GATE CSE
First time here? Checkout the FAQ!
x
+3 votes
136 views

s

asked in Calculus by (51 points)   | 136 views

1 Answer

+1 vote
Best answer

​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 smiley ​,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)
answered by Veteran (16k points)  
selected by


Top Users Jul 2017
  1. Bikram

    3782 Points

  2. manu00x

    2464 Points

  3. Debashish Deka

    1832 Points

  4. joshi_nitish

    1494 Points

  5. Arnab Bhadra

    1096 Points

  6. Arjun

    1054 Points

  7. Hemant Parihar

    1050 Points

  8. Shubhanshu

    972 Points

  9. Ahwan

    876 Points

  10. akash.dinkar12

    642 Points


23,953 questions
30,895 answers
70,108 comments
29,272 users