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.

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. Find the bounds on the integrals.

The plot shows a circle centered at (2,0) of radius 2 and a limacon centered on the positive x axis with the outer loop going through x = 3, and the inner loop going through x = 1. The circle and limacon intersect at x = 1. The region between them looks like a crescent moon.

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 \(r=4\cos\theta\).

Volume

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

Find the volume of the solid above the crescent shape, below the surface \(z=\dfrac{1}{\sqrt{x^2+y^2}}\).

The plot shows a solid over the crescent of the previous plot with vertical sides and a top which slopes down toward the positive x axis.

In polar coordinates, the surface becomes \(z=\dfrac{1}{r}\). So the volume below the paraboloid is \[\begin{aligned} \text{Volume} &=\int_{-\pi/3}^{\pi/3}\int_{1+2\cos\theta}^{4\cos\theta} \dfrac{1}{r}\,r\,dr\,d\theta =\int_{-\pi/3}^{\pi/3} \left[\rule{0pt}{10pt}r\right]_{1+2\cos\theta}^{4\cos\theta}\,d\theta \\ &=\int_{-\pi/3}^{\pi/3} \left(2\cos\theta-1\right)\,d\theta =\left[\rule{0pt}{10pt}2\sin\theta-\theta\right]_{-\pi/3}^{\pi/3} \\ &=2\left(2\sin\dfrac{\pi}{3}-\dfrac{\pi}{3}\right) =2\sqrt{3}-\dfrac{2\pi}{3} \end{aligned}\]

Area

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

Find the area of the crescent shape.

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: \[\begin{aligned} A&=\dfrac{1}{2}\int_{-\pi/3}^{\pi/3} (6(1+\cos(2\theta))-1-4\cos\theta)\,d\theta \\ &=\dfrac{1}{2}\int_{-\pi/3}^{\pi/3} \left[5+6\cos(2\theta)-4\cos\theta\right]\,d\theta \\ &=\dfrac{1}{2} \left[\rule{0pt}{10pt}5\theta+3\sin(2\theta)-4\sin\theta\right]_{-\pi/3}^{\pi/3} \\ &=5\dfrac{\pi}{3}+3\sin\left(2\dfrac{\pi}{3}\right)-4\sin\dfrac{\pi}{3} =\dfrac{5}{3}\pi-\,\dfrac{1}{2}\sqrt{3} \end{aligned}\]

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

Mass

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

Find the mass of the crescent shape if its mass density is \(\delta=\dfrac{1}{\sqrt{x^2+y^2}}=\dfrac{1}{r}\).

The mass is: \[\begin{aligned} M&=\iint_R \delta(r,\theta)\,r\,dr\,d\theta =\int_{-\pi/3}^{\pi/3}\int_{1+2\cos\theta}^{4\cos\theta} \dfrac{1}{r}\,r\,dr\,d\theta \\ &=\int_{-\pi/3}^{\pi/3}\int_{1+2\cos\theta}^{4\cos\theta} 1\,dr\,d\theta =\int_{-\pi/3}^{\pi/3} \left[\rule{0pt}{10pt}r{}\right]_{r=1+2\cos\theta}^{4\cos\theta}\,d\theta \\ &=\int_{-\pi/3}^{\pi/3} (2\cos\theta-1)\,d\theta =\left[\rule{0pt}{10pt}2\sin\theta-\theta\right]_{-\pi/3}^{\pi/3} \\ &=2\left(\sqrt{3}-\,\dfrac{\pi}{3}\right)=2\sqrt{3}-\,\dfrac{2\pi}{3} \end{aligned}\]

Center of Mass

Recall that the center of mass of a plate occupying a region \(R\) with density \(\delta(x,y)\) is: \[ \bar{x}=\dfrac{M_x}{M} \qquad \text{and} \qquad \bar{y}=\dfrac{M_y}{M} \] 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 \[\begin{aligned} M_x&=\iint_R x\,\delta(x,y)\,dA =\iint_R r\cos\theta\,\delta(r,\theta)\,r\,dr\,d\theta \\[10pt] M_y&=\iint_R y\,\delta(x,y)\,dA =\iint_R r\sin\theta\,\delta(r,\theta)\,r\,dr\,d\theta \\ \end{aligned}\]

It is important to remember that we cannot compute \(\bar{r}=\dfrac{M_r}{M}\) and \(\bar{\theta}=\dfrac{M_\theta}{M}\) where \[\text{\Large\textcolor{red}Wrong}\qquad M_r=\iint_R r\,\delta(r,\theta)\,dA \qquad \text{and} \qquad M_\theta=\iint_R \theta\,\delta(r,\theta)\,dA \qquad \text{\Large\textcolor{red}Wrong} \] 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 \[ \bar r=\sqrt{\bar x^2+\bar y^2} \qquad \tan\bar\theta=\dfrac{\bar y}{\bar x} \]

Find the center of mass of the crescent shape if its mass density is \(\delta=\dfrac{1}{\sqrt{x^2+y^2}}=\dfrac{1}{r}\).

