Double Integrals in Polar Coordinates

Direct Knowlege Contributor David A. Smith
  • By David A. Smith, Founder & CEO, Direct Knowledge
  • David Smith has a B.S. and M.S. in Mathematics and has enjoyed teaching calculus, linear algebra, and number theory at both Tarrant County College and the University of Texas at Arlington. David is the Founder and current CEO of Direct Knowledge.

Fubini’s Theorem in Polar Coordinates

The polar conversion formulas are used to convert from rectangular to polar coordinates: \begin{equation} x=r \cos \theta, \quad y=r \sin \theta, \quad r=\sqrt{x^2+y^2}, \quad \tan \theta =\frac{y}{x}. \end{equation}

Theorem. (Fubini’s Theorem in Polar Coordinates) If $f$ is continuous in the polar region $R$ described by $$ 0\leq r_1(\theta )\leq r\leq r_2(\theta ) \qquad \qquad \alpha \leq \theta \leq \beta $$ (with $0\leq \beta -\alpha \leq 2\pi $), then \begin{equation} \iint_R f(r,\theta )d A=\int_{\alpha }^{\beta } \int_{r_1(\theta )}^{r_2(\theta )}f(r,\theta )r \, dr \, d\theta . \end{equation}

Example. Evaluate \begin{equation} \iint_R \left(3x+4y^2\right)\, dA \end{equation} where $R$ is the upper-half plane bounded by the circles $x^2+y^2=1$ and $x^2+y^2=4.$

Solution. The region of integration $R$ is described as $$ R=\left\{(x,y) \mid y\geq 0,1\leq x^2+y^2\leq 4\right\}. $$ It is an upper ring with polar coordinates given by $1\leq r\leq 2,$ $ 0\leq \theta \leq \pi .$ Therefore \begin{align} \iint_R\left( 3x+4y^2\right) \, d A & =\int_0^{\pi }\int_1^2\left(3r \cos \theta +4r^2\sin ^2 \theta \right) r \, dr \, d\theta \\ & =\int_0^{\pi } \left(7 \cos \theta +15\sin ^2\theta \right)\, d\theta =\frac{15\pi }{2}\end{align} by using the trigonometric identity $\sin^2 \theta=\frac{1-\cos (2\theta)}{2}.$

Example. Evaluate \begin{equation} \int_0^{\pi /2}\int_1^3 r e^{-r^2} \, dr \, d\theta. \end{equation}

Solution. In polar coordinates the region of integration is described as \begin{equation} R=\{(r,\theta) \mid 0\leq \theta \leq \frac{\pi}{2}, 1\leq r \leq 3\}. \end{equation} Using Fubini’s Theorem in polar coordinates the iterated integral is evaluated as \begin{equation}\int_0^{\pi /2}\int_1^3 r e^{-r^2}drd\theta =\int_0^{\pi /2} \frac{e^8-1}{2 e^9 } \, d\theta =\frac{\left(e^8-1\right) \pi }{4 e^9}. \end{equation}

Example. Evaluate \begin{equation} \int_0^{\pi /2}\int_1^2 \sqrt{4-r^2}r \,dr \, d\theta. \end{equation}

Solution. In polar coordinates the region of integration is described as \begin{equation} R=\{(r,\theta) \mid 0\leq \theta \leq \frac{\pi}{2}, 1\leq r \leq 2\}.\end{equation} Using polar coordinates the iterated integral is evaluated as
\begin{equation} \int_0^{\pi /2}\int_1^2 \sqrt{4-r^2}r drd\theta =\int_0^{\pi /2} \sqrt{3} \, d\theta =\frac{\sqrt{3} \pi }{2}. \end{equation}

Example. Evaluate \begin{equation} \int_0^{\pi }\int_0^4 r^2 \sin ^2\theta \, dr \, d\theta. \end{equation}

