Change the order of integration and hence show that ∬dx dy/[ ( 1+e^y) √(1-x^2-y^2)] = (𝝿 /2) log(2e/(1+e) where x-0 to x=1 and y=0 to y=√(1-x^2) DOUBLE INTEGRALS #duoble #integral #sum

PROBLEM : Change the order of integration and hence show that ∬dx dy/[ ( 1+e^y) √(1-x^2-y^2)]   = (𝝿 /2)   log(2e/(1+e) where x-0 to x=1 and y=0 to y=√(1-x^2)

SOLUTION :

 

*   Evaluate ∬ f(x,y) dxdy where f(x,y)= (2y-1)/ x+1 , and E is the region bounded by x=0,y=0, y=2x-        4.

 *   In the integral ∬ (4-y) dydx, change the order of integration and evaluate the integral where x=2 to            x=4 and y=4/x to y=(20-4x)/(8-x)

 *  Sketch the region of integration and evaluate ∬ xsiny dydx where x=0 to x=𝛑 and y-0 to y=x.

 *  Sketch the region of integration and write an equivalent double integral with the order of integration

     reversed for ∬ 3y dxdy where y=0 to y=1 and x= -√(1-y^2) to y=√(1-y ^2).

 *Evaluate ∬ √(4x^2-y^2) dxdy where E is the region bounded by the lines y=0,y=x and            x=1.

 *  Evaluate  ∬f(x,y)dxdy , where f(x,y)=x²+y² and E={(x,y)/y=x²,x=2,y=1}   

 *  Evaluate ∬ xydxdy where E is the region bounded by xy=1,y=0,y=x,x=2.