Wize University Calculus 3 Textbook > Multiple Integrals

Double Integrals in Polar Coordinates

The polar coordinate system is a substitute to the Cartesian xy-coordinate system. In this system, rather than specifying the coordinate of a point with x and y coordinates in 2D space, we can use the radius and the angle to define the coordinate of a point uniquely. The radius and angle with respect to the positive x-axis are shown below:

Image illustrating Double Integrals in Polar Coordinates

The following equations relates the polar coordinates system with Cartesian coordinates system:

\begin{array}{c} x=r\cos\theta\\[10pt] y=r\sin\theta\\[10pt] x^2+y^2=(r\cos\theta)^2+(r\sin\theta)^2= r^2 \end{array}

The general form of a double integral from rectangular coordinates to polar coordinates is in the form of:

\iint_Df(x,y)dA=\int_\alpha^\beta\int_a^bf(r\cos\theta,r\sin\theta)rdrd\theta

where r changes from a to b and $\theta$ changes from $\alpha$ to $\beta$ .

Evaluate the double integral

I=\int_{ }^{ }\int_D^{ }\sin(x^2+y^2)dA

where $D$ is a region inside a circle with a radius of 3 in the first quadrant.

Image illustrating Example: Double Integrals in Polar Coordinates (1)

\displaystyle I=\iint_{D}\sin(x^2+y^2)dA\newline{}\newline{} ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ =\int_0^{\frac{\pi}{2}}\int_0^3\sin(r^2)rdrd\theta~~~~{\scriptsize\color{magenta}~~~Let~u=r^2.~Then,~\frac{du}{2}=rdr}\newline{}\newline{} ~~~~~~~~~ =\frac{1}{2}\int_0^{\frac{\pi}{2}}\int_0^3\Big(\sin{u}~du\Big)d\theta\newline{}\newline{} ~~~~~~~~ =\frac{1}{2}\int_0^{\frac{\pi}{2}}d\theta~\cdot~(-\cos{u})\Big|_0^3\newline{}\newline{} ~~~~~~~~ =\frac{1}{2}\int_0^{\frac{\pi}{2}}d\theta~\cdot~(-\cos{r^2})\Big|_0^3\newline{}\newline{} ~~~~~~~~~~~~~ =\frac{1}{2}\int_0^{\frac{\pi}{2}}d\theta~\cdot~\Big(-\cos{9}+1\Big)\newline{}\newline{} ~~ =\frac{1}{2}(1-\cos9)\int_0^{\frac{\pi}{2}}d\theta\newline{}\newline{} =\frac{1}{2}(1-\cos9)\Big|_0^{\frac{\pi}{2}}~~~~~\newline{}\newline{} =\frac{\pi}{4}(1-\cos9)~~~~~~~

Evaluate $\displaystyle\int_D\int{}2xy~dA,~$ where D is the portion of the region between the circles of radius 2 and radius 5 centered at the origin that lies in the first quadrant.

We can determine the domain, D, in terms of polar coordinates. So,

2\leq{r}\leq{5},~~0\leq{\theta}\leq{\frac{\pi}{2}}

Then,

\displaystyle\int_D\int2xy~dA=\int_{0}^{\frac{\pi}{2}}\int_{2}^{5}2(rcos\theta)(rsin\theta)rdrd\theta\newline{}\newline{} ~~~~~~~~~ =\int_{0}^{\frac{\pi}{2}}\int_{2}^{5}r^3\sin2\theta~drd\theta\newline{}\newline{} ~~~~~~~~~~~~~~ =\int_{0}^{\frac{\pi}{2}}\sin2\theta~d\theta~\cdot~\Bigg(\frac{r^4}{4}\Bigg)\Bigg|_2^5\newline{}\newline{} ~~~~~~~~~~~~~~~~~~~~~~ =\int_{0}^{\frac{\pi}{2}}\sin2\theta~d\theta~\cdot~\Bigg(\frac{625}{4}~-~4\Bigg)\newline{}\newline{} ~~ =\frac{-609}{8}~\cdot~\big(cos2\theta\big)\bigg|_0^{\frac{\pi}{2}}\newline{}\newline{} ~~~~~~~~~~~~~~~~~~~~~~~ =\frac{-609}{8}\Bigg[\cos\bigg({2\frac{\pi}{2}}\bigg)~-~\cos(0) \Bigg]\newline{}\newline{} =\frac{609}{4}~~~~~~~~~~~~~~~~~~~

What is the volume of the region under the sphere $x^2+y^2+z^2=9,~$ above the plane $z=0,~$ and inside $~x^2+y^2=5?$

Your answer

I don't know

Find $\displaystyle\int_{D}^{}\int_{}^{}\sqrt{1+4x^2+4y^2}dA~$ where D is the lower half of $x^2+y^2=16$

Your answer

I don't know

Double Integrals in Polar Coordinates

Create a free account to get started

Create a free account to get started

Create a free account to get started

Create a free account to get started

Create a free account to get started