21. Multiple Integrals in Curvilinear Coordinates

a. Integrating in Polar Coordinates

4. Applications

Integrals in polar coordinates are particularly useful when the integrand and/or the region of integration are easily expressed in terms of polar coordinates. We previously did Applications of Double Integrals in rectangular coordinates. All of those applications can also be done using polar coordinates, and on the next page we will see that it is often useful to convert from rectangular to polar coordinates.

All the examples on this page of applications of double integrals in polar coordinates will use the crescent shape inside the circle \(r=4\cos\theta\) but outside the limacon \(r=1+2\cos\theta\) as shown in the pink region of the plot.

The \(\theta\)-bounds on the integrals are the angles of intersection. So we equate the values of \(r\): \[\begin{aligned} 4\cos\theta&=1+2\cos\theta \qquad 2\cos\theta=1 \\ \cos\theta&=\dfrac{1}{2} \qquad \theta=\pm \dfrac{\pi}{3} \end{aligned}\] The lower bound (smaller value) of \(r\) is the limacon \(r=1+2\cos\theta\) and the upper bound (larger value) is the circle \(r=4\cos\theta\).



Recall that the volume under a surface \(z=f(x,y)\) above a region \(R\) in the \(xy\)-plane is: \[ \text{Volume}=\iint\limits_R f(x,y)\,dA \] When using polar coordinates, as in the examples below, the function and differential must be expressed in polar coordinates \(r\) and \(\theta\): \[ \text{Volume}=\iint_R f(r,\theta)\,dA =\iint_R f(r,\theta)\,r\,dr\,d\theta \]

In polar coordinates, the surface becomes \(z=\dfrac{1}{r}\). So the volume below the paraboloid is \[\begin{aligned} \text{Volume} &=\int_{-\pi/3}^{\pi/3}\int_{1+2\cos\theta}^{4\cos\theta} \dfrac{1}{r}\,r\,dr\,d\theta =\int_{-\pi/3}^{\pi/3} \left[r\dfrac{}{}\right]_{1+2\cos\theta}^{4\cos\theta}\,d\theta \\ &=\int_{-\pi/3}^{\pi/3} \left(2\cos\theta-1\right)\,d\theta =\left[2\sin\theta-\theta\dfrac{}{}\right]_{-\pi/3}^{\pi/3} \\ &=2\left(2\sin\dfrac{\pi}{3}-\dfrac{\pi}{3}\right) =2\sqrt{3}-\dfrac{2\pi}{3} \end{aligned}\]


Recall that the area of a 2D region \(R\) is: \[ \text{Area}=\iint\limits_R 1\,dA =\iint\limits_R r\,dr\,d\theta \]

The area is: \[\begin{aligned} A &=\int_{-\pi/3}^{\pi/3}\int_{1+2\cos\theta}^{4\cos\theta} r\,dr\,d\theta =\int_{-\pi/3}^{\pi/3} \left[\dfrac{r^2}{2}\right]_{r=1+2\cos\theta}^{4\cos\theta}\,d\theta\\ &=\dfrac{1}{2}\int_{-\pi/3}^{\pi/3} \left[(4\cos\theta)^2-(1+2\cos\theta)^2\right]\,d\theta \\ &=\dfrac{1}{2}\int_{-\pi/3}^{\pi/3} (12\cos^2\theta-1-4\cos\theta)\,d\theta \end{aligned}\] To complete the integral, we use the identity \(\displaystyle \cos^2\theta=\dfrac{1+\cos(2\theta)}{2}\). So: \[\begin{aligned} A&=\dfrac{1}{2}\int_{-\pi/3}^{\pi/3} (6(1+\cos(2\theta))-1-4\cos\theta)\,d\theta \\ &=\dfrac{1}{2}\int_{-\pi/3}^{\pi/3} \left[5+6\cos(2\theta)-4\cos\theta\right]\,d\theta \\ &=\dfrac{1}{2} \left[5\theta+3\sin(2\theta)-4\sin\theta\dfrac{}{}\right]_{-\pi/3}^{\pi/3} \\ &=5\dfrac{\pi}{3}+3\sin\left(2\dfrac{\pi}{3}\right)-4\sin\dfrac{\pi}{3} =\dfrac{5}{3}\pi-\,\dfrac{1}{2}\sqrt{3} \end{aligned}\]

