Learning Objectives
- Evaluate a triple integral by changing to cylindrical coordinates.
Review of Cylindrical Coordinates
As we have seen earlier, in two-dimensional space [latex]\mathbb{R}^2[/latex], a point with rectangular coordinates [latex](x, y)[/latex] can be identified with [latex]{({r},{\theta})}[/latex] in polar coordinates and vice versa, where [latex]{x} = {r}{\cos}{\theta}, {y} = {r}{\sin}{\theta}, {r} = {x^2}+{y^2}[/latex] and [latex]{\tan}{\theta} = {({\frac{y}{x}})}[/latex] are the relationships between the variables.
In three-dimensional space [latex]\mathbb{R}^3[/latex], a point with rectangular coordinates [latex](x, y, z)[/latex] can be identified with cylindrical coordinates [latex]{({r},{\theta},{z})}[/latex] and vice versa. We can use these same conversion relationships, adding [latex]z[/latex] as the vertical distance to the point from the [latex]xy[/latex]-plane as shown in the following figure.
To convert from rectangular to cylindrical coordinates, we use the conversion [latex]{x} = {r}{\cos}{\theta}[/latex] and [latex]{y} = {r}{\sin}{\theta}[/latex]. To convert from cylindrical to rectangular coordinates, we use [latex]{r^2} = {x^2} + {y^2}[/latex] and [latex]{\theta} = {\tan^{-1}}{({\frac{y}{x}})}[/latex] (noting that we may need to add [latex] \pi [/latex] to arrive at the appropriate quadrant). The [latex]z[/latex]-coordinate remains the same in both cases.
In the two-dimensional plane with a rectangular coordinate system, when we say [latex]x=k[/latex] (constant) we mean an unbounded vertical line parallel to the [latex]y[/latex]-axis and when [latex]y=l[/latex] (constant) we mean an unbounded horizontal line parallel to the [latex]x[/latex]-axis. With the polar coordinate system, when we say [latex]r=c[/latex] (constant), we mean a circle of radius [latex]c[/latex] units and when [latex]{\theta} = {\alpha}[/latex] (constant) we mean an infinite ray making an angle [latex]{\alpha}[/latex] with the positive [latex]x[/latex]-axis.
Similarly, in three-dimensional space with rectangular coordinates [latex](x, y, z)[/latex], the equations [latex]x=k[/latex], [latex]y=l[/latex], and [latex]z=m[/latex], where [latex]k[/latex], [latex]l[/latex], and [latex]m[/latex] are constants, represent unbounded planes parallel to the [latex]yz[/latex]-plane, [latex]xz[/latex]-plane and [latex]xy[/latex]-plane, respectively. With cylindrical coordinates [latex]{({r},{\theta},{z})}[/latex] by [latex]{r} = {c}, {\theta} = {\alpha},[/latex] and [latex]z=m[/latex], where [latex]{c}, {\alpha},[/latex] and [latex]m[/latex] are constants, we mean an unbounded vertical cylinder with the [latex]z[/latex]-axis as its radial axis; a plane making a constant angle [latex]{\alpha}[/latex] with the [latex]xy[/latex]-plane; and an unbounded horizontal plane parallel to the [latex]xy[/latex]-plane, respectively. This means that the circular cylinder [latex]x^{2}+y^{2}=c^{2}[/latex] in rectangular coordinates can be represented simply as [latex]r=c[/latex] in cylindrical coordinates. (Refer to Cylindrical and Spherical Coordinates for more review.)
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 Table 5.1. These equations will become handy as we proceed with solving problems using triple integrals.
Circular cylinder | Circular cone | Sphere | Paraboloid | |
---|---|---|---|---|
Rectangular | [latex]x^{2}+y^{2}=c^{2}[/latex] | [latex]z^{2}=c^{2}(x^{2}+y^{2})[/latex] | [latex]x^{2}+y^{2}+z^{2}=c^{2}[/latex] | [latex]z=c(x^{2}+y^{2})[/latex] |
Cylindrical | [latex]r=c[/latex] | [latex]z=cr[/latex] | [latex]r^{2}+z^{2}=c^{2}[/latex] | [latex]z=cr^{2}[/latex] |
Table 5.1 Equations of Some Common Shapes
As before, we start with the simplest bounded region [latex]B[/latex] in [latex]\mathbb{R}^3[/latex], to describe in cylindrical coordinates, in the form of a cylindrical box, [latex]{B} = {\left \{ {({r},{\theta},{z})}{\mid}{a} \ {\leq} \ {r} \ {\leq} \ {b,{\alpha}} \ {\leq} \ {\theta} \ {\leq} \ {{\beta},{c}} \ {\leq} \ {z} \ {\leq} \ {d} \right \}}[/latex] (Figure 2). Suppose we divide each interval into [latex]l[/latex], [latex]m[/latex], and [latex]n[/latex] subdivisions such that [latex]{\Delta}{r} = {\frac{b-a}{l}}, {\Delta}{\theta} = {\frac{{\beta}-{\alpha}}{m}},[/latex] and [latex]{\Delta}{z} = {\frac{d-c}{n}}[/latex]. Then we can state the following definition for a triple integral in cylindrical coordinates.
