Triple Integrals in Cylindrical Coordinates

Learning Objectives

Evaluate a triple integral by changing to cylindrical coordinates.

Review of Cylindrical Coordinates

As we have seen earlier, in two-dimensional space $\mathbb{R}^2$ , a point with rectangular coordinates $(x, y)$ can be identified with ${({r},{\theta})}$ in polar coordinates and vice versa, where ${x} = {r}{\cos}{\theta}, {y} = {r}{\sin}{\theta}, {r} = {x^2}+{y^2}$ and ${\tan}{\theta} = {({\frac{y}{x}})}$ are the relationships between the variables.

In three-dimensional space $\mathbb{R}^3$ , a point with rectangular coordinates $(x, y, z)$ can be identified with cylindrical coordinates ${({r},{\theta},{z})}$ and vice versa. We can use these same conversion relationships, adding $z$ as the vertical distance to the point from the $xy$ -plane as shown in the following figure.

A box [latex]E[/latex] where the projection [latex]D[/latex] in the [latex]xy[/latex]-plane is of Type I.

Figure 1. Cylindrical coordinates are similar to polar coordinates with a vertical $z$ coordinate added.

To convert from rectangular to cylindrical coordinates, we use the conversion ${x} = {r}{\cos}{\theta}$ and ${y} = {r}{\sin}{\theta}$ . To convert from cylindrical to rectangular coordinates, we use ${r^2} = {x^2} + {y^2}$ and ${\theta} = {\tan^{-1}}{({\frac{y}{x}})}$ (noting that we may need to add $\pi$ to arrive at the appropriate quadrant). The $z$ -coordinate remains the same in both cases.

In the two-dimensional plane with a rectangular coordinate system, when we say $x=k$ (constant) we mean an unbounded vertical line parallel to the $y$ -axis and when $y=l$ (constant) we mean an unbounded horizontal line parallel to the $x$ -axis. With the polar coordinate system, when we say $r=c$ (constant), we mean a circle of radius $c$ units and when ${\theta} = {\alpha}$ (constant) we mean an infinite ray making an angle ${\alpha}$ with the positive $x$ -axis.

Similarly, in three-dimensional space with rectangular coordinates $(x, y, z)$ , the equations $x=k$ , $y=l$ , and $z=m$ , where $k$ , $l$ , and $m$ are constants, represent unbounded planes parallel to the $yz$ -plane, $xz$ -plane and $xy$ -plane, respectively. With cylindrical coordinates ${({r},{\theta},{z})}$ by ${r} = {c}, {\theta} = {\alpha},$ and $z=m$ , where ${c}, {\alpha},$ and $m$ are constants, we mean an unbounded vertical cylinder with the $z$ -axis as its radial axis; a plane making a constant angle ${\alpha}$ with the $xy$ -plane; and an unbounded horizontal plane parallel to the $xy$ -plane, respectively. This means that the circular cylinder $x^{2}+y^{2}=c^{2}$ in rectangular coordinates can be represented simply as $r=c$ 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	$x^{2}+y^{2}=c^{2}$	$z^{2}=c^{2}(x^{2}+y^{2})$	$x^{2}+y^{2}+z^{2}=c^{2}$	$z=c(x^{2}+y^{2})$
Cylindrical	$r=c$	$z=cr$	$r^{2}+z^{2}=c^{2}$	$z=cr^{2}$

Table 5.1 Equations of Some Common Shapes

As before, we start with the simplest bounded region $B$ in $\mathbb{R}^3$ , to describe in cylindrical coordinates, in the form of a cylindrical box, ${B} = {\left \{ {({r},{\theta},{z})}{\mid}{a} \ {\leq} \ {r} \ {\leq} \ {b,{\alpha}} \ {\leq} \ {\theta} \ {\leq} \ {{\beta},{c}} \ {\leq} \ {z} \ {\leq} \ {d} \right \}}$ (Figure 2). Suppose we divide each interval into $l$ , $m$ , and $n$ subdivisions such that ${\Delta}{r} = {\frac{b-a}{l}}, {\Delta}{\theta} = {\frac{{\beta}-{\alpha}}{m}},$ and ${\Delta}{z} = {\frac{d-c}{n}}$ . 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.

Figure 2. A cylindrical box $B$ described by cylindrical coordinates.

definition

Consider the cylindrical box (expressed in cylindrical coordinates)

${B} = {\left \{ {({r},{\theta},{z})}{\mid}{a} \ {\leq} \ {r} \ {\leq} \ {b,{\alpha}} \ {\leq} \ {\theta} \ {\leq} \ {{\beta},{c}} \ {\leq} \ {z} \ {\leq} \ {d} \right \}}$ .