More generally, the area between two polar curves \(r=g(\theta)\) and \(r=h(\theta)\) which intersect at \(\theta=\alpha\) and \(\theta=\beta\) is: \[\begin{aligned} A&=\iint\limits_R 1\,dA=\int_\alpha^\beta\int_{g(\theta)}^{h(\theta)} r\,dr\,d\theta =\int_\alpha^\beta \left[\dfrac{r^2}{2}\right]_{g(\theta)}^{h(\theta)}\,d\theta \\ &=\dfrac{1}{2}\int_\alpha^\beta (h(\theta)^2-g(\theta)^2)\,d\theta \end{aligned}\] which is a formula you may have learned in Calculus 2. Notice that the second line in the solution was just this formula.


Recall that the mass of a plate occupying a 2D region, \(R\), is the integral of the density, \(\delta(x,y)\). After converting to polar coordinates, the density becomes \(\delta(r,\theta)\). So the mass is: \[ M=\iint\limits_R \delta(x,y)\,dA =\iint\limits_R \delta(r,\theta)\,r\,dr\,d\theta \]

The mass is: \[\begin{aligned} M&=\iint\limits_R \delta(r,\theta)\,r\,dr\,d\theta =\int_{-\pi/3}^{\pi/3}\int_{1+2\cos\theta}^{4\cos\theta} \dfrac{1}{r}\,r\,dr\,d\theta \\ &=\int_{-\pi/3}^{\pi/3}\int_{1+2\cos\theta}^{4\cos\theta} 1\,dr\,d\theta =\int_{-\pi/3}^{\pi/3} \left[r\dfrac{}{}\right]_{r=1+2\cos\theta}^{4\cos\theta}\,d\theta \\ &=\int_{-\pi/3}^{\pi/3} (2\cos\theta-1)\,d\theta =\left[2\sin\theta-\theta\dfrac{}{}\right]_{-\pi/3}^{\pi/3} \\ &=2\left(\sqrt{3}-\,\dfrac{\pi}{3}\right)=2\sqrt{3}-\,\dfrac{2\pi}{3} \end{aligned}\]

Center of Mass

Recall that the center of mass of a plate occupying a region \(R\) with density \(\delta(x,y)\) is given by \[ \bar{x}=\dfrac{M_x}{M} \qquad \text{and} \qquad \bar{y}=\dfrac{M_y}{M} \] where the denominator is the mass, \(M\) of the plate and numerators are the \(x\) and \(y\) \(1^\text{st}\)-moments of the mass, given by \[\begin{aligned} M_x&=\iint\limits_R x\,\delta(x,y)\,dA =\iint\limits_R r\cos\theta\,\delta(r,\theta)\,r\,dr\,d\theta \\ M_y&=\iint\limits_R y\,\delta(x,y)\,dA =\iint\limits_R r\sin\theta\,\delta(r,\theta)\,r\,dr\,d\theta \\ \end{aligned}\]

It is important to remember that we cannot compute \(\bar{r}=\dfrac{M_r}{M}\) and \(\bar{\theta}=\dfrac{M_\theta}{M}\) where \[\text{\Large\textcolor{red}Wrong}\qquad M_r=\iint\limits_R r\,\delta(r,\theta)\,dA \qquad \text{and} \qquad M_\theta=\iint\limits_R \theta\,\delta(r,\theta)\,dA \qquad \text{\Large\textcolor{red}Wrong} \] 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 polar coordinates, \((\bar r,\bar\theta)\) of the center of mass, we find them from \[ \bar r=\sqrt{\bar x^2+\bar y^2} \qquad \tan\bar\theta=\dfrac{\bar y}{\bar x} \]

