OBJECTIVES

When you finish this section, you should be able to:

1. Find a double integral using polar coordinates (p. 924)
2. Find area and volume using polar coordinates (p. 925)

NEED TO REVIEW?

Polar coordinates are discussed in Section 9.4, pp. 661-668.

THEOREM Iterated Double Integral in Polar Coordinates

Let the function \(z=f(x,y)\) be continuous over a closed, bounded region \(R\). Suppose \(R\) is enclosed by the rays \(\theta =\alpha\) and \(\theta =\beta\) and the curves \(r=r_{1}(\theta)\) and \( r=r_{2}(\theta)\), where \(0\leq r_{1}\leq r_{2}.\) If \(r=\) \(r_{1}(\theta)\) and \(r=r_{2}(\theta)\) are continuous for \(\alpha \leq \theta \leq \beta\), then \[\bbox[5px, border:1px solid black, #F9F7ED]{ \displaystyle\iint\limits_{\kern-3ptR}f(x,y) \,{\it dA}=\displaystyle\iint\limits_{\kern-3ptR}f(r\cos\theta,r\sin\theta)\, {\it dA} =\int_{\alpha }^{\beta }\int_{r_{1}(\theta) }^{r_{2}(\theta) }f(r\cos\theta,r\sin\theta)\, r\,dr\,d\theta } \]

Finding a Double Integral Using Polar Coordinates

Find \(\displaystyle\iint\limits_{\kern-3ptR}\cos \theta \,{\it dA}\), where \(R\) is the region enclosed by the rays \(\theta =0\) and \(\theta =\dfrac{\pi }{4}\) and the circle \(r=4\cos \theta\).

Figure 26

Solution We begin with a graph of the region \(R\). See Figure 26. Notice that \(\theta\) varies from \(0\) to \(\dfrac{\pi }{4}\) and \(r\) varies from \(r_{1}=0\) to the circle \(r_{2}=4\cos \theta\). Then \[ \begin{array} \displaystyle\iint\limits_{\kern-3ptR}\cos \theta \,{\it dA} &=&\int_{0}^{\pi /4}\int_{0}^{4\cos \theta }r\cos \theta \,dr\,d\theta =\int_{0}^{\pi /4}\left[ \dfrac{r^{2}}{2}\cos \theta \right] _{0}^{4\cos \theta }d\theta =\int_{0}^{\pi /4}8\cos ^{3}\theta \,d\theta \notag \\[4pt] &=&8\int_{0}^{\pi /4}(1-\sin ^{2}\theta )\cos \theta \,d\theta = 8\int_{0}^{\sqrt{2}/2}(1-u^{2})\, du\\[-8pt] && \qquad \qquad \qquad \quad \underset{\color{#0066A7}{\hbox{\(u=\sin \theta ;\, {du}=\cos \theta\, d\theta\)}}}{\color{#0066A7}{\uparrow }}\\[8pt] &=&8\left[ u-\dfrac{u^{3}}{3}\right] _{0}^{\sqrt{2}/2}=8\left[ \dfrac{\sqrt{2} }{2}-\dfrac{\sqrt{2}}{12}\right] =\dfrac{10\sqrt{2}}{3} \end{array} \]

NEED TO REVIEW?

Integrals containing trigonometric functions are discussed in Section 7.2, pp. 480-486.

NOW WORK

Problems 9 and 15.

NOTE

The order \({\it dy}\,{\it dx}\) indicates the integration is done with respect to \(y\) first, then with respect to \(x\).

Finding an Iterated Integral Using Polar Coordinates

Find the iterated integral \(\int_{0}^{2}\int_{0}^{\sqrt{4-x^{2}} }(x^{2}+y^{2})\,{\it dy}\,{\it dx}\).

