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^2`circle 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`