Triple Integrals in Alternative Coordinate Systems: Learn It 2

Giselle Miller

Triple Integrals in Alternative Coordinate Systems: Learn It 2

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]

5.1 Equations of Some Common Shapes

When to Use Cylindrical Coordinates

Cylindrical coordinate systems work well for solids that are symmetric around an axis, such as cylinders and cones. Look for problems involving circular cross-sections in the [latex]xy[/latex]-plane or equations containing [latex]x^2 + y^2[/latex].

Setting Up Triple Integrals in Cylindrical Coordinates

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]

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}}[/latex], [latex]{\Delta}{\theta} = {\frac{{\beta}-{\alpha}}{m}}[/latex], and [latex]{\Delta}{z} = {\frac{d-c}{n}}[/latex].

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. — Figure 2. A cylindrical box [latex]B[/latex] described by cylindrical coordinates.

Just as we defined triple integrals in rectangular coordinates using Riemann sums, we can define them in cylindrical coordinates. The key difference is that we must account for how volume elements change in this coordinate system. When we divide the cylindrical box into small pieces, each piece has volume [latex]r\Delta r\Delta\theta\Delta z[/latex], where the factor of [latex]r[/latex] accounts for the way “wedges” get wider as we move away from the [latex]z[/latex]-axis.

Consider a cylindrical box [latex]{B} = {{ {({r},{\theta},{z})}{\mid}{a} \ {\leq} \ {r} \ {\leq} \ {b,{\alpha}} \ {\leq} \ {\theta} \ {\leq} \ {{\beta},{c}} \ {\leq} \ {z} \ {\leq} \ {d} }}[/latex]. If the function [latex]{f}{({r},{\theta},{z})}[/latex] is continuous on [latex]B[/latex], we can define the triple integral in cylindrical coordinates as the limit of a triple Riemann sum:

[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].

Notice the extra factor of [latex]r^*_{ijk}[/latex] in the sum. This comes from the geometry of cylindrical coordinates and is crucial for obtaining correct results.

converting triple integrals to cylindrical coordinates

If [latex]g(x, y, z)[/latex] is a continuous function in rectangular coordinates, then we can convert to cylindrical coordinates using [latex]x = r\cos\theta[/latex], [latex]y = r\sin\theta[/latex], and [latex]z = z[/latex]:

[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}[/latex].

The volume element in cylindrical coordinates is [latex]dV = r dr d\theta dz[/latex].

Don’t Forget the Extra [latex]r[/latex]!

When converting from rectangular to cylindrical coordinates, the volume element [latex]dV[/latex] becomes [latex]r \, dr \, d\theta \, dz[/latex]. The extra factor of [latex]r[/latex] comes from the Jacobian of the transformation and is essential for correct integration.

Fubini’s Theorem in Cylindrical Coordinates

All the properties of double integrals work well in triple integrals, whether in rectangular coordinates or cylindrical coordinates. They also hold for iterated integrals.

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.

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]

Triple Integrals over General Cylindrical Regions

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, is:

[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.

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.

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. — Figure 3. Setting up a triple integral in cylindrical coordinates over a cylindrical region.

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:

[latex]{dz} \ {dr} \ {{d}{\theta}}[/latex]
[latex]{dr} \ {dz} \ {{d}{\theta}}[/latex]

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. — Figure 4. Setting up a triple integral in cylindrical coordinates over a conical region.