We have already found the mass is \[ M=\int_{-\pi/3}^{\pi/3}\int_{1+2\cos\theta}^{4\cos\theta} \dfrac{1}{r}\,r\,dr\,d\theta =2\sqrt{3}-\,\dfrac{2\pi}{3} \] By symmetry (since the region and density are the same above and below the \(x\)-axis): \[ M_y=\iint\limits_R y\,\delta(x,y)\,dA=0 \qquad \text{and} \qquad \bar{y}=\dfrac{M_y}{M}=0 \] We compute the \(x\) moment of mass as \[\begin{aligned} M_x&=\iint\limits_R x\,\delta(x,y)\,dA =\iint\limits_R r\cos\theta\,\delta(r,\theta)\,r\,dr\,d\theta \\ &=\int_{-\pi/3}^{\pi/3}\int_{1+2\cos\theta}^{4\cos\theta} r\cos\theta\,dr\,d\theta =\int_{-\pi/3}^{\pi/3} \cos\theta\left[\dfrac{r^2}{2}\right]_{r=1+2\cos\theta}^{4\cos\theta}\,d\theta \\ &=\dfrac{1}{2}\int_{-\pi/3}^{\pi/3} \cos\theta\left[(4\cos\theta)^2-(1+2\cos\theta)^2\right]\,d\theta \\ &=\dfrac{1}{2}\int_{-\pi/3}^{\pi/3} (12\cos^3\theta-\cos\theta-4\cos^2\theta)\,d\theta \\ \end{aligned}\] To complete the integral, we use the identities \(\displaystyle \cos^2\theta=1-\sin^2\theta\) and \(\displaystyle \cos^2\theta=\dfrac{1+\cos(2\theta)}{2}\). So: \[\begin{aligned} M_x&=\dfrac{1}{2}\int_{-\pi/3}^{\pi/3} \left(12(1-\sin^2\theta)\cos\theta -\cos\theta-2[1+\cos(2\theta)]\right)\,d\theta \\ &=\dfrac{1}{2}\int_{-\pi/3}^{\pi/3} \left(11\cos\theta-12\sin^2\theta\cos\theta-2 -2\cos(2\theta)\right)\,d\theta \\ &=\dfrac{1}{2} \left[11\sin\theta-4\sin^3\theta-2\theta -\sin(2\theta)\dfrac{}{}\right]_{-\pi/3}^{\pi/3} \\ &=11\sin\dfrac{\pi}{3}-4\sin^3\dfrac{\pi}{3}-2\dfrac{\pi}{3} -\sin(\dfrac{2\pi}{3}) \\ &=11\dfrac{\sqrt{3}}{2}-4\dfrac{3\sqrt{3}}{8}-\,\dfrac{2\pi}{3} -\,\dfrac{\sqrt{3}}{2} \\ &=7\dfrac{\sqrt{3}}{2}-\,\dfrac{2\pi}{3} \end{aligned}\] Thus: \[ \bar{x}=\dfrac{M_x}{M} =\dfrac{7\dfrac{\sqrt{3}}{2}-\,\dfrac{2\pi}{3}}{2\sqrt{3}-\,\dfrac{2\pi}{3}} =\dfrac{21\sqrt{3}-4\pi}{12\sqrt{3}-4\pi} \] Thus, the center of mass of the region is: \[ (\bar{x},\bar{y})_\text{cm} =\left(\dfrac{21\sqrt{3}-4\pi}{12\sqrt{3}-4\pi},0\right) \approx(2.897,0) \]

It is interesting to note that the center of mass of the crescent is not actually within the crescent. However, it is within the convex hull of the crescent. By definition, the convex hull of any set \(R\) is the smallest convex set containing \(R\). For the crescent, the convex hull is the pink and yellow region shown in the plot.
Make the plot have CM

Convex hull of the crescent region showing
the center of mass for density \(\delta=\dfrac{1}{r}\)

Average Value

Recall that the average value of a function over a 2D region, \(R\), is: \[ f_\text{ave}=\dfrac{1}{A}\iint\limits_R f(x,y)\,dA =\dfrac{1}{A}\iint\limits_R f(r,\theta)\,r\,dr\,d\theta \]

The average value of the density is \[ \delta_\text{ave}=\dfrac{1}{A}\iint\limits_R \delta(r,\theta)\,r\,dr\,d\theta \] We recognize the numerator is the mass which was previously computed to be: \[ M=\iint\limits_R \delta(r,\theta)\,r\,dr\,d\theta =2\sqrt{3}-\,\dfrac{2\pi}{3} \] The denominator is the area which was previously computed to be: \[ A=\iint\limits_R r\,dr\,d\theta =\dfrac{5}{3}\pi-\,\dfrac{1}{2}\sqrt{3} \] So the average value is: \[ \delta_\text{ave}=\dfrac{M}{A} =\dfrac{2\sqrt{3}-\,\dfrac{2\pi}{3}} {\dfrac{5}{3}\pi-\,\dfrac{1}{2}\sqrt{3}} =\dfrac {12\sqrt{3}-4\pi} {10\pi-3\sqrt{3}} \]


