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)