Use polar co ordinates to evaluate `int int (x^2+y^2)^2/(x^2y^2)` π π π π over yhe area Common to circle `x^2+y^2=ax "and" x^2+y^2=by, a>b>0`

#### Solution

Let I = `int int (x^2+y^2)^2/(x^2y^2)` π
π π
π

Region of integration is : Area common to the circle

`x^2+y^2=ax "and" x^2+y^2=by

To change the Cartesian coordinates to polar coordinates

Put x= r cosπ½ and y = r sin π½

Circles : r=acos π½ πππ
r=asin π½

The function becomes : f(x,y) `= (x^2+y^2)^2/(x^2y^2)=r^4/(r^4sin^2thetacos^2theta)=4/(sin^2 2theta)=f(r,theta)`

Intersection of both circles is at angle = `tan^(-1) a/b`

Divide the region into two equal halves.

For one region ,

π≤π≤πππππ½

π≤π½≤πΆ

For another region ,

π≤π≤ππππ π½

πΆ≤π½≤`pi/2`

`therefore "I"=int_0^\alphaint_0^(bsintheta)(4rdrd theta)/(sin^2 2theta) + int_0^(acostheta) int_alpha^(pi/2)(4rdrd theta)/(sin^2 2theta)`

`therefore "I"=int_0^alpha 4/(sin^2 2theta)[r^2/2]_0^(bsin theta)d theta+int_0^(pi/2)4/(sin^2 2theta)[r^2/2]_0^(acostheta)d theta`

`=1/2b^2int_0^alphasec^2theta d theta+a^2/2int_alpha^(pi/2)cosec^2theta d theta`

`=1/2b^2tanalpha+a^2/2cotalpha`

`=(ab)/2+(ab)/2`

∴ I = ab