Solution. In polar coordinates the region of integration is described as \begin{equation} R=\{(r,\theta) \mid 0\leq \theta \leq \pi, 0\leq r \leq 4\}. \end{equation} Using polar coordinates the iterated integral is evaluated as
\begin{equation} \int_0^{\pi }\int_0^4 r^2 \sin ^2\theta dr \, d\theta =\int_0^{\pi } \frac{64 \sin ^2\theta }{3} \, d\theta =\frac{32 \pi }{3}
\end{equation} by using the trigonometric identity $\sin^2 \theta=\frac{1-\cos (2\theta)}{2}.$

Example. Evaluate \begin{equation} \int_0^{\pi /2}\int_1^3 r^2 \cos ^2\theta \, dr \, d\theta . \end{equation}

Solution. In polar coordinates the region is described as
\begin{equation} R=\{(r,\theta) \mid 0\leq \theta \leq \frac{\pi}{2}, 1\leq r \leq 3\}. \end{equation} Using polar coordinates the iterated integral is evaluated as \begin{equation} \int_0^{\pi /2}\int_1^3 r^2 \cos ^2\theta \, dr \, d\theta =\int_0^{\pi /2} \frac{26 \cos ^2\theta }{3} \, d\theta =\frac{13 \pi }{6}
\end{equation} by using the trigonometric identity $\cos^2 \theta=\frac{1+\cos (2\theta)}{2}.$

Example. Evaluate \begin{equation} \int_0^{\pi }\int_0^{1+\sin \theta } \, dr \, d\theta . \end{equation}

Solution. In polar coordinates the region is described as \begin{equation}
R=\{(r,\theta) \mid 0\leq \theta \leq \pi, 0\leq r \leq 1+\sin \theta\}
\end{equation} Using polar coordinates the iterated integral is evaluated as
\begin{equation} \int_0^{\pi }\int_0^{1+\sin \theta } drd\theta =\int_0^{\pi } (\sin \theta +1) \, d\theta =2+\pi. \end{equation}

Example. Evaluate \begin{equation} \int_{-3}^3\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}}\ln \left(x^2+y^2+9\right) \, dy \, dx. \end{equation}

Solution. In polar coordinates \begin{align} \int_{-3}^3\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}}\ln \left(x^2+y^2+9\right) \, dy \, dx & =\int_0^{2\pi }\int_0^3\ln \left(r^2+9\right) rdrd\theta \\ & =\int_0^{2\pi }\left[-\frac{9}{2} (1+\ln (9)-2 \ln (18))\right]d\theta \\ & = (-9 \pi (1+\ln (9)-2 \ln (18))) \\ & \approx 73.0473. \end{align}

Example. Evaluate \begin{equation} \int_0^2\int_0^{\sqrt{2x-x^2}}\frac{x-y}{x^2+y^2} \, dy \, dx \end{equation}

Solution. In polar coordinates \begin{align} \int_0^2\int_0^{\sqrt{2x-x^2}}\frac{x-y}{x^2+y^2} \, dy \, dx & =\int_0^{\pi /2}\int_0^{2\cos \theta }(\sin \theta -\cos \theta ) \, dr \, d\theta \\ & =\int_0^{\pi /2}2 \cos \theta (\sin \theta -\cos (\theta )) \, d\theta \\ & =\frac{2-\pi }{2}. \\ & \approx -0.570796 \end{align}

Example. Use a double integral to find the area enclosed by one loop of the four leaved rose $r=\cos 2\theta .$

Solution. From the sketch of the curve we see that a loop is given by the region
\begin{equation} R=\left\{(r,\theta )\mid\frac{-\pi }{4}\leq \theta \leq \frac{\pi }{4}\, , 0\leq r\leq \cos 2\theta \right\}. \end{equation} So the area is \begin{equation} \iint_R \, d A=\int_{-\pi /4}^{\pi /4} \int_0^{\cos 2\theta }r dr \, d\theta =\frac{1}{4}\int{-\pi /4}^{\pi /4} (1+\cos 4\theta ) \, d\theta =\frac{\pi }{8}. \end{equation}

Example. Find the volume of the bounded solid region bounded below by the rectangle $R:$ $ 0\leq x\leq 1,$ $ 0\leq y\leq 1$ in the $x y$ plane and above by the graph of $$ z=f(x,y)=\sqrt{x+y}. $$

