Gate_EE_2006

Question

Answer 1

​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) .

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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}

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​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}
 

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=> A + B = 0

= 0 (Ans)