The cone is of radius 1 where it meets the paraboloid. Since [latex]{z} = {2} - {x^2} - {y^2} = {2} - {{r}^{2}}[/latex] and [latex]{z} = {\sqrt{{x^2}+{y^2}}} = {r}[/latex] (assuming [latex]r[/latex] is nonnegative), we have [latex]2-r^{2}=r[/latex]. Solving, we have [latex]r^{2}+r-2=(r+2)(r-1)=0[/latex]. Since [latex]r\geq 0[/latex], we have [latex]r=1[/latex]. Therefore [latex]z=1[/latex]. So the intersection of these two surfaces is a circle of radius 1 in the plane [latex]z=1[/latex]. The cone is the lower bound for [latex]z[/latex] and the paraboloid is the upper bound. The projection of the region onto the [latex]xy[/latex]-plane is the circle of radius 1 centered at the origin.
Thus, we can describe the region as

[latex]{E} = {\left \{ {({r},{\theta},{z})}{\mid}{0} \ {\leq} \ {\theta} \ {\leq} \ {{{2}{\pi}},{0}} \ {\leq} \ {r} \ {\leq} \ {{1},{r}} \ {\leq} \ {z} \ {\leq} \ {{2}-{r^2}} \right \}}[/latex].

Hence the integral for the volume is

[latex]V=\displaystyle\int_{\theta=0}^{\theta=2\pi}{\displaystyle\int^{r = 1}_{r = 0}} \ {\displaystyle\int^{z=2-r^2}_z} r \ {dz} \ {dr} \ d\theta[/latex].
We can also write the cone surface as [latex]r=z[/latex] and the paraboloid as [latex]r^{2}=2-z[/latex]. The lower bound for [latex]r[/latex] is zero, but the upper bound is sometimes the cone and the other times it is the paraboloid. The plane [latex]z=1[/latex] divides the region into two regions. Then the region can be described as
[latex]\begin{aligned} {E} & = {\left \{ {({r},{\theta},{z})}{\mid}{0} \ {\leq} \ {\theta} \ {\leq} \ {{{2}{\pi}},{0}} \ {\leq} \ {z} \ {\leq} \ {{1},{0}} \ {\leq} \ {r} \ {\leq} \ {z} \right \}} \\ & \ {\cup} \ {\left \{ {({r},{\theta},{z})}{\mid}{0} \ {\leq} \ {\theta} \ {\leq} \ {{{2}{\pi}},{1}} \ {\leq} \ {z} \ {\leq} \ {{2},{0}} \ {\leq} \ {r} \ {\leq} \ {\sqrt{{2}-{z}}} \right \}}. \end{aligned}[/latex]

Now the integral for the volume becomes

[latex]{V} = {\displaystyle\int^{{\theta} = {{2}{\pi}}}_{{\theta} = {0}}} \ {\displaystyle\int^{{z} = {1}}_{{z} = {0}}} \ {\displaystyle\int^{{r} = {z}}_{{r} = {0}}}{r}{dr}{dz}{{d}{\theta}} + {\displaystyle\int^{{\theta} = {{2}{\pi}}}_{{\theta} = {0}}} \ {\displaystyle\int^{{z} = {2}}_{{z} = {1}}} \ {\displaystyle\int^{{r} = {\sqrt{{2}-{z}}}}_{{r} = {0}}}{r}{dr}{dz}{{d}{\theta}}.[/latex]

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:

[latex]{dz} \ {dr} \ {{d}{\theta}}[/latex]
[latex]{dr} \ {dz} \ {{d}{\theta}}[/latex].

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. — Figure 5. Finding a cylindrical volume with a triple integral in cylindrical coordinates.

Note that the equation for the sphere is
[latex]{x^2} + {y^2} + {z^2} = {4} \ {\text{or}} \ {r^2} + {z^2} = {4}[/latex]

and the equation for the cylinder is

[latex]{x^2} + {y^2} = {1} \ {\text{or}} \ {r^2} = {1}[/latex].

Thus, we have for the region [latex]E[/latex]

[latex]{E} = {\left \{ {({r},{\theta},{z})}{\mid}{0} \ {\leq} \ {z} \ {\leq} \ {{\sqrt{{4}-{r^2}}},{0}} \ {\leq} \ {r} \ {\leq} \ {{1},{0}} \ {\leq} \ {\theta} \ {\leq} \ {{2}{\pi}} \right \}}[/latex].