definition
Consider the cylindrical box (expressed in cylindrical coordinates)
[latex]{B} = {\left \{ {({r},{\theta},{z})}{\mid}{a} \ {\leq} \ {r} \ {\leq} \ {b,{\alpha}} \ {\leq} \ {\theta} \ {\leq} \ {{\beta},{c}} \ {\leq} \ {z} \ {\leq} \ {d} \right \}}[/latex].
If the function [latex]{f}{({r},{\theta},{z})}[/latex] is continuous on [latex]B[/latex] and if [latex]{({{r}^{*}_{ijk}},{{\theta}^{*}_{ijk}},{{z}^{*}_{ijk}})}[/latex] is any sample point in the cylindrical subbox [latex]{{B}_{ijk}} = {[{{r}_{i-1}},{{r}_{i}}]} {\times} {[{{\theta}_{j-1}},{{\theta}_{j}}]} {\times} {[{{z}_{k-1}},{{z}_{k}}]}[/latex] (Figure 2), then we can define the triple integral in cylindrical coordinates as the limit of a triple Riemann sum, provided the following limit exists:
[latex]\displaystyle\lim_{l,m,n\to\infty}\displaystyle\sum_{i=1}^l\displaystyle\sum_{j=1}^m\displaystyle\sum_{k=1}^n{f}(r^{*}_{ijk},\theta^*_{ijk},z^*_{ijk})r^*_{ijk}\Delta{r}\Delta\theta\Delta{z}[/latex].
Note that if [latex]g(x, y, z)[/latex] is the function in rectangular coordinates and the box [latex]B[/latex] is expressed in rectangular coordinates, then the triple integral [latex]\underset{B}{\displaystyle\iiint}{g}{(x,y,z)}{dV}[/latex] is equal to the triple integral [latex]\underset{B}{\displaystyle\iiint}{g}{({{r}{\cos}{\theta}},{{r}{\sin}{\theta}},{z})}{r}{dr}{{d}{\theta}}{dz}[/latex] and we have
[latex]\underset{B}{\displaystyle\iiint}{g}{(x,y,z)}{dV} = \underset{B}{\displaystyle\iiint}{g}{({{r}{\cos}{\theta}},{{r}{\sin}{\theta}},{z})}{r}{dr}{{d}{\theta}}{dz} = \underset{B}{\displaystyle\iiint}{f}{({r},{\theta},{z})}{r}{dr}{{d}{\theta}}{dz}[/latex].
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:
theorem: fubini’s theorem in cylindrical coordinates
Suppose that [latex]g(x, y, z)[/latex] is continuous on a rectangular box [latex]B[/latex], which when described in cylindrical coordinates looks like [latex]{B} = {\left \{ {({r},{\theta},{z})}{\mid}{a} \ {\leq} \ {r} \ {\leq} \ {b,{\alpha}} \ {\leq} \ {\theta} \ {\leq} \ {{\beta},{c}} \ {\leq} \ {z} \ {\leq} \ {d} \right \}}[/latex].
Then [latex]{g}{({x},{y},{z})} = {g}{({{r}{\cos}{\theta}},{{r}{\sin}{\theta}},{z})} = {f}{({r},{\theta},{z})}[/latex] and
[latex]\underset{B}{\displaystyle\iiint}{g}{({x},{y},{z})}{dV} = {\displaystyle\int^{d}_{c}}{\displaystyle\int^{\beta}_{\alpha}}{\displaystyle\int^{b}_{a}}{f}{({r},{\theta},{z})}{r}{dr}{{d}{\theta}}{dz}[/latex].
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.
Example: evaluating a triple integral over a cylindrical box
Evaluate the triple integral [latex]\underset{B}{\displaystyle\iiint}{({zr}{\sin}{\theta})}{r}{dr}{{d}{\theta}}{dz}[/latex] where the cylindrical box [latex]B[/latex] is [latex]{B} = {\left \{ {({r},{\theta},{z})}{\mid}{0} \ {\leq} \ {r} \ {\leq} \ {2,0} \ {\leq} \ {\theta} \ {\leq} \ {{{\pi}/2},{0}} \ {\leq} \ {z} \ {\leq} \ {4} \right \}}[/latex].
try it
Evaluate the triple integral [latex]{\displaystyle\int^{{\theta} = {\pi}}_{{\theta} = {0}}} \ {\displaystyle\int^{r = 1}_{r = 0}} \ {\displaystyle\int^{z = 4}_{z = 0}} \ {rz} \ {\sin} \ {{\theta}{r}} \ {dz} \ {{d}{\theta}}[/latex].