Figure 27 \(0\leq y\leq \sqrt{4-x^2}\),
\(0\leq x\leq 2\) \(0\leq r\leq 2\),
\(0\leq \theta \leq \dfrac{\pi}{2}\)

Solution The region \(R\) is enclosed by the graph of \(y=\sqrt{ 4-x^{2}}\), (a portion of a circle of radius 2), \(0\leq x\leq 2\), and the positive \(x\)- and \(y\)-axes. See Figure 27. In polar coordinates, this region \(R\) is given by \(0\leq r\leq 2\) and \(0\leq \theta \leq \dfrac{\pi }{2} \). Then \[ \begin{eqnarray*} \int_{0}^{2}\int_{0}^{\sqrt{4-x^{2}}}(x^{2}+y^{2})\,{\it dy}\,{\it dx}&=&\displaystyle\iint\limits_{\kern-3ptR}(x^{2}+y^{2}) \, {\it dA}\\[4pt] &=&\displaystyle\iint\limits_{\kern-3ptR}r^{2}r\,{\it dr}\,d\theta\qquad{\color{#0066A7}{\hbox{$x^2 + y^2=r^2;\quad {dA}=r\,{dr}\,d\theta$}}}\\[4pt] &=&\int_{0}^{\pi /2}\int_{0}^{2}r^{3}\,dr\,d\theta =\int_{0}^{\pi /2}\left[\dfrac{r^{4}}{4}\right] _{0}^{2}\,d\theta\\[4pt] &=&\int_{0}^{\pi /2}4\,d\theta = \big[ 4\theta \big] _{0}^{\pi /2}=2\pi \end{eqnarray*} \]

NOW WORK

Problems 25 and 29.

NEED TO REVIEW?

Elementary functions are discussed in Section 7.6, p. 508.

Finding a Double Integral Using Polar Coordinates

Find \(\displaystyle\iint\limits_{\kern-3ptR}e^{x^{2}+y^{2}}\,d\!A\), where \(R\) is the region in the first quadrant inside the circle \(x^{2}+y^{2}=a^{2}\).

Figure 28 \(0\leq r\leq a,\) \(0\leq \theta \leq \dfrac{\pi }{2}\)

Solution In rectangular coordinates, \(\iint \limits_{\kern-3ptR}e^{x^{2}+y^{2}}\,d\!A\) only can be approximated using numerical techniques. But the presence of \(x^{2}+y^{2}\) in the integrand and the fact that the region \(R\) has part of a circle as its boundary suggest using polar coordinates. Figure 28 shows the region \(R.\) We use polar coordinates to obtain

NOW WORK

Problem 35.

Finding the Volume of a Solid

Find the volume \(V\) of the solid under the circular paraboloid \( z=12-3x^{2}-3y^{2}\) and above the \(xy\)-plane.

Solution We begin by graphing the paraboloid for \(z\geq 0,\) as shown in Figure 29.

Figure 29

The region \(R\) is determined by the intersection of the surface \(z\) and the \( xy\)-plane. So, we let \(z=0\) and simplify the equation. \begin{eqnarray*} 12-3x^{2}-3y^{2} &=&0 \\[3pt] 3x^{2}+3y^{2} &=&12 \\[3pt] x^{2}+y^{2} &=&4 \end{eqnarray*}

So \(R\) is the region enclosed by the circle \(x^{2}+y^{2}=4\).

To convert the problem to polar coordinates, we let \(x=r\cos \theta\) and \( y=r\sin \theta.\) Then the paraboloid is \(z=12-3x^{2}-3y^{2}=12-3(x^{2}+y^{2}) =12-3r^{2},\) and the region \(R\) is enclosed by the circle \(r^{2}=4.\) So, \(R\) is given by \(0\leq r\leq 2\) and \(0\leq \theta \leq 2\pi.\)

Then \[ \begin{eqnarray*} V &=&\displaystyle\iint\limits_{\kern-3ptR}( 12-3x^{2}-3y^{2})\, {\it dA}\underset{\underset{\color{#0066A7}{\hbox{$\begin{array}{c}3x^{2}+3y^{2}=3r^{2}\\ {dA}=r\,{dr}\,d\theta\end{array}$}}} {\color{#0066A7}{\uparrow}}}{=} \int\nolimits_{0}^{2\pi }\!\!\!\int_{0}^{2}( 12-3r^{2})\, r\,{\it dr}\,d\theta = \int_{0}^{2\pi }\!\!\!\int_{0}^{2}( 12r-3r^{3})\, dr\, d\theta \\[-15.5pt] &=&\int_{0}^{2\pi }\left[ 6r^{2}-3\dfrac{r^{4}}{4}\right] _{0}^{2}~d\theta = \int_{0}^{2\pi }12~d\theta =\big[ 12\theta \big] _{0}^{2\pi }=24\pi \hbox{ cubic units} \end{eqnarray*} \]

NOW WORK

Problem 43.