Recall that the centroid of a region \(R\) is given by \[ \bar{x}=\dfrac{1}{A}\iint\limits_R x\,dA \qquad \text{and} \qquad \bar{y}=\dfrac{1}{A}\iint\limits_R y\,dA \] In polar coordinates, we express \(x\) and \(y\) in terms of \(r\) and \(\theta\), so that the formulas become \[\begin{aligned} \bar{x}&=\dfrac{1}{A}\iint\limits_R r\cos\theta\,dA =\dfrac{1}{A}\iint\limits_R r^2\cos\theta\,dr\,d\theta \\ \bar{y}&=\dfrac{1}{A}\iint\limits_R r\sin\theta\,dA =\dfrac{1}{A}\iint\limits_R r^2\sin\theta\,dr\,d\theta \end{aligned}\]

We already know that the area of the crescent is \[ A=\dfrac{5}{3}\pi-\,\dfrac{1}{2}\sqrt{3} \] By symmetry (since the region is the same above and below the \(x\)-axis): \[ A_y=\iint\limits_R y\,dA=0 \qquad \text{and} \qquad \bar{y}=\dfrac{A_y}{A}=0 \] Now we compute the \(x\) moment as \[\begin{aligned} A_x&=\iint\limits_R r\cos\theta\,dA =\int_{-\pi/3}^{\pi/3}\int_{1+2\cos\theta}^{4\cos\theta} r^2\cos\theta\,dr\,d\theta \\ &=\int_{-\pi/3}^{\pi/3} \left[\dfrac{r^3}{3}\cos\theta\right]_{r=1+2\cos\theta}^{4\cos\theta}\,d\theta \\ &=\dfrac{1}{3}\int_{-\pi/3}^{\pi/3} (4\cos\theta)^3\cos\theta-(1+2\cos\theta)^3\cos\theta\,d\theta \\ &=\dfrac{1}{3}\int_{-\pi/3}^{\pi/3} 64\cos^4\theta-(8\cos^4\theta+12\cos^3\theta+6\cos^2\theta+\cos\theta)\,d\theta \\ &=\dfrac{1}{3}\int_{-\pi/3}^{\pi/3} (56\cos^4\theta-12\cos^3\theta-6\cos^2\theta-\cos\theta)\,d\theta \end{aligned}\] We now use the identities \(\displaystyle \cos^2\theta=1-\sin^2\theta\) and \(\displaystyle \cos^2\theta=\dfrac{1+\cos(2\theta)}{2}\) repeatedly to obtain: \[ A_x=4\pi+\dfrac{1}{4}\sqrt{3} \] Thus the \(x\)-component of the centroid is: \[ \bar{x}=\dfrac{A_x}{A} =\dfrac{4\pi+\dfrac{1}{4}\sqrt{3}} {\dfrac{5}{3}\pi-\,\dfrac{1}{2}\sqrt{3}} =\dfrac {48\pi+3\sqrt{3}} {20\pi-6\sqrt{3}} \] Thus, the centroid of the region is: \[ (\bar{x},\bar{y})_\text{centroid} =\left(\dfrac{48\pi+3\sqrt{3}}{20\pi-6\sqrt{3}},0\right) \approx(2.975,0) \]

Like the center of mass, the centroid of the crescent is not within the crescent, but is within the convex hull of the crescent.

Convex hull of the crescent region
showing the centroid

Here is a blow-up of the edge of the crescent showing the centroid with an \(\times\) and the center of mass with an \(\circ\). Notice that the center of mass is to the left of the centroid: \[ \bar{x}_\text{cm} < \bar{x}_\text{centroid} \] This is because the density \(\delta=\dfrac{1}{r}\) decreases to the right, pulling the centroid to the left.
PY: Fix the plots to have both centroid and center of mass.

Enlargement of the edge of the crescent,
with the centroid and the center of mass

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Supported in part by NSF Grant #1123255