Triple Integrals In Cylindrical Coordinates
w/ Step-by-Step Examples!

There are many applications of triple integrals that are best expressed in non-Cartesian coordinates. In particular, evaluating triple integrals in cylindrical coordinates is sometimes preferable and expected.

For example, you may want to study the fluid flow in a cylindrical pipe or heat flow in a metal rod, all of which are best evaluated in cylindrical coordinates.

Cylindrical Coordinates

So, how do we change from Cartesian coordinates to cylindrical coordinates?

By using the following formulas:

Triple Integral

And the formula for triple integration in cylindrical coordinates is:

\begin{equation}
\iiint_{S} f(x, y, z) d V=\int_{c}^{d} \int_{\alpha}^{\beta} \int_{a}^{b} f(r, \theta, z) r d r d \theta d z
\end{equation}

Where \(S\) is the cylindrical wedge.

\begin{equation}
S=\{(r, \theta, z): a \leq r \leq b, \alpha \leq \theta \leq \beta, c \leq z \leq d\}
\end{equation}

Recall that area in polar coordinates is expressed as \(dA = rdrd\theta \).

Thus, for triple integrals representing volume, we say \(dV = rdrd\theta dz\), not forgetting the Jacobian transformation, or Jacobian determinant, of \(r\) that enables us to transform from rectangular coordinates to cylindrical coordinates.

And notice that all we are doing is replacing the \(x\) and \(y\) coordinates with polar coordinates \(r\) and theta and leaving the \(z\) coordinate unchanged, as so nicely stated by Oregon State Mathematics.

And don’t forget that with all multiple integrals, we can switch the order of integration. (i.e., \(rdrd\theta dz\) or \(rdzd\theta dr\) , etc.)

Therefore, just like we learned about how to evaluate triple integrals in rectangular coordinates, there are six forms for evaluating a triple integral in cylindrical coordinates as well.

Simple, right?

Example

Let’s look at an example.

Evaluate \(\int_{0}^{2} \int_{0}^{\sqrt{4-x^{2}}} \int_{0}^{2 x y}\left(x^{2}+y^{2}\right) d z d y d x\)

Notice that directly integrating this iterated triple integral would be downright torturous.

Therefore, let’s convert this integral into cylindrical coordinates, given that the integrand and limits of integration resemble a disk.

If we focus on the limit bounds, we notice that:

\begin{equation}
0 \leq z \leq 2 x y, 0 \leq y \leq \sqrt{4-x^{2}} \text {, and } 0 \leq x \leq 2
\end{equation}

This means that if we can rewrite our inequalities as follows:

Hemisphere Cartesian Polar Coordinates

\begin{equation}
z \leq 2(r \cos \theta)(r \sin \theta) \text { and } z \geq 0, \text { then } \mathrm{z} \text { varies from } 0 \text { to } z=2 r^{2} \cos \theta \sin \theta
\end{equation}

\(y \leq \sqrt{4-x^{2}}\) represents a half-circle with radius 2
$$
\begin{aligned}
&y^{2}=4-x^{2} \\
&x^{2}+y^{2}=4 \\
&r^{2}=4 \\
&r=2
\end{aligned}
$$

But since \(0 \le x \le 2\), this means that our angle of rotation is limited to the first quadrant only.

Therefore, our half-circle becomes a quarter-circle in which \(0 \le \theta \le \frac{\pi }{2}\).

Polar Coordinates Shaded Region

So, our three bounds are:

\begin{equation}
0 \leq z \leq 2 r^{2} \cos \theta \sin \theta \quad, \quad 0 \leq r \leq 2 \quad, \quad 0 \leq \theta \leq \frac{\pi}{2}
\end{equation}

Now we must change our integrand from \({x^2} + {y^2}\) into polar coordinates, which is \({r^2}\).

And we are ready to drop everything into our formula for triple integration in cylindrical coordinates.

However, since our z-bounds have two variables, we must put our order of integration as \(rdzdrd\theta \).

First, we will integrate with respect to \(z\):

\begin{equation}
\int_{0}^{\frac{\pi}{2}} \int_{0}^{2} \int_{0}^{2 r^{2} \sin \theta \cos \theta}\left(r^{2}\right) r d z d r d \theta
\end{equation}

\begin{equation}
=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2} \int_{0}^{2 r^{2} \sin \theta \cos \theta}\left(r^{3}\right) d z d r d \theta
\end{equation}

\begin{equation}
=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2}\left(\left[r^{3} z\right]_{0}^{2 r^{2} \sin \theta \cos \theta}\right) d r d \theta
\end{equation}

\begin{equation}
=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2}\left(2 r^{5} \sin \theta \cos \theta\right) d r d \theta
\end{equation}

Next, we will integrate with respect to \(r\).

\begin{equation}
\int_{0}^{\frac{\pi}{2}}\left(\int_{0}^{2}\left(2 r^{5} \sin \theta \cos \theta\right) d r\right) d \theta
\end{equation}

\begin{equation}
=\int_{0}^{\frac{\pi}{2}}\left(\left[\frac{1}{3} r^{6} \sin \theta \cos \theta\right]_{0}^{2}\right) d \theta
\end{equation}

\begin{equation}
=\int_{0}^{\frac{\pi}{2}}\left(\frac{64}{3} \sin \theta \cos \theta\right) d \theta
\end{equation}

Lastly, we will integrate with respect to theta using either u-substitution or double-angle identity.

\begin{equation}
\int_{0}^{\frac{\pi}{2}}\left(\frac{64}{3} \sin \theta \cos \theta\right) d \theta
\end{equation}

\begin{equation}
=-\left.\frac{32}{3}\left(\cos ^{2} \theta\right)\right|_{0} ^{\frac{\pi}{2}}
\end{equation}

\begin{equation}
=\frac{32}{3}
\end{equation}

Nice.

So, together we will walk through several examples of evaluating a triple integral in cylindrical coordinates and find new limits of integration when we learn how to transform a Cartesian iterated triple integral into cylindrical coordinates.

Let’s get to it!

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