# Use Polar Co Ordinates to Evaluate β« β« ( X 2 + Y 2 ) 2 X 2 Y 2 ππ ππ Over Yhe Area Common to Circle X 2 + Y 2 = a X and X 2 + Y 2 = B Y , a > B > 0 - Applied Mathematics 2

Sum

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 cosπ½ and y = r sin π½
Circles : r=acos π½ πππ r=asin π½
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≤πππππ½
π≤π½≤πΆ
For another region ,
π≤π≤ππππ π½
πΆ≤π½≤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

Concept: Application of Double Integrals to Compute Area
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