Solution. The volume is \begin{align} \int_0^1\int_0^1\sqrt{x+y}dy\, dx & =\int_0^1 \left(\frac{2}{3} \left(-x^{3/2}+(1+x)^{3/2}\right)\right) \, dx \\ & =\frac{8}{15} \left(-1+2 \sqrt{2}\right). \end{align}

Example. Find the volume of the bounded solid region below by the rectangle $R:$ $ 1\leq x\leq 2,$ $1\leq y\leq e$ in the $x y-\text{plane}$ and above by the graph of $$ z=f(x,y)=x \ln (x y). $$

Solution. The volume is \begin{align} \int_1^2\int_1^ex \ln (x y)dy\, dx & =\int_1^2 (x (1+(-1+e) \ln x)) \, dx \\ & =-\frac{3}{4} (-3+e)+2(-1+e) \ln 2. \end{align} Note in this example we used the formulas \begin{equation}
\int \ln udu=u \ln u-u+C, u>0 \quad \text{and} \quad \int u \ln udu=\frac{-1}{4}u^2+\frac{1}{2}u^2\ln u. \end{equation}

Example. Find the volume of the bounded solid region bounded by the plane $z=0$ and the paraboloid $z=1-x^2-y^2.$

Solution. If we put $z=0$ in the equation of the paraboloid, we get $x^2+y^2=1.$ This means that the plane intersects the paraboloid in the circle $x^2+y^2=1,$ so the solid lies under the paraboloid and above the circular disk $R$ given by $x^2+y^2\leq 1.$ In polar coordinates the region of integration $R$ is given by $0\leq r\leq 1,$ $0\leq \theta \leq 2\pi .$ Since $1-x^2-y^2=1-r^2,$ the volume is
\begin{equation} V=\iint_R \left(1-x^2-y^2\right)d A=\int_0^{2\pi}\int_0^1\left(1-r^2\right)r \, dr \, d\theta = \frac{\pi}{4} \int_0^{2\pi} \, d\theta = \frac{\pi }{2}. \end{equation}

Example. Find the volume of the bounded solid region under the paraboloid $z=x^2+y^2,$ above the $x y-\text{plane},$ and inside the cylinder $x^2+y^2=2x$

Solution. The solid lies above the disk $R$ whose boundary circle has equation $x^2+y^2=2x$ or $(x-1)^2+y^2=1.$ In polar coordinates the boundary is $r^2=2 r \cos \theta $ or $r=2 \cos \theta .$ Thus the region of integration is
\begin{equation} R=\left\{(r,\theta )\mid -\frac{\pi }{2}\leq \theta \leq \frac{\pi }{2}, \, 0\leq r\leq 2 \cos \theta \right\} \end{equation} and so the volume is \begin{equation} V=\iint_R \left(x^2+y^2\right)d A=\int_{-\pi /2}^{\pi /2}\int_0^{2 \cos \theta }r^2rdrd\theta =4\int_{-\pi /2}^{\pi /2}\cos ^4\theta d\theta =\frac{3\pi }{2}\end{equation} by using the trigonometric identity $\cos^4 \theta=\left(\frac{1+\cos(2\theta)}{2}\right)^2.$

Example. Find the volume of the bounded solid common to the cylinder $x^2+y^2=2$ and the ellipsoid $3x^2+3y^2+z^2=7.$

Solution. In polar coordinates \begin{align} V & =2\int_0^{2\pi }\int_0^{\sqrt{2}}\sqrt{7-3r^2}r \, dr \, d\theta =2\int_0^{2\pi }\left(-\frac{1}{9}+\frac{7 \sqrt{7}}{9}\right)\, d\theta \\ & =\frac{4\pi}{9} \left(-1+7 \sqrt{7}\right). \\ & \approx 24.4629. \end{align}

Example. Find the volume of the solid bounded above by the cone $z=x^2+y^2,$ below by the plane $z=0,$ and on both sides by the cylinder $x^2+y^2=y.$

Solution. In polar coordinates \begin{equation} V=\int_0^{\pi }\int_0^{\sin \theta }r^3 \, dr \, d\theta =\int_0^{\pi }\frac{\sin ^4 \theta }{4} \, d\theta =\frac{3 \pi }{32} \approx 0.294524. \end{equation}

Exercises on Double Integrals in Polar Coordinates

Exercise. Evaluate $\int_0^{\pi }\int_0^{1+\sin \theta } \, dr \, d\theta .$

Exercise. Sketch the region of integration and then evaluate the iterated integral in polar coordinates.

