# Evaluate ∫ a √ 2 0 ∫ √ a 2 − Y 2 Y Log ( X 2 + Y 2 ) Dxdy by Changing to Polar Coordinates . - Applied Mathematics 2

Evaluate int_0^(a/sqrt2) int_y^(sqrt(a2-y^2)) log (x^2+y^2) "dxdy by changing to polar Coordinates".

#### Solution

let I =int_0^(a/sqrt2) int_y^(sqrt(a2-y^2)) log (x^2+y^2)

Region of integration :  y<=x<= sqrt(a^2-y^2)

0<= y <= a/sqrt2

The line x=y is inclined at 45° to the +ve x-axis.

Curves : (i) x=y, y=0, y=a/sqrt2

(ii) x=sqrt(a^2-y^2)

x^2+y^2=a^2circle with centre (0,0) and radius a.

Cartesian coordinates → Polar coordinates

(x,y) →  (r,𝜽)

Put x = r cos 𝜽 and y = r sin 𝜽

f(x,y)=log(x^2+y^2) = log r^2=2 log r=f(r,θ)

Limits changes to :    0<= r <= a

0 <= θ <= pi/4

∴ I= int_0^(pi/4) int_0^a 2log r.r dr dθ

= 2 int_0^(pi/4)[log r r^2/2 - r^2/4] _0^a dθ

=2 int_0^(pi/4) [log a a^2/2 - a^2/4]dθ

∴ I= [log a.a^2/2 - a^2/4] xx pi/4

Concept: Linear Differential Equation with Constant Coefficient‐ Complementary Function
Is there an error in this question or solution?