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Section 4-6 : Triple Integrals in Cylindrical Coordinates

In this section we want do take a look at triple integrals done completely in Cylindrical Coordinates. Recall that cylindrical coordinates are really nothing more than an extension of polar coordinates into three dimensions. The following are the conversion formulas for cylindrical coordinates.

\[x = r\cos \theta \hspace{0.25in}y = r\sin \theta \hspace{0.25in}z = z\]

In order to do the integral in cylindrical coordinates we will need to know what \(dV\) will become in terms of cylindrical coordinates. We will be able to show in the Change of Variables section of this chapter that,

\[dV = r\;dz\,dr\,d\theta \]

The region, \(E\), over which we are integrating becomes,

\[\begin{align*}E & = \left\{ {\left( {x,y,z} \right)|\left( {x,y} \right) \in D,\,\,\,{u_1}\left( {x,y} \right) \le z \le {u_2}\left( {x,y} \right)} \right\}\\ & = \left\{ {\left( {r,\theta ,z} \right)|\alpha \le \theta \le \beta ,\,\,{h_1}\left( \theta \right) \le r \le {h_2}\left( \theta \right),\,\,\,{u_1}\left( {r\cos \theta ,r\sin \theta } \right) \le z \le {u_2}\left( {r\cos \theta ,r\sin \theta } \right)} \right\}\end{align*}\]

Note that we’ve only given this for \(E\)’s in which \(D\) is in the \(xy\)-plane. We can modify this accordingly if \(D\) is in the \(yz\)-plane or the \(xz\)-plane as needed.

In terms of cylindrical coordinates a triple integral is,

\[\iiint\limits_{E}{{f\left( {x,y,z} \right)\,dV}} = \int_{{\,\alpha }}^{{\,\beta }}{{\int_{{{h_1}\left( \theta \right)}}^{{{h_2}\left( \theta \right)}}{{\int_{{{u_1}\left( {r\cos \theta ,r\sin \theta } \right)}}^{{{u_2}\left( {r\cos \theta ,r\sin \theta } \right)}}{{r\,f\left( {r\cos \theta ,r\sin \theta ,z} \right)\,dz}}\,dr}}\,d\theta }}\]

Don’t forget to add in the \(r\) and make sure that all the \(x\)’s and \(y\)’s also get converted over into cylindrical coordinates.

Let’s see an example.

Example 1 Evaluate \( \displaystyle \iiint\limits_{E}{{y\,dV}}\) where \(E\) is the region that lies below the plane \(z = x + 2\) above the \(xy\)-plane and between the cylinders \({x^2} + {y^2} = 1\) and \({x^2} + {y^2} = 4\).
Show Solution

There really isn’t too much to do with this one other than do the conversions and then evaluate the integral.

We’ll start out by getting the range for \(z\) in terms of cylindrical coordinates.

\[0 \le z \le x + 2\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}0 \le z \le r\cos \theta + 2\]

Remember that we are above the \(xy\)-plane and so we are above the plane \(z = 0\)

Next, the region \(D\) is the region between the two circles \({x^2} + {y^2} = 1\) and \({x^2} + {y^2} = 4\) in the \(xy\)-plane and so the ranges for it are,

\[0 \le \theta \le 2\pi \hspace{0.25in}\hspace{0.25in}1 \le r \le 2\]

Here is the integral.

\[\begin{align*}\iiint\limits_{E}{{y\,dV}} & = \int_{0}^{{2\pi }}{{\int_{1}^{2}{{\int_{0}^{{r\cos \theta + 2}}{{\left( {r\sin \theta } \right)r\,dz}}\,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\int_{1}^{2}{{{r^2}\sin \theta \left( {r\cos \theta + 2} \right)\,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\int_{1}^{2}{{\frac{1}{2}{r^3}\sin \left( {2\theta } \right) + 2{r^2}\sin \theta \,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\left. {\left( {\frac{1}{8}{r^4}\sin \left( {2\theta } \right) + \frac{2}{3}{r^3}\sin \theta } \right)} \right|_1^2\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\frac{{15}}{8}\sin \left( {2\theta } \right) + \frac{{14}}{3}\sin \theta \,d\theta }}\\ & = \left. {\left( { - \frac{{15}}{{16}}\cos \left( {2\theta } \right) - \frac{{14}}{3}\cos \theta } \right)} \right|_0^{2\pi }\\ & = 0\end{align*}\]

Just as we did with double integral involving polar coordinates we can start with an iterated integral in terms of \(x\), \(y\), and \(z\) and convert it to cylindrical coordinates.

Example 2 Convert \(\displaystyle \int_{{\, - 1}}^{{\,1}}{{\int_{{\,0}}^{{\,\sqrt {1 - {y^2}} }}{{\int_{{\,{x^2} + {y^2}}}^{{\,\sqrt {{x^2} + {y^2}} }}{{xyz\,dz}}\,dx}}\,dy}}\) into an integral in cylindrical coordinates.
Show Solution

Here are the ranges of the variables from this iterated integral.

\[\begin{array}{c} - 1 \le y \le 1\\ 0 \le x \le \sqrt {1 - {y^2}} \\ {x^2} + {y^2} \le z \le \sqrt {{x^2} + {y^2}} \end{array}\]

The first two inequalities define the region \(D\) and since the upper and lower bounds for the \(x\)’s are \(x = \sqrt {1 - {y^2}} \) and \(x = 0\) we know that we’ve got at least part of the right half a circle of radius 1 centered at the origin. Since the range of \(y\)’s is \( - 1 \le y \le 1\) we know that we have the complete right half of the disk of radius 1 centered at the origin. So, the ranges for \(D\) in cylindrical coordinates are,

\[\begin{array}{c}\displaystyle - \frac{\pi }{2} \le \theta \le \frac{\pi }{2}\\ 0 \le r \le 1\end{array}\]

All that’s left to do now is to convert the limits of the \(z\) range, but that’s not too bad.

\[{r^2} \le z \le r\]

On a side note notice that the lower bound here is an elliptic paraboloid and the upper bound is a cone. Therefore, \(E\) is a portion of the region between these two surfaces.

The integral is,

\[\begin{align*}\int_{{\, - 1}}^{{\,1}}{{\int_{{\,0}}^{{\,\sqrt {1 - {y^2}} }}{{\int_{{\,{x^2} + {y^2}}}^{{\,\sqrt {{x^2} + {y^2}} }}{{xyz\,dz}}\,dx}}\,dy}} & = \int_{{\, - {\pi }/{2}\;}}^{{\,{\pi }/{2}\;}}{{\int_{{\,0}}^{{\,1}}{{\int_{{\,{r^2}}}^{{\,r}}{{r\left( {r\cos \theta } \right)\left( {r\sin \theta } \right)z\,dz}}\,dr}}\,d\theta }}\\ & = \int_{{\, - {\pi }/{2}\;}}^{{\,{\pi }/{2}\;}}{{\int_{{\,0}}^{{\,1}}{{\int_{{\,{r^2}}}^{{\,r}}{{z{r^3}\cos \theta \sin \theta \,dz}}\,dr}}\,d\theta }}\end{align*}\]