If the function ${f}{({r},{\theta},{z})}$ is continuous on $B$ and if ${({{r}^{*}_{ijk}},{{\theta}^{*}_{ijk}},{{z}^{*}_{ijk}})}$ is any sample point in the cylindrical subbox ${{B}_{ijk}} = {[{{r}_{i-1}},{{r}_{i}}]} {\times} {[{{\theta}_{j-1}},{{\theta}_{j}}]} {\times} {[{{z}_{k-1}},{{z}_{k}}]}$ (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:

$\displaystyle\lim_{l,m,n\to\infty}\displaystyle\sum_{i=1}^l\displaystyle\sum_{j=1}^m\displaystyle\sum_{k=1}^n{f}(r^{*}_{ijk},\theta^*_{ijk},z^*_{ijk})r^*_{ijk}\Delta{r}\Delta\theta\Delta{z}$ .

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 $\underset{B}{\displaystyle\iiint}{g}{(x,y,z)}{dV}$ is equal to the triple integral $\underset{B}{\displaystyle\iiint}{g}{({{r}{\cos}{\theta}},{{r}{\sin}{\theta}},{z})}{r}{dr}{{d}{\theta}}{dz}$ and we have

$\underset{B}{\displaystyle\iiint}{g}{(x,y,z)}{dV} = \underset{B}{\displaystyle\iiint}{g}{({{r}{\cos}{\theta}},{{r}{\sin}{\theta}},{z})}{r}{dr}{{d}{\theta}}{dz} = \underset{B}{\displaystyle\iiint}{f}{({r},{\theta},{z})}{r}{dr}{{d}{\theta}}{dz}$ .

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 $g(x, y, z)$ is continuous on a rectangular box $B$ , which when described in cylindrical coordinates looks like ${B} = {\left \{ {({r},{\theta},{z})}{\mid}{a} \ {\leq} \ {r} \ {\leq} \ {b,{\alpha}} \ {\leq} \ {\theta} \ {\leq} \ {{\beta},{c}} \ {\leq} \ {z} \ {\leq} \ {d} \right \}}$ .

Then ${g}{({x},{y},{z})} = {g}{({{r}{\cos}{\theta}},{{r}{\sin}{\theta}},{z})} = {f}{({r},{\theta},{z})}$ and

$\underset{B}{\displaystyle\iiint}{g}{({x},{y},{z})}{dV} = {\displaystyle\int^{d}_{c}}{\displaystyle\int^{\beta}_{\alpha}}{\displaystyle\int^{b}_{a}}{f}{({r},{\theta},{z})}{r}{dr}{{d}{\theta}}{dz}$ .

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 $\underset{B}{\displaystyle\iiint}{({zr}{\sin}{\theta})}{r}{dr}{{d}{\theta}}{dz}$ where the cylindrical box $B$ is ${B} = {\left \{ {({r},{\theta},{z})}{\mid}{0} \ {\leq} \ {r} \ {\leq} \ {2,0} \ {\leq} \ {\theta} \ {\leq} \ {{{\pi}/2},{0}} \ {\leq} \ {z} \ {\leq} \ {4} \right \}}$ .

Show Solution

As stated in Fubini’s theorem, we can write the triple integral as the iterated integral

$\underset{B}{\displaystyle\iiint}{({zr}{\sin}{\theta})}{r}{dr}{{d}{\theta}}{dz} = {\displaystyle\int^{{\theta} = {\pi}/{2}}_{{\theta} = {0}}}{\displaystyle\int^{r = 2}_{r = 0}}{\displaystyle\int^{z = 4}_{z = 0}}{({zr}{\sin}{\theta})}{r}{dz}{dr}{{d}{\theta}}$ .

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:

${\displaystyle\int^{{\theta} = {\pi}/{2}}_{{\theta} = {0}}}{\displaystyle\int^{r = 2}_{r = 0}}{\displaystyle\int^{z = 4}_{z = 0}}{({zr}{\sin}{\theta})}{r}{dz}{dr}{{d}{\theta}} \\ \\ = {\left ( {\displaystyle\int^{{\pi}/2}_{0}}{\sin}{\theta}{{d}{\theta}} \right )}{\left ( {\displaystyle\int^{2}_{0}}{r^2}{dr} \right )}{\left ( {\displaystyle\int^{4}_{0}}{z}{dz} \right )} = {\left ( {-cos}{\theta}{{\mid}^{{\pi}/{2}}_{0}} \right )}{\left ( {\frac{r^3}{3}}{\bigg\vert}^{2}_{0} \right )}{\left ( {\frac{z^2}{2}}{\bigg\vert}^{4}_{0} \right )} = {\frac{64}{3}}.$

try it

Evaluate the triple integral ${\displaystyle\int^{{\theta} = {\pi}}_{{\theta} = {0}}} \ {\displaystyle\int^{r = 1}_{r = 0}} \ {\displaystyle\int^{z = 4}_{z = 0}} \ {rz} \ {\sin} \ {{\theta}{r}} \ {dz} \ {{d}{\theta}}$ .

