# 21. Multiple Integrals in Curvilinear Coordinates

## b. Integrating in Cylindrical Coordinates

## 3. Applications

Integrals in cylindrical coordinates are particularly useful when the integrand and/or the limits of integration are axially symmetric about the \(z\)-axis. We previously did Applications of Triple Integrals in rectangular coordinates. All of those applications can also be done using cylindrical coordinates and on the next page we will see that it is often useful to convert from rectangular to cylindrical coordinates.

All the examples on this page of applications of triple integrals in cylindrical coordinates will use the solid between the paraboloids given by \(z=x^2+y^2\) and \(z=8-x^2-y^2\) as shown in the plot. The first paraboloid opens upward with its vertex at the origin. The second one opens downward with its vertex at \((0,0,8)\). To find the intersection of the two paraboloids, we equate them: \[ x^2+y^2=8-x^2-y^2 \qquad \Longrightarrow \qquad x^2+y^2=4 \] The solution is a circle of radius \(r=2\). Equivalently, in cylindrical coordinates, the paraboloids are \(z=r^2\) and \(z=8-r^2\). We equate them: \[ r^2=8-r^2 \qquad \Longrightarrow \qquad r^2=4 \] Again the radius is \(r=2\). So the bounds are \[ 0 \le \theta \le 2\pi \qquad 0 \le r \le 2 \qquad r^2 \le z \le 8-r^2 \]

### Volume

To find the volume of a 3D region \(R\) in cylindrical coordinates, the differential of volume \(dV\) must be expressed in cylindrical coordinates: \[ \text{Volume}=\iiint_R 1\,dV =\iiint_R r\,dr\,d\theta\,dz \]

### Mass

To find the mass of a 3D region \(R\) in cylindrical coordinates, the density \(\delta\) must also be expressed in cylindrical coordinates: \[ M=\iiint_R \delta(x,y,z)\,dV =\iiint_R \delta(r,\theta,z)\,r\,dr\,d\theta\,dz \]

### Center of Mass

To find the center of mass of a 3D region \(R\) in cylindrical coordinates, the coordinates \(x\), \(y\) and \(z\) must also be expressed in cylindrical coordinates. So the center of mass is: \[ \bar{x}=\dfrac{M_x}{M} \quad \text{and} \quad \bar{y}=\dfrac{M_y}{M} \quad \text{and} \quad \bar{z}=\dfrac{M_z}{M} \] where the mass of the region is \[ M=\iiint_R \delta(x,y,z)\,dV =\iiint_R \delta(r,\theta,z)\,r\,dr\,d\theta\,dz \] and the \(x\), \(y\) and \(z\) \(1^\text{st}\)-moments of the mass, respectively, are \[\begin{aligned} M_x&=\iiint_R x\,\delta(x,y,z)\,dV =\iiint_R r\cos\theta\,\delta(r,\theta,z)\,r\,dr\,d\theta\,dz \\ M_y&=\iiint_R y\,\delta(x,y,z)\,dV =\iiint_R r\sin\theta\,\delta(r,\theta,z)\,r\,dr\,d\theta\,dz \\ M_z&=\iiint_R z\,\delta(x,y,z)\,dV =\iiint_R z\,\delta(r,\theta,z)\,r\,dr\,d\theta\,dz \end{aligned}\]

It is important to remember that we cannot compute \(\bar{r}=\dfrac{M_r}{M}\) and \(\bar{\theta}=\dfrac{M_\theta}{M}\) where \[\text{\Large\textcolor{red}Wrong}\qquad \begin{array}{rl} M_r&=\displaystyle\iiint_R r\,\delta(r,\theta,z)\,dV \\[8pt] M_\theta&=\displaystyle\iiint_R \theta\,\delta(r,\theta,z)\,dV \end{array} \qquad \text{\Large\textcolor{red}Wrong} \] because moments are only defined for rectangular coordinates since the balance beam derivation (\(\text{Torque} = \text{Force} \times \text{Lever Arm}\)) does not work for the \(r\) and \(\theta\) directions. If we want the cylindrical coordinates, \((\bar r,\bar\theta,\bar z)\) of the center of mass, we must first find the rectangular components and then convert to cylindrical.

### Average Value

To find the average value of a function over a 3D region in cylindrical coordinates, the function must be evaluated in cylindrical coordinates before integrating: \[ f_\text{ave}=\dfrac{1}{V}\iiint_R f(x,y,z)\,dV =\dfrac{1}{V}\iiint_R f(r,\theta,z)\,r\,dr\,d\theta\,dz \]

### Centroid

The centroid of a region \(R\) in 3D is the same as the center of mass but with constant density which we take as \(\delta=1\). Then the mass reduces to the volume and the moments of mass become moments of volume. Thus the centroid is: \[ \bar{x}=\dfrac{V_x}{V} \quad \text{and} \quad \bar{y}=\dfrac{V_y}{V} \quad \text{and} \quad \bar{z}=\dfrac{V_z}{V} \] where \(V\) is the volume and the \(x\), \(y\) and \(z\) \(1^\text{st}\)-moments of the volume, respectively, are \[\begin{aligned} V_x&=\iiint_R x\,dV =\iiint_R r\cos\theta\,r\,dr\,d\theta\,dz \\ V_y&=\iiint_R y\,dV =\iiint_R r\sin\theta\,r\,dr\,d\theta\,dz \\ V_z&=\iiint_R z\,dV =\iiint_R z\,r\,dr\,d\theta\,dz \end{aligned}\]

For the region between the paraboloids considered in the other examples, we can use symmetry to recognize that the centroid is obviously \((0,0,4)\). So in the following example we only consider the bottom half of that shape.

### 4-Volume (Optional)

If \(f(x,y,z)\) and \(g(x,y,z)\) are functions on a region \(R\) in \(3\)-space and \(f(x,y,z) \ge g(x,y,z)\), then the 4-volume above the 4-dimensional region \(R\) between \(w=g(x,y,z)\) and \(w=f(x,y,z)\) is: \[ V_4=\iiint_R f(x,y,z)-g(x,y,z)\,dV \] In cylindrical coordinates, we have:

If \(f(r,\theta,z)\) and \(g(r,\theta,z)\) are functions on a region \(R\) in \(3\)-space and \(f(r,\theta,z) \ge g(r,\theta,z)\), then the 4-volume above the 4-dimensional region \(R\) between \(w=g(r,\theta,z)\) and \(w=f(r,\theta,z)\) is \[ V_4=\iiint_R (f(r,\theta,z)-g(r,\theta,z))\,r\,dr\,d\theta\,dz \]

PY: Add an exercise with parts for volume, mass, center of mass, centroid and average value.## Heading

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