Watch the following video to see the worked solution to the above Try It
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 [latex]{f}{({r},{\theta},{z})}[/latex] over a general solid region [latex]{E} = {\left \{ {({r},{\theta},{z})}{\mid}{({r},{\theta})} \ {\in} \ {{D},{{u_1}({r},{\theta})}} \ {\leq} \ {z} \ {\leq} \ {u_2}{({r},{\theta})} \right \}}[/latex] in [latex]\mathbb{R}^3[/latex], where [latex]D[/latex] is the projection of [latex]E[/latex] onto the [latex]{r}{\theta}[/latex]-plane,
[latex]\underset{E}{\displaystyle\iiint}{f}{({r},{\theta},{z})}{r}{dr}{{d}{\theta}}{dz} = \underset{D}{\displaystyle\iint}{\left [ \displaystyle\int^{{u_2}{({r},{\theta})}}_{{u_1}{({r},{\theta})}}{f}{({r},{\theta},{z})} \ {dz} \right ]}{r}{dr}{{d}{\theta}}[/latex].
In particular, if [latex]{D} = {\left \{ {({r},{\theta})}{\mid}{g_1} \ {(\theta)} \ {\leq} \ {r} \ {\leq} \ {g_2} \ {(\theta)}, {\alpha} \ {\leq} \ {\theta} \ {\leq} \ {\beta} \right \}}[/latex], then we have
[latex]\underset{E}{\displaystyle\iiint}{f}{({r},{\theta},{z})}{r}{dr}{{d}{\theta}} = {\displaystyle\int^{{\theta} = {\beta}}_{{\theta} = {\alpha}}} \ {\displaystyle\int^{{r} = {g_2}{(\theta)}}_{{r} = {g_1}{(\theta)}}} \ {\displaystyle\int^{{z} = {u_2}{({r},{\theta})}}_{{z} = {u_1}{({r},{\theta})}}}{f}{({r},{\theta},{z})}{r}{dz}{dr}{{d}{\theta}}[/latex].
Similar formulas exist for projections onto the other coordinate planes. We can use polar coordinates in those planes if necessary.
Example: setting up a triple integral in cylindrical coordinates over a general region
Consider the region [latex]E[/latex] inside the right circular cylinder with equation [latex]{r} = {2}{\sin}{\theta}[/latex], bounded below by the [latex]{r}{\theta}[/latex]-plane and bounded above by the sphere with radius 4 centered at the origin (Figure 3). Set up a triple integral over this region with a function [latex]{f}{({r},{\theta},{z})}[/latex] in cylindrical coordinates.
try it
Consider the region [latex]E[/latex] inside the right circular cylinder with equation [latex]{r} = {2}{\sin} \ {\theta}[/latex], bounded below by the [latex]{r} \ {\theta}[/latex]-plane and bounded above by [latex]z=4-y[/latex]. Set up a triple integral with a function [latex]{f}{({r},{\theta},{z})}[/latex] in cylindrical coordinates.
Example: setting up a triple integral in two ways
Let [latex]E[/latex] be the region bounded below by the cone [latex]{z} = {\sqrt{{x^2}+{y^2}}}[/latex] and above by the paraboloid [latex]z=2-x^{2}-y^{2}[/latex] (Figure 4). Set up a triple integral in cylindrical coordinates to find the volume of the region, using the following orders of integration:
1. [latex]{dz} \ {dr} \ {{d}{\theta}}[/latex]
2. [latex]{dr} \ {dz} \ {{d}{\theta}}[/latex]
try it
Redo the previous example with the order of integration [latex]{{d}{\theta}} \ {dz} \ {dr}[/latex].
Example: finding a volume with triple integrals in two ways
Let [latex]E[/latex] be the region bounded below by the [latex]{r}{\theta}[/latex]-plane, above by the sphere [latex]x^{2}+y^{2}+z^{2}=4[/latex], and on the sides by the cylinder [latex]x^{2}+y^{2}=1[/latex] (Figure 5). 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. [latex]{dz} \ {dr} \ {{d}{\theta}}[/latex]
2. [latex]{dr} \ {dz} \ {{d}{\theta}}[/latex].
try it
Redo the previous example with the order of integration [latex]{{d}{\theta}} \ {dz} \ {dr}[/latex].