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): \[ M_y=\iint_R y\,\delta(x,y)\,dA=0 \qquad \text{and} \qquad \bar{y}=\dfrac{M_y}{M}=0 \] We compute the \(x\) moment of mass as \[\begin{aligned} M_x&=\iint_R x\,\delta(x,y)\,dA =\iint_R r\cos\theta\,\delta(r,\theta)\,r\,dr\,d\theta \\ &=\int_{-\pi/3}^{\pi/3}\int_{1+2\cos\theta}^{4\cos\theta} r\cos\theta\,dr\,d\theta =\int_{-\pi/3}^{\pi/3} \cos\theta\left[\dfrac{r^2}{2}\right]_{r=1+2\cos\theta}^{4\cos\theta}\,d\theta \\ &=\dfrac{1}{2}\int_{-\pi/3}^{\pi/3} \cos\theta\left[(4\cos\theta)^2-(1+2\cos\theta)^2\right]\,d\theta \\ &=\dfrac{1}{2}\int_{-\pi/3}^{\pi/3} (12\cos^3\theta-\cos\theta-4\cos^2\theta)\,d\theta \\ \end{aligned}\] 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: \[\begin{aligned} M_x&=\dfrac{1}{2}\int_{-\pi/3}^{\pi/3} \left(12(1-\sin^2\theta)\cos\theta -\cos\theta-2[1+\cos(2\theta)]\right)\,d\theta \\ &=\dfrac{1}{2}\int_{-\pi/3}^{\pi/3} \left(11\cos\theta-12\sin^2\theta\cos\theta-2 -2\cos(2\theta)\right)\,d\theta \\ &=\dfrac{1}{2} \left[\rule{0pt}{10pt}11\sin\theta-4\sin^3\theta-2\theta -\sin(2\theta)\right]_{-\pi/3}^{\pi/3} \\ &=11\sin\dfrac{\pi}{3}-4\sin^3\dfrac{\pi}{3}-2\dfrac{\pi}{3} -\sin(\dfrac{2\pi}{3}) \\ &=11\dfrac{\sqrt{3}}{2}-4\dfrac{3\sqrt{3}}{8}-\,\dfrac{2\pi}{3} -\,\dfrac{\sqrt{3}}{2} \\ &=7\dfrac{\sqrt{3}}{2}-\,\dfrac{2\pi}{3} \end{aligned}\] 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.

The center of mass is shown as a blue star.

The plot shows the crescent of the first plot with a star just inside of the large loop of the limacon. It also has a vertical line at x = 1 which connects the tips of the crescent. The piece of the circle to the right of this line is the convex hull of the crescent.
Convex hull of the crescent region showing
the center of mass for density \(\delta=\dfrac{1}{r}\)

Centroid

Recall that the centroid of a region \(R\) is the same as the center of mass but with constant density which we take as \(\delta=1\). Then the mass reduces to the area and the moments of mass become moments of area. Thus the centroid is: \[ \bar{x}=\dfrac{1}{A}\iint_R x\,dA \qquad \text{and} \qquad \bar{y}=\dfrac{1}{A}\iint_R y\,dA \] In polar coordinates, we express \(x\) and \(y\) in terms of \(r\) and \(\theta\), so that the formulas become \[\begin{aligned} \bar{x}&=\dfrac{1}{A}\iint_R r\cos\theta\,dA =\dfrac{1}{A}\iint_R r^2\cos\theta\,dr\,d\theta \\[10pt] \bar{y}&=\dfrac{1}{A}\iint_R r\sin\theta\,dA =\dfrac{1}{A}\iint_R r^2\sin\theta\,dr\,d\theta \end{aligned}\]

Find the centroid of the crescent shape.

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): \[ A_y=\iint_R y\,dA=0 \qquad \text{and} \qquad \bar{y}=\dfrac{A_y}{A}=0 \] Now we compute the \(x\) moment as \[\begin{aligned} A_x&=\iint_R r\cos\theta\,dA =\int_{-\pi/3}^{\pi/3}\int_{1+2\cos\theta}^{4\cos\theta} r^2\cos\theta\,dr\,d\theta \\ &=\int_{-\pi/3}^{\pi/3} \left[\dfrac{r^3}{3}\cos\theta\right]_{r=1+2\cos\theta}^{4\cos\theta}\,d\theta \\ &=\dfrac{1}{3}\int_{-\pi/3}^{\pi/3} (4\cos\theta)^3\cos\theta-(1+2\cos\theta)^3\cos\theta\,d\theta \\ &=\dfrac{1}{3}\int_{-\pi/3}^{\pi/3} 64\cos^4\theta-(8\cos^4\theta+12\cos^3\theta+6\cos^2\theta+\cos\theta)\,d\theta \\ &=\dfrac{1}{3}\int_{-\pi/3}^{\pi/3} (56\cos^4\theta-12\cos^3\theta-6\cos^2\theta-\cos\theta)\,d\theta \end{aligned}\] 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.