Show Solution

$\frac{16}{3}$

Watch the following video to see the worked solution to the above Try It

You can view the transcript for “CP 5.27” here (opens in new window).

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 ${f}{({r},{\theta},{z})}$ over a general solid region ${E} = {\left \{ {({r},{\theta},{z})}{\mid}{({r},{\theta})} \ {\in} \ {{D},{{u_1}({r},{\theta})}} \ {\leq} \ {z} \ {\leq} \ {u_2}{({r},{\theta})} \right \}}$ in $\mathbb{R}^3$ , where $D$ is the projection of $E$ onto the ${r}{\theta}$ -plane,

$\underset{E}{\displaystyle\iiint}{f}{({r},{\theta},{z})}{r}{dr}{{d}{\theta}}{dz} = \underset{D}{\displaystyle\iint}{\left [ \displaystyle\int^{{u_2}{({r},{\theta})}}_{{u_1}{({r},{\theta})}}{f}{({r},{\theta},{z})} \ {dz} \right ]}{r}{dr}{{d}{\theta}}$ .

In particular, if ${D} = {\left \{ {({r},{\theta})}{\mid}{g_1} \ {(\theta)} \ {\leq} \ {r} \ {\leq} \ {g_2} \ {(\theta)}, {\alpha} \ {\leq} \ {\theta} \ {\leq} \ {\beta} \right \}}$ , then we have

$\underset{E}{\displaystyle\iiint}{f}{({r},{\theta},{z})}{r}{dr}{{d}{\theta}} = {\displaystyle\int^{{\theta} = {\beta}}_{{\theta} = {\alpha}}} \ {\displaystyle\int^{{r} = {g_2}{(\theta)}}_{{r} = {g_1}{(\theta)}}} \ {\displaystyle\int^{{z} = {u_2}{({r},{\theta})}}_{{z} = {u_1}{({r},{\theta})}}}{f}{({r},{\theta},{z})}{r}{dz}{dr}{{d}{\theta}}$ .

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 $E$ inside the right circular cylinder with equation ${r} = {2}{\sin}{\theta}$ , bounded below by the ${r}{\theta}$ -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 ${f}{({r},{\theta},{z})}$ 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.

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

Show Solution

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 ${z} = {\sqrt{{16}-{r^2}}}$ . Then the limits for $r$ are from $0$ to ${r} = {2}{\sin}{\theta}$ . Finally, the limits for ${\theta}$ are from $0$ to ${\pi}$ . Hence the region is

${E} = {\left \{ {({r},{\theta},{z})}{\mid}{0} \ {\leq} \ {\theta} \ {\leq} \ {{\pi},{0}} \ {\leq} \ {r} \ {\leq} \ {{{2}{\sin}{\theta}},{0}} \ {\leq} \ {z} \ {\leq} \ {\sqrt{{16} - {r^2}}} \right \}}$ .

Therefore, the triple integral is

$\underset{E}{\displaystyle\iiint}{f}{({r},{\theta},{z})}{r}{dz}{dr}{{d}{\theta}} = {\displaystyle\int^{{\theta} = {\pi}}_{{\theta} = {0}}} \ {\displaystyle\int^{{r} = {2}{\sin}{\theta}}_{{r} = {0}}} \ {\displaystyle\int^{{z} = {\sqrt{{16} - {r^2}}}}_{{z} = {0}}}{f}{({r},{\theta},{z})}{r}{dz}{dr}{{d}{\theta}}$ .

try it

Consider the region $E$ inside the right circular cylinder with equation ${r} = {2}{\sin} \ {\theta}$ , bounded below by the ${r} \ {\theta}$ -plane and bounded above by $z=4-y$ . Set up a triple integral with a function ${f}{({r},{\theta},{z})}$ in cylindrical coordinates.

Show Solution

$\underset{E}{\displaystyle\iiint}f(r,\theta,z)r \ dz \ d\theta=\displaystyle\int_{\theta=0}^{\theta=\pi}\displaystyle\int_{r=0}^{r=2\sin\theta}\displaystyle\int_{z=0}^{z=4-r\sin\theta}f(r,\theta,z)r \ dz \ dr \ d\theta.$

Example: setting up a triple integral in two ways

Let $E$ be the region bounded below by the cone ${z} = {\sqrt{{x^2}+{y^2}}}$ and above by the paraboloid $z=2-x^{2}-y^{2}$ (Figure 4). Set up a triple integral in cylindrical coordinates to find the volume of the region, using the following orders of integration:

1. ${dz} \ {dr} \ {{d}{\theta}}$

2. ${dr} \ {dz} \ {{d}{\theta}}$

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.