$(1) \quad \displaystyle \int_0^{\pi /2}\int_1^3r e^{-r^2} drd\theta $

$(2) \quad \displaystyle \int_0^{\pi /2}\int_1^2\sqrt{4-r^2} drd\theta$

$(3) \quad \displaystyle \int_0^{\pi }\int_0^4r^2\sin ^2\theta dr \, d\theta $

$(4) \quad \displaystyle \int_0^{\pi /2}\int_1^3r ^2\cos ^2\theta drd\theta $

$(5) \quad \displaystyle \int_0^{\pi}\int_0^{1+\sin \theta } dr \, d\theta $

Exercise. Sketch the enclosed region and then use an iterated integral to find the area of the region.

$(1) \quad r=2 \cos \theta $

$(2) \quad r=4 \cos 3\theta $

$(3) \quad r=1$ and $r=2\sin \theta $

$(4) \quad r=1$ and $r=1+\cos \theta $

Exercise. Use polar coordinates to evaluate the iterated integrals.

$(1) \quad \displaystyle \int_0^1\int_x^{\sqrt{x}}\left(x^2+y^2\right)^{3/2} dy dx$

$(2) \quad \displaystyle \int_0^2\int_0^{\sqrt{2x-x^2}}\frac{x-y}{x^2+y^2} dy dx.$

Exercise. Evaluate the following iterated integrals by converting to polar coordinates.

$(1) \quad \displaystyle \int_0^3\int_0^{\sqrt{9-x^2}}x dy dx$

$(2) \quad \displaystyle \int_0^2\int_0^{\sqrt{4-y^2}}e^{x^2+y^2} dx dy$

$(3) \quad \displaystyle \int_0^3\int_0^{\sqrt{9-x^2}}x dy dx$

$(4) \quad \displaystyle \int_{-3}^3\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}}\ln \left(x^2+y^2+9\right) dy dx$

$(5) \quad \displaystyle \int_0^2\int_y^{\sqrt{8-y^2}}\frac{1}{\sqrt{1+x^2+y^2}}dx dy$

$(6) \quad \displaystyle \int_0^3\int_0^{\sqrt{9-x^2}}\cos \left(x^2+y^2\right) dy dx$

$(7) \quad \displaystyle \int_0^4\int_0^{\sqrt{4y-y^2}}\frac{1}{\sqrt{x^2+y^2}} dx dy$

Exercise. Find the volume of the solid region common to the cylinder $x^2+y^2=2$ and the ellipsoid $3x^2+3y^2+z^2=7.$

Exercise. Find the volume of the ice cream cone bounded by the hemisphere $z=\sqrt{8-x^2-y^2}$ and the cone $z=\sqrt{x^2+y^2}.$

Exercise. Find the volume of the solid region bounded above by the cone $z=x^2+y^2,$ below by the plane $z=0,$ and on both sides by the cylinder $x^2+y^2=y.$

Exercise. Find the volume of the given solid region common to the cylinder $x^2+y^2=2$ and the ellipsoid $3x^2+3y^2+z^2=7.$

Exercise. Find the volume of the given solid region bounded above by the cone $z=x^2+y^2,$ below by the plane $z=0,$ and on both sides by the cylinder $x^2+y^2=y.$

Exercise. For a constant $a$ where $0\leq a\leq R,$ the plane $z=R-a$ cuts off a $\texttt{“}$cap$\texttt{“}$ from the hemisphere $z=\sqrt{R^2-x^2-y^2}.$ Use a double integral in polar coordinates to find the volume of the cap. For what values of $a$ is the volume of the cap half the volume of the hemisphere?