For the crescent, the convex hull is the pink and yellow region shown in the plot.

The centroid is shown as a green circle.

The plot shows the crescent of the first plot along with its convex hull. There is a small circle just inside of the large loop of the limacon.
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 center of mass to the left.

The center of mass is shown as a blue star.

The centroid is shown as a green circle.

The plot shows the limacon edge of the crescent. To the left of this edge is the small circle and to the left of that is the star.
Enlargement of the edge of the crescent,
with the centroid and the center of mass

Average Value

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

Compute the average density on the crescent shape if the density is \(\delta=\dfrac{1}{\sqrt{x^2+y^2}}=\dfrac{1}{r}\).

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

The exercises on the rest of this page refer to the quarter annulus which is the region between the circles \(r=3\) and \(r=5\) within the first quadrant. Find the limits of integration.

The plot shows the region between the circles of radii 3 and 5, but only in the first quadrant.

\(3 \le r \le 5\)
\(0 \le \theta \le \dfrac{\pi}{2}\)

This is a polar rectangle in the first quadrant: \[ 3 \le r \le 5 \qquad \text{and} \qquad 0 \le \theta \le \dfrac{\pi}{2} \]

mj 

Find the volume of the solid above the quarter annulus below the surface \(z=\sqrt{x^2+y^2}\).

The plot show a solid above the region between the circles. It has vertical sides and a top which slants downward toward the origin.

The volume below a function \(f\) is: \[ V=\iint_R f\,dA \] Express the function in polar coordinates and use \(dA=r\,dr\,d\theta\).

\(V=\dfrac{49\pi}{3}\)

In polar coordinates the surface is \(z=\sqrt{x^2+y^2}=r\). Using the limits found in the previous problem, the volume under this surface is: \[\begin{aligned} V &= \iint_R f\,dA = \int_0^{\pi/2} \int_3^5 r\cdot r\,dr\,d\theta \\ &= \dfrac{\pi}{2} \left[\dfrac{r^3}{3}\right]_3^5 = \dfrac{\pi}{6}(125-27) = \dfrac{49\pi}{3} \end{aligned}\]

mj 

Find the area of the quarter annulus.

\(A=4\pi\)

We use the limits found in a previous problem: \[\begin{aligned} A &= \iint_R 1\,dA = \int_0^{\pi/2} \int_3^5 r\,dr\,d\theta \\ &= \dfrac{\pi}{2} \left[\dfrac{r^2}{2}\right]_3^5 = \dfrac{\pi}{4}(25-9) = 4\pi \end{aligned}\]

mj 

Of course we can do this by high school geometry. The area of the annulus is \(A=\pi 5^2-\pi 3^2=16\pi\). So the area of a quarter annulus is \(A=\dfrac{1}{4}16\pi=4\pi\).

Find the mass of the quarter annulus if the mass density is \(\delta=x\).

If the density is \(\delta\), the mass is \(\displaystyle M=\iint_R \delta\, dA\).

\(M = \dfrac{98}{3}\)

In polar coordinates, \(x=r\cos\theta\), so \[\begin{aligned} M&=\iint_R \delta\, dA=\int_0^{\pi/2} \int_3^5 r\cos\theta \,r \,dr\, d\theta \\ &= \left[\rule{0pt}{10pt}\sin\theta\right]_0^{\pi/2} \left[\dfrac{r^3}{3}\right]_3^5 = \dfrac{1}{3}(1-0)(125-27) = \dfrac{98}{3} \end{aligned}\]

mj 

Find the center of mass of the quarter annulus if the mass density is \(\delta=x\).