Show Solution

1. The cone is of radius 1 where it meets the paraboloid. Since ${z} = {2} - {x^2} - {y^2} = {2} - {{r}^{2}}$ and ${z} = {\sqrt{{x^2}+{y^2}}} = {r}$ (assuming $r$ is nonnegative), we have $2-r^{2}=r$ . Solving, we have $r^{2}+r-2=(r+2)(r-1)=0$ . Since $r\geq 0$ , 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 $xy$ -plane is the circle of radius 1 centered at the origin.

Thus, we can describe the region as

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

Hence the integral for the volume is

$V=\displaystyle\int_{\theta=0}^{\theta=2\pi}{\displaystyle\int^{r = 1}_{r = 0}} \ {\displaystyle\int^{z=2-r^2}_z} r \ {dz} \ {dr} \ d\theta$ .

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

$\begin{aligned} {E} & = {\left \{ {({r},{\theta},{z})}{\mid}{0} \ {\leq} \ {\theta} \ {\leq} \ {{{2}{\pi}},{0}} \ {\leq} \ {z} \ {\leq} \ {{1},{0}} \ {\leq} \ {r} \ {\leq} \ {z} \right \}} \\ & \ {\cup} \ {\left \{ {({r},{\theta},{z})}{\mid}{0} \ {\leq} \ {\theta} \ {\leq} \ {{{2}{\pi}},{1}} \ {\leq} \ {z} \ {\leq} \ {{2},{0}} \ {\leq} \ {r} \ {\leq} \ {\sqrt{{2}-{z}}} \right \}}. \end{aligned}$

Now the integral for the volume becomes

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

try it

Redo the previous example with the order of integration ${{d}{\theta}} \ {dz} \ {dr}$ .

Show Solution

$E=\{(r,\theta,z)|0\leq\theta\leq2\pi,0\leq{z}\leq1,z\leq{r}\leq2-x^2\}\text{ and }V=\displaystyle\int_{r=0}^{r=1}\displaystyle\int_{z=r}^{z=2-r^2}\displaystyle\int_{\theta=0}^{\theta=2\pi}r \ dr \ d\theta \ dz \ dr.$

Example: finding a volume with triple integrals in two ways

Let $E$ be the region bounded below by the ${r}{\theta}$ -plane, above by the sphere $x^{2}+y^{2}+z^{2}=4$ , and on the sides by the cylinder $x^{2}+y^{2}=1$ (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. ${dz} \ {dr} \ {{d}{\theta}}$

2. ${dr} \ {dz} \ {{d}{\theta}}$ .

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.

Show Solution

1. Note that the equation for the sphere is

${x^2} + {y^2} + {z^2} = {4} \ {\text{or}} \ {r^2} + {z^2} = {4}$

and the equation for the cylinder is

${x^2} + {y^2} = {1} \ {\text{or}} \ {r^2} = {1}$ .

Thus, we have for the region $E$

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

Hence the integral for the volume is

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

2. Since the sphere is $x^{2}+y^{2}+z^{2}=4$ , which is $r^{2}+z^{2}=4$ , and the cylinder is $x^{2}+y^{2}=1$ , which is $r^{2}=1$ , we have $1+z^{2}=4$ , that is, $z^{2}=3$ . Thus we have two regions, since the sphere and the cylinder intersect at ${\left ( {1},{\sqrt{3}} \right )}$ in the $rz$ -plane

${E_1} = {\left \{ {({r},{\theta},{z})}{\mid}{0} \ {\leq} \ {r} \ {\leq} \ {{\sqrt{{4}-{r^2}}},{\sqrt{3}}} \ {\leq} \ {z} \ {\leq} \ {{2},{0}} \ {\leq} \ {\theta} \ {\leq} \ {{2}{\pi}} \right \}}$

and

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

Hence the integral for the volume is

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

try it

Redo the previous example with the order of integration ${{d}{\theta}} \ {dz} \ {dr}$ .

Show Solution

$E=\{(r,\theta,z)|0\leq\theta\leq2\pi,0\leq{r}\leq1,z\leq{z}\leq\sqrt{4-r^2}\}\text{ and }V=\displaystyle\int_{r=0}^{r=1}\displaystyle\int_{z=r}^{z=\sqrt{4-r^2}}\displaystyle\int_{\theta=0}^{\theta=2\pi}r \ dr \ d\theta \ dz \ dr.$

Module 5: Multiple Integration