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asked in Calculus | 149 views

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​We have to integrate the Surface Sxy = 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 - x2 -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)