\((\bar x,\bar y)=\left(\dfrac{51\pi}{49},\dfrac{102}{49}\right) \approx(3.27,2.08)\)

We first compute the \(x\)-moment of mass: \[\begin{aligned} M_x &=\iint_R x\,\delta\,dA =\int_0^{\pi/2} \int_3^5 r^2\cos^2\theta \, r\, dr\, d\theta \\ &=\int_0^{\pi/2} \dfrac{1+\cos2\theta}{2} \,d\theta \int_3^5 r^3 \,dr =\left[\dfrac{\theta}{2}+\dfrac{\sin2\theta}{4}\right]_0^{\pi/2} \left[\dfrac{r^4}{4}\right]_3^5 \\ &=\dfrac{1}{4}(\dfrac{\pi}{4}+0-0)(625-81) = 34\pi \end{aligned}\]

We next compute the \(y\)-moment of mass: \[\begin{aligned} M_y &=\iint_R y\,\delta\,dA =\int_0^{\pi/2} \int_3^5 r^2\sin\theta\cos\theta \, r\, dr\, d\theta \\ &= \left[\dfrac{\sin^2\theta}{2}\right]_0^{\pi/2} \left[\dfrac{r^4}{4}\right]_3^5 =\dfrac{1}{8}(1-0)(625-81) = 68 \end{aligned}\]

Since we previously computed \(M=\dfrac{98}{3}\), we conclude: \[\begin{aligned} \bar x &= \dfrac{M_x}{M}=34\pi\cdot\dfrac{3}{98}=\dfrac{51\pi}{49}\approx 3.27 \\ \bar y &= \dfrac{M_y}{M}=68\cdot\dfrac{3}{98}=\dfrac{102}{49}\approx 2.08 \end{aligned}\] Thus the center of mass is at: \[ (\bar x,\bar y)=\left(\dfrac{51\pi}{49},\dfrac{102}{49}\right)\approx(3.27,2.08) \]

The plot shows the region between the circles with a star at x = 3.27 and y = 2.08.

mj 

Find the centroid of the quarter annulus.

\((\bar x,\bar y)=\left(\dfrac{49}{6\pi},\dfrac{49}{6\pi}\right) \approx(2.6,2.6)\)

We first compute the \(x\)-moment of area: \[\begin{aligned} A_x &=\iint_R x\,dA =\int_0^{\pi/2} \int_3^5 r\cos\theta \, r\, dr\, d\theta \\ &= \left[\rule{0pt}{10pt}\sin\theta\right]_0^{\pi/2} \left[\dfrac{r^3}{3}\right]_3^5 =\dfrac{1}{3}(1-0)(125-27) = \dfrac{98}{3} \end{aligned}\]

We next compute the \(y\)-moment of area: \[\begin{aligned} A_y &=\iint_R y\,dA =\int_0^{\pi/2} \int_3^5 r\sin\theta \, r\, dr\, d\theta \\ &= \left[\rule{0pt}{10pt}-\cos\theta\right]_0^{\pi/2} \left[\dfrac{r^3}{3}\right]_3^5 =\dfrac{1}{3}(0--1)(125-27) = \dfrac{98}{3} \end{aligned}\]

Since the area is \(A=4\pi\), we conclude: \[\begin{aligned} \bar x &= \dfrac{A_x}{A}=\dfrac{98}{3}\dfrac{1}{4\pi}=\dfrac{49}{6\pi}\approx 2.6 \\ \bar y &= \dfrac{A_y}{A}=\dfrac{98}{3}\dfrac{1}{4\pi}=\dfrac{49}{6\pi}\approx 2.6 \end{aligned}\] Thus the center of mass is at: \[ (\bar x,\bar y)=\left(\dfrac{49}{6\pi},\dfrac{49}{6\pi}\right)\approx(2.6,2.6) \] By symmetry we ahould have known that \(\bar x=\bar y\).

The plot shows the region between the circles with a small circle at x = 2.6 and y = 2.6.

mj 

This plot shows the center of mass as well as the centroid.
The center of mass is plotted as a blue star.
The centroid at is plotted as a green circle.
Notice that the center of mass is further to the right and below the centroid because the density, \(\delta=x\), is bigger on the right and the right side is below the center.

The plot shows the region between the circles with a star at x = 3.27 and y = 2.08 and a small circle at x = 2.6 and y = 2.6.

Find the average value of the function \(f(x,y)=\dfrac{1}{\sqrt{x^2+y^2}}\) on the quarter annulus.

\(f_{\text{ave}}=\dfrac{1}{4}\)

In polar coordinates, \(f(r,\theta)=\dfrac{1}{r}\), so \[\begin{aligned} \iint_R f\,dA &= \int_0^{\pi/2} \int_3^5 \dfrac{1}{r}\cdot r\, dr\, d\theta = \dfrac{\pi}{2} \int_3^5 1\, dr = \dfrac{\pi}{2}(5-3) = \pi \end{aligned}\] Thus \[ f_{\text{ave}}=\dfrac{1}{A}\iint_R f\,dA =\dfrac{1}{4\pi}\cdot \pi=\dfrac{1}{4} \]

mj 

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