MULTIPLE INTEGRALS

Evaluate the double integral

I=

∫∫

R

(

a+

√

x^2 +y^2

)

dx dy,

whereRis the region bounded by the circlex^2 +y^2 =a^2.

In Cartesian coordinates, the integral may be written

I=

∫a

−a

dx

∫√a (^2) −x 2
−
√
a^2 −x^2
dy

(

a+

√

x^2 +y^2

)

,

and can be calculated directly. However, because of the circular boundary of the integration
region, a change of variables to plane polar coordinatesρ,φis indicated. The relationship
between Cartesian and plane polar coordinates is given byx=ρcosφandy=ρsinφ.
Using (6.12) we can therefore write

I=

∫∫

R′

(a+ρ)

∣

∣

∣∣∂(x, y)

∂(ρ, φ)

∣

∣

∣∣dρ dφ,

whereR′is the rectangular region in theρφ-plane whose sides areρ=0,ρ=a,φ=0
andφ=2π. The Jacobian is easily calculated, and we obtain

J=

∂(x, y)

∂(ρ, φ)

=

∣∣

∣

∣

cosφ sinφ

−ρsinφρcosφ

∣∣

∣

∣=ρ(cos

(^2) φ+sin (^2) φ)=ρ.
So the relationship between the area elements in Cartesian and in plane polar coordinates is
dx dy=ρdρdφ.
Therefore, when expressed in plane polar coordinates, the integral is given by
I=

∫∫

R′

(a+ρ)ρdρdφ

=

∫ 2 π

0

dφ

∫a

0

dρ(a+ρ)ρ=2π

[

aρ^2

2

+

ρ^3

3

]a

0

=

5 πa^3

3

.

6.4.2 Evaluation of the integralI=

∫∞

−∞e

−x^2 dx

By making a judicious change of variables, it is sometimes possible to evaluate

an integral that would be intractable otherwise. An important example of this

method is provided by the evaluation of the integral

I=

∫∞

−∞

e−x

2

dx.

Its value may be found by first constructingI^2 , as follows:

I^2 =

∫∞

−∞

e−x

2

dx

∫∞

−∞

e−y

2

dy=

∫∞

−∞

dx

∫∞

−∞

dy e−(x

(^2) +y (^2) )

∫∫
R
e−(x
(^2) +y (^2) )
dx dy,