) = ( − cos θ | 0 π / 2 ) ( r 3 3 | 0 2 ) ( z 2 2 | 0 4 ) = 64 3 .
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 . ∫ θ = 0 θ = π ∫ r = 0 r = 1 ∫ z = 0 z = 4 ( r z sin θ ) r d z d r d θ .
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 ) 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 ) >E = < ( r , θ , z ) | ( r , θ ) ∈ D , u 1 ( r , θ ) ≤ z ≤ u 2 ( r , θ ) >in ℝ 3 , ℝ 3 , where D D is the projection of E E onto the r θ 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 . ∭ 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 .
∭ 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 . ∭ 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.
Consider the region E E inside the right circular cylinder with equation r = 2 sin θ , r = 2 sin θ , bounded below by the r θ r θ -plane and bounded above by the sphere with radius 4 4 centered at the origin (Figure 5.52). Set up a triple integral over this region with a function f ( r , θ , z ) f ( r , θ , z ) in cylindrical coordinates.
First, identify that the equation for the sphere is r 2 + z 2 = 16 . r 2 + z 2 = 16 . We can see that the limits for z z are from 0 0 to z = 16 − r 2 . z = 16 − r 2 . Then the limits for r r are from 0 0 to r = 2 sin θ . r = 2 sin θ . Finally, the limits for θ θ are from 0 0 to π . π . Hence the region is
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 . ∭ 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 .
Consider the region E E inside the right circular cylinder with equation r = 2 sin θ , r = 2 sin θ , bounded below by the r θ r θ -plane and bounded above by z = 4 − y . z = 4 − y . Set up a triple integral with a function f ( r , θ , z ) f ( r , θ , z ) in cylindrical coordinates.
Let E E be the region bounded below by the cone z = x 2 + y 2 z = x 2 + y 2 and above by the paraboloid z = 2 − x 2 − y 2 . z = 2 − x 2 − y 2 . (Figure 5.53). Set up a triple integral in cylindrical coordinates to find the volume of the region, using the following orders of integration:
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." width="524" height="340" />
V = ∫ θ = 0 θ = 2 π ∫ r = 0 r = 1 ∫ z = r z = 2 − r 2 r d z d r d θ . V = ∫ θ = 0 θ = 2 π ∫ r = 0 r = 1 ∫ z = r z = 2 − r 2 r d z d r d θ .
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 . 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 .
Redo the previous example with the order of integration d θ d z d r . d θ d z d r .
