Numerical Methods: Simpsons 1/3 rule for double integral PLEASE DO NOT TAKE THIS

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Numerical Methods: Simpsons 1/3 rule for double integral

PLEASE DO NOT TAKE THIS QUESTION IF YOU ARE NOT KNOWLEDGABLE ON THE SUBJECT. I am trying to learn, not get free answers, and would greatly appreacite a thorough explanation of the below code.

Question: We are learning Simpons 1/3 rule for double integrals and our professor gives us some code to look at in Fortan that will aid us in later assignments. I have no idea what he is doing in the last do loop where he says "First integrate w.r.t to x". Can someone please explain this section of code.

subroutine SimpsonDIntRuleForFunc0(a,b,m,c,d,n)
   IMPLICIT NONE
   INTEGER :: i,n,m,j
   REAL :: x(m+1), y(n+1), INTa, a, b, c, d, h, k, fy0, fy1, fy2,SimpFXY

   h= (b-a)/m
   k= (d-c)/n

   x(1) = a
   DO i= 1, m
       x(i+1) = x(i)+h
   ENDDO

      print*, "x values, m = ",m
      do i = 1, m
        print*,(x(i))
        end do

   y(1) = c
   DO i= 1, n
       y(i+1) = y(i)+k
   ENDDO

        print*, "y values n= ",n
      do i = 1, n
        print*,(y(i))
        end do

   INTa = 0
   Do i= 1,n-1,2
  
       ! First integrate the function w.r.t. x using Simpson's 1/3 rule.
       fy0 = 0
       fy1 = 0
       fy2 = 0
       DO j = 1,m-1,2
             fy0 = fy0 + SimpFXY(x(j),y(i)) + 4*SimpFXY(x(j+1),y(i)) + SimpFXY(x(j+2),y(i))
             fy1 = fy1 + SimpFXY(x(j),y(i+1)) + 4*SimpFXY(x(j+1),y(i+1)) + SimpFXY(x(j+2),y(i+1))
             fy2 = fy2 + SimpFXY(x(j),y(i+2)) + 4*SimpFXY(x(j+1),y(i+2)) + SimpFXY(x(j+2),y(i+2))
       ENDDO

       ! Integration step w.r.t. y using Simpson's 1/3 rule.
       INTa = INTa + fy0 + 4*fy1 + fy2
   ENDDO

   INTa = INTa*h*k/9

   print*, 'The value of the integral is: ', INTa

END subroutine SimpsonDIntRuleForFunc0

FUNCTION SimpFXY(x,y)
   IMPLICIT NONE
   real::x,y,SimpFXY,e
    e = 2.7182818284590452
SimpFXY = 8 * e**(-x**2 - y**4)

END FUNCTION SimpFXY

Explanation / Answer

IMPLICIT NONE
   INTEGER :: i,n,m,j
   REAL :: x(m+1), y(n+1), INTa, a, b, c, d, h, k, fy0, fy1, fy2,SimpFXY

   h= (b-a)/m
   k= (d-c)/n

   x(1) = a
   DO i= 1, m
       x(i+1) = x(i)+h
   ENDDO

      print*, "x values, m = ",m
      do i = 1, m
        print*,(x(i))
        end do

   y(1) = c
   DO i= 1, n
       y(i+1) = y(i)+k
   ENDDO

        print*, "y values n= ",n
      do i = 1, n
        print*,(y(i))
        end do

   INTa = 0
   Do i= 1,n-1,2
  
       ! First integrate the function w.r.t. x using Simpson's 1/3 rule.
       fy0 = 0
       fy1 = 0
       fy2 = 0
       DO j = 1,m-1,2
             fy0 = fy0 + SimpFXY(x(j),y(i)) + 4*SimpFXY(x(j+1),y(i)) + SimpFXY(x(j+2),y(i))
             fy1 = fy1 + SimpFXY(x(j),y(i+1)) + 4*SimpFXY(x(j+1),y(i+1)) + SimpFXY(x(j+2),y(i+1))
             fy2 = fy2 + SimpFXY(x(j),y(i+2)) + 4*SimpFXY(x(j+1),y(i+2)) + SimpFXY(x(j+2),y(i+2))
       ENDDO

       ! Integration step w.r.t. y using Simpson's 1/3 rule.
       INTa = INTa + fy0 + 4*fy1 + fy2
   ENDDO

   INTa = INTa*h*k/9

   print*, 'The value of the integral is: ', INTa

END subroutine SimpsonDIntRuleForFunc0

FUNCTION SimpFXY(x,y)
   IMPLICIT NONE
   real::x,y,SimpFXY,e
    e = 2.7182818284590452
SimpFXY = 8 * e**(-x**2 - y**4)

END FUNCTION SimpFXY

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