Finding the Volume of a Solid

Find the volume \(V\) of the solid under the surface \(z=x^{2}+y^{2}\), above the \(xy\)-plane, and inside the cylinder \(x^{2}+y^{2}=2y\).

Figure 30

Solution See Figure 30. The volume we seek lies above the region \(R\) in the \(xy\)-plane enclosed by the circle \(x^{2}+y^{2}=2y\), whose center is at \((0,1,0)\) and whose radius is 1. To convert to polar coordinates, we let \(x=r\cos \theta\) and \(y=r\sin \theta\). Then the region \(R\) is enclosed by \[ \begin{eqnarray*} x^{2}+y^{2} &=&2y \\[3pt] r^{2}\cos ^{2}\theta +r^{2}\sin ^{2}\theta &=&2r\sin \theta \\[3pt] r^{2} &=&2r\sin \theta \\[3pt] r &=&2\sin \theta \end{eqnarray*} \]

The region \(R\) is bounded by \(0\leq r\leq 2\sin \theta\) and \(0\leq \theta \leq \pi .\)

The volume \(V\) under \(z=x^{2}+y^{2}=r^{2}\), above the \(xy\)-plane, and over \(R\) is \[ \begin{eqnarray*} V &=&\displaystyle\iint\limits_{\kern-3ptR} (x^2+y^2)\,dA =\displaystyle\iint\limits_{\kern-3ptR}r^{2}r\,{\it dr}\,d\theta =\int_{0}^{\pi }\int_{0}^{2\sin \theta }r^{3}\,dr\,d\theta =\int_{0}^{\pi }\left[ \dfrac{r^{4}}{4}\right] _{0}^{2\sin \theta }d\theta \\[4pt] &=&4\int_{0}^{\pi }\sin ^{4}\theta \,d\theta =4\int_{0}^{\pi }\left( \dfrac{ 1-\cos (2\theta) }{2}\right) ^{2} d\theta \\[4pt] &=&\int_{0}^{\pi }[ 1-2\cos (2\theta) +\cos ^{2}(2\theta) ] \,d\theta =\int_{0}^{\pi }\left[ 1-2\cos (2\theta) +\dfrac{1+\cos ( 4\theta ) }{2}\right]\! d\theta \\[4pt] &=&\left[ \theta -\sin (2\theta) +\dfrac{1}{2}\theta +\dfrac{1}{ 2}\cdot \dfrac{\sin ( 4\theta )}{4} \right] _{0}^{\pi }=\dfrac{3\pi }{2} \hbox{ cubic units} \end{eqnarray*} \]

NEED TO REVIEW?

Integrals containing trigonometric functions are discussed in Section 7.2, pp. 480-486.

NOW WORK

Problem 45.

Finding the Area Enclosed by a Cardioid

Use double integration to find the area enclosed by the cardioid \(r=a(1-\cos \theta)\).

Figure 31 The upper half of the cardioid is swept out by the rays going from \(\theta =0\) to \(\theta =\pi\).

Solution The cardioid \(r=a(1-\cos \theta)\) is shown in Figure 31. Using symmetry, the area enclosed by the cardioid is twice the area of the shaded region \(R\) in Figure 31. So, \[ \begin{equation*} A=2\displaystyle\iint\limits_{\kern-3ptR}r\,dr\,d\theta \end{equation*} \]

where the region \(R\) is given by \(0\leq r\leq a( 1-\cos \theta )\) and \(0\leq \theta \leq \pi.\) Then \begin{eqnarray*} A &=&2\int_{0}^{\pi }\int_{0}^{a(1-\cos \theta )}r\,dr\,d\theta =2\int_{0}^{\pi }\left[ \dfrac{r^{2}}{2}\right] _{0}^{a( 1-\cos \theta ) }d\theta\\[4pt] &=&\int_{0}^{\pi }a^{2}( 1-\cos \theta ) ^{2}\,d\theta =a^{2}\int_{0}^{\pi }( 1-2\cos \theta +\cos ^{2}\theta )\, d\theta\\[4pt] &=&a^{2}\!\left[ \theta -2\sin \theta +\left( \dfrac{\theta }{2}+\dfrac{\sin (2\theta) }{4}\right) \right] _{0}^{\pi }=\dfrac{3}{2}\pi a^{2} \hbox{ square units} \end{eqnarray*}

NOW WORK

Problem 49.