Hence the integral for the volume is

[latex]\begin{aligned} {V}{(E)} & = {\displaystyle\int^{{\theta} = {{2}{\pi}}}_{{\theta} = {0}}} \ {\displaystyle\int^{{r} = {1}}_{{r} = {0}}} \ {\displaystyle\int^{{z} = {\sqrt{{4}-{r^2}}}}_{{z} = {0}}}{r}{dz}{dr}{{d}{\theta}} \\ \\ & = {\displaystyle\int^{{\theta} = {{2}{\pi}}}_{{\theta} = {0}}} \ {\displaystyle\int^{{r} = {1}}_{{r} = {0}}}{\left [ {rz}{\big\vert}^{{z} = {\sqrt{{4}-{r^2}}}}_{{z} = {0}} \right ]}{dr}{{d}{\theta}} = {\int\limits^{{\theta} = {{2}{\pi}}}_{{\theta} = {0}}} \ {\int\limits^{{r} = {1}}_{{r} = {0}}}{\left ( {r}{\sqrt{{4}-{r^2}}} \right )}{dr}{{d}{\theta}} \\ \\ & = {\displaystyle\int^{{2}{\pi}}_{0}}{\left ( {\frac{8}{3}} - {\sqrt{3}} \right )}{{d}{\theta}} = {{2}{\pi}}{\left ( {\frac{8}{3}} - {\sqrt{3}} \right )} \ {\text{cubic units.}} \end{aligned}[/latex]
Since the sphere is [latex]x^{2}+y^{2}+z^{2}=4[/latex], which is [latex]r^{2}+z^{2}=4[/latex], and the cylinder is [latex]x^{2}+y^{2}=1[/latex], which is [latex]r^{2}=1[/latex], we have [latex]1+z^{2}=4[/latex], that is, [latex]z^{2}=3[/latex]. Thus we have two regions, since the sphere and the cylinder intersect at [latex]{\left ( {1},{\sqrt{3}} \right )}[/latex] in the [latex]rz[/latex]-plane
[latex]{E_1} = {\left \{ {({r},{\theta},{z})}{\mid}{0} \ {\leq} \ {r} \ {\leq} \ {{\sqrt{{4}-{r^2}}},{\sqrt{3}}} \ {\leq} \ {z} \ {\leq} \ {{2},{0}} \ {\leq} \ {\theta} \ {\leq} \ {{2}{\pi}} \right \}}[/latex]

and

[latex]{E_2} = {\left \{ {({r},{\theta},{z})}{\mid}{0} \ {\leq} \ {r} \ {\leq} \ {{1},{0}} \ {\leq} \ {z} \ {\leq} \ {{\sqrt{3}},{0}} \ {\leq} \ {\theta} \ {\leq} \ {{2}{\pi}} \right \}}.[/latex]

Hence the integral for the volume is

[latex]\begin{aligned} {{V}{(E)}} & = {\displaystyle\int^{{\theta} = {{2}{\pi}}}_{{\theta} = {0}}} \ {\displaystyle\int^{{z} = {2}}_{{z} = {\sqrt{3}}}} \ {\displaystyle\int^{{r} = {\sqrt{{4}-{r^2}}}}_{{r} = {0}}}{r}{dr}{dz}{{d}{\theta}} + {\displaystyle\int^{{\theta} = {{2}{\pi}}}_{{\theta} = {0}}} \ {\displaystyle\int^{{z} = {\sqrt{3}}}_{{z} = {0}}} \ {\displaystyle\int^{{r} = {1}}_{{r} = {0}}}{r}{dr}{dz}{{d}{\theta}} \\ & = {{\sqrt{3}}{\pi}} + {\left ( {\frac{16}{3}} - {{3}{\sqrt{3}}} \right )}{\pi} = {{2}{\pi}}{\left ( {\frac{8}{3}} - {\sqrt{3}} \right )} \ {\text{cubic units.}} \end{aligned}[/latex]