devotion & mathematics

  PROBLEM   :     Evaluate ∬xy(x+y) dxdy where E is the region bounded by y=x^2 and y=x. 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.  *    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)  *    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 ...

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

CONTENTS   : 1. Definition of a function on a bounded region. 2. Definition of integration over bounded region. 3. few problems on integration over some non rectangular bounded regions 1. DEFINITION :                              Let E be a bounded region in R^2. Let R be the rectangular region enclosing E.                             We define a function F(x,y) over R as follows :                             F(x,y) = f(x,y) ,∀ x in E and F(x,y) = 0 ∀ x not in E. 2. DEFINITION :                           A function f(x,y) over a bounded region E is said to be integrable if F(x,y) is                           integrable over R....

 PROBLEM : 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.  SOLUTION:    *    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)  *    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= -...

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. ...

 PROBLEM :    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) 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.  *    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)   *   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...

  *   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.  *    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)  *    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 li...

 PROBLEM :  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). 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.  *    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)  *    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)  *...

PROBLEM : Sketch the region of integration and evaluate ∬ xsiny dydx where x=0                            to x=𝛑 and y-0 to y=x 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.  *    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)  *    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 write an equivalent double integral with the order of integration      reversed for ∬ 3y dxdy where y=0 to y=...