We have to integrate the Surface S_{xy} = 2x + 5y - 3 , over the circumference of circle
(x+1)^{2} + (y-1)^{2} = (√2)^{2} ------------- (1)
{Thanks to Praveen Sir for pointing this out ,Equation given in question is wrong..}
Centre of circle is (-1, 1) & radius is √2 . => This means x goes from (-1 - √2) to (-1 + √2) .
Now, $\int_{(-1 \ - \ \sqrt2)}^{(-1\ +\ \sqrt2)} ( 2x \ + \ 5y\ - \ 3)\ dx$
At circle circumference , y = $\pm$ √(1 - x^{2} -2x) + 1 {from (1)}
So, we have two possible values of y here , as x goes from (-1 - √2) to (-1 + √2)
putting the value of y in terms of x, we get -
$\int_{(-1 \ - \ \sqrt2)}^{(-1\ +\ \sqrt2)} ( 2x \ + \ 5\sqrt(1 -x^{2} - 2x) \ +5 - \ 3)\ dx$
+ $\int_{(-1 \ - \ \sqrt2)}^{(-1\ +\ \sqrt2)} ( 2x \ - \ 5\sqrt(1 -x^{2} - 2x) \ +5 - \ 3)\ dx$
= A + B {saying first integral as A and 2nd as B}
I am showing how to solve A ,
A = $\left [ x^{2} + (-\frac{5}{2})(\frac{(1 - x^{2} - 2x)^{\frac{3}{2}}}{x + 1}) \ +\ 2x \right ]_{(-1 \ - \ \sqrt2)}^{(-1\ +\ \sqrt2)}$
=> A = 0 {Similarly B comes 0}
=> A + B = 0
= 0 (Ans)