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Integration in cylindrical coordinates

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 [link] . These equations will become handy as we proceed with solving problems using triple integrals.

Equations of some common shapes
Circular cylinder Circular cone Sphere Paraboloid
Rectangular x 2 + y 2 = c 2 z 2 = c 2 ( x 2 + y 2 ) x 2 + y 2 + z 2 = c 2 z = c ( x 2 + y 2 )
Cylindrical r = c z = c r r 2 + z 2 = c 2 z = c r 2

As before, we start with the simplest bounded region B in 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 } ( [link] ). Suppose we divide each interval into l , m and n subdivisions such that Δ r = b a l , Δ θ = β α m , and Δ z = d c n . Then we can state the following definition for a triple integral in cylindrical coordinates.

In cylindrical box is shown with its projection onto the polar coordinate plane with inner radius a, outer radius b, and sides defined by theta = alpha and beta. The cylindrical box B starts at height c and goes to height d with the rest of the values the same as the projection onto the plane.
A cylindrical box B described by cylindrical coordinates.

Definition

Consider the cylindrical box (expressed in cylindrical coordinates)

B = { ( r , θ , z ) | a r b , α θ β , c z d } .

If the function f ( r , θ , z ) is continuous on B and if ( 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 ] ( [link] ), 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 .

Note that if g ( x , y , z ) is the function in rectangular coordinates and the box B is expressed in rectangular coordinates, then the triple integral 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 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 .

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:

Fubini’s theorem in cylindrical coordinates

Suppose that g ( x , y , z ) is continuous on a rectangular box B , which when described in cylindrical coordinates looks like 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 ) 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 .

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.

Evaluating a triple integral over a cylindrical box

Evaluate the triple integral B ( z r sin θ ) r d r d θ d z where the cylindrical box B is 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 .

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 .
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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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