Home page for accesible maths 4 Double integrals: general regions and change of variable 4.2 Double integrals over triangular and more general regions 4.4 Change of variable for double integrals: the general case

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4.3 Double integrals using polar coordinates

Polar coordinates are useful for describing circles, and regions involving circles. The circle with centre

O

and radius

a

is given by:

r=a,\;0\leq\theta\leq 2\pi

. The interior of this circle (the disc of radius

a

) is described by:

0\leq r\leq a,\quad 0\leq\theta\leq 2\pi.

The upper half of this disc (the part with

y\geq 0

) is given by

The region described in rectangular coordinates by

x\geq 0,\quad y\geq 0,\quad a^{2}\leq x^{2}+y^{2}\leq b^{2}

is given by

The circle with centre $(a,0)$ and radius $a$ passes through $O$ . As the diagram shows, it is given by $r=2a\cos\theta$ and $-\pi/2\leq\theta\leq\pi/2$ . Its interior is given by: $0\leq r\leq 2a\cos\theta$ , for the same range of $\theta$ .

Recall that the length of the arc of a circle of radius $r$ between two radii at angle $\theta$ is $r\theta$ (see the examples on path length in section 2).

The area of the small region determined by $r$ , $r+\delta r$ and $\theta,\theta+\delta\theta$ is approximately $r\>\delta r\>\delta\theta$ . The contribution of this small region to the volume below $z=f(x,y)$ is approximately this area times the value of the function (which gives the height of the surface), that is, $r\>\delta r\>\delta\theta\>f(r\cos\theta,r\sin\theta)$ .

We combine small regions and pass to the limit. The conclusion is:

\int\!\int_{A}f(x,y)\>dx\>dy=\int\!\int_{A^{\prime}}f(r\cos\theta,r\sin\theta)% \>r\>dr\>d\theta,

where $A^{\prime}$ is the same region as $A$ , expressed in terms of $r$ and $\theta$ .

Example 4.1

The area of a disc. Let $A$ be the disc of radius $a$ . It is given by $0\leq r\leq a$ , $0\leq\theta\leq 2\pi$ . So its area is

$\displaystyle{\int\!\int_{A}1\>dx\>dy=}$

Example 4.2

Find $I=\displaystyle{\int\!\int_{A}y\>dx\>dy}$ , where $A$ is the half-disc given by $x^{2}+y^{2}\leq a^{2},\;y\geq 0$ .

Example 4.3

Find $I=\displaystyle{\int\!\int_{A}xy\>dx\>dy}$ , where $A$ is the region given by $a^{2}\leq x^{2}+y^{2}\leq b^{2}$ , $x\geq 0,\;y\geq 0$ .

In polar coordinates, the region is given by:

Example 4.4

Find $I=\displaystyle{\int\!\int_{D}(x^{2}+y^{2})^{-1/2}\>dx\>dy}$ , where $D$ is the disc with centre $(a,0)$ and radius $a$ .

As seen above, the disc is given by $0\leq r\leq 2a\cos\theta$ , $-\pi/2\leq\theta\leq\pi/2$ . Also, $(x^{2}+y^{2})^{-1/2}=1/r$ . So

The “probability integral”

The following famous integral (of a function of one variable) is sometimes called the probability integral. It is important in statistics. It can be evaluated by the ingenious method of considering a double integral that equals $I^{2}$ and transforming to polar coordinates.

Proposition 4.5 We have $\displaystyle{\int_{-\infty}^{\infty}e^{-x^{2}}\>dx=\sqrt{\pi}}$ .

Proof. Denote the integral by $I$ . Then

\int\!\int_{\mathbb{R}^{2}}e^{-x^{2}-y^{2}}\>dx\>dy=\int_{-\infty}^{\infty}e^{% -y^{2}}\left(\int_{-\infty}^{\infty}e^{-x^{2}}\>dx\right)\;dy=\int_{-\infty}^{% \infty}I\>e^{-y^{2}}\;dy=I^{2}.

The plane $\mathbb{R}^{2}$ is given by: $0\leq r<\infty,\;0\leq\theta\leq 2\pi$ . So

I^{2}=\int_{0}^{2\pi}\int_{0}^{\infty}e^{-r^{2}}\>r\>dr\>d\theta.

Now

$\displaystyle{\int_{0}^{R}r\,e^{-r^{2}}\>dr=}$

so $\int_{0}^{\infty}r\,e^{-r^{2}}\>dr=\frac{1}{2}$ . Hence

I^{2}=\int_{0}^{2\pi}\mbox{$\frac{1}{2}$}\>d\theta=\pi,

so $I=\sqrt{\pi}$ .