Math Problem Statement

Find the Riemann sum approximation of the double integral

D (6x − 9y) dA where D is the rectangle D = [0, 4] ✕ [1, 5]. Use a regular partition by dividing D into 16 equal size subrectangles and taking sample points at the center of each subrectangle.

Solution

To compute the Riemann sum approximation of the double integral

D(6x9y)dA\int \int_D (6x - 9y) \, dA

where D=[0,4]×[1,5]D = [0, 4] \times [1, 5], we'll follow these steps:

Step 1: Partitioning the Region DD

The region DD is a rectangle with dimensions [0,4][0, 4] in the xx-direction and [1,5][1, 5] in the yy-direction. We are asked to divide this rectangle into 16 subrectangles, which means we will have 4 subdivisions in both the xx- and yy-directions.

So:

  • Δx=404=1\Delta x = \frac{4 - 0}{4} = 1
  • Δy=514=1\Delta y = \frac{5 - 1}{4} = 1

Each subrectangle has dimensions 1×11 \times 1.

Step 2: Sample Points at the Center of Each Subrectangle

The sample points for the Riemann sum are taken at the center of each subrectangle. The center of each subrectangle is:

(xi,yj)=(xi+Δx2,yj+Δy2)\left( x_i, y_j \right) = \left( x_i + \frac{\Delta x}{2}, y_j + \frac{\Delta y}{2} \right)

The coordinates of the centers of the 16 subrectangles are:

(x,y)=(0.5,1.5),(0.5,2.5),(0.5,3.5),(0.5,4.5)(x, y) = (0.5, 1.5), (0.5, 2.5), (0.5, 3.5), (0.5, 4.5) (x,y)=(1.5,1.5),(1.5,2.5),(1.5,3.5),(1.5,4.5)(x, y) = (1.5, 1.5), (1.5, 2.5), (1.5, 3.5), (1.5, 4.5) (x,y)=(2.5,1.5),(2.5,2.5),(2.5,3.5),(2.5,4.5)(x, y) = (2.5, 1.5), (2.5, 2.5), (2.5, 3.5), (2.5, 4.5) (x,y)=(3.5,1.5),(3.5,2.5),(3.5,3.5),(3.5,4.5)(x, y) = (3.5, 1.5), (3.5, 2.5), (3.5, 3.5), (3.5, 4.5)

Step 3: Evaluate the Function at Each Sample Point

The function we're integrating is f(x,y)=6x9yf(x, y) = 6x - 9y. Now, let's evaluate this function at each of the center points:

  1. f(0.5,1.5)=6(0.5)9(1.5)=313.5=10.5f(0.5, 1.5) = 6(0.5) - 9(1.5) = 3 - 13.5 = -10.5
  2. f(0.5,2.5)=6(0.5)9(2.5)=322.5=19.5f(0.5, 2.5) = 6(0.5) - 9(2.5) = 3 - 22.5 = -19.5
  3. f(0.5,3.5)=6(0.5)9(3.5)=331.5=28.5f(0.5, 3.5) = 6(0.5) - 9(3.5) = 3 - 31.5 = -28.5
  4. f(0.5,4.5)=6(0.5)9(4.5)=340.5=37.5f(0.5, 4.5) = 6(0.5) - 9(4.5) = 3 - 40.5 = -37.5
  5. f(1.5,1.5)=6(1.5)9(1.5)=913.5=4.5f(1.5, 1.5) = 6(1.5) - 9(1.5) = 9 - 13.5 = -4.5
  6. f(1.5,2.5)=6(1.5)9(2.5)=922.5=13.5f(1.5, 2.5) = 6(1.5) - 9(2.5) = 9 - 22.5 = -13.5
  7. f(1.5,3.5)=6(1.5)9(3.5)=931.5=22.5f(1.5, 3.5) = 6(1.5) - 9(3.5) = 9 - 31.5 = -22.5
  8. f(1.5,4.5)=6(1.5)9(4.5)=940.5=31.5f(1.5, 4.5) = 6(1.5) - 9(4.5) = 9 - 40.5 = -31.5
  9. f(2.5,1.5)=6(2.5)9(1.5)=1513.5=1.5f(2.5, 1.5) = 6(2.5) - 9(1.5) = 15 - 13.5 = 1.5
  10. f(2.5,2.5)=6(2.5)9(2.5)=1522.5=7.5f(2.5, 2.5) = 6(2.5) - 9(2.5) = 15 - 22.5 = -7.5
  11. f(2.5,3.5)=6(2.5)9(3.5)=1531.5=16.5f(2.5, 3.5) = 6(2.5) - 9(3.5) = 15 - 31.5 = -16.5
  12. f(2.5,4.5)=6(2.5)9(4.5)=1540.5=25.5f(2.5, 4.5) = 6(2.5) - 9(4.5) = 15 - 40.5 = -25.5
  13. f(3.5,1.5)=6(3.5)9(1.5)=2113.5=7.5f(3.5, 1.5) = 6(3.5) - 9(1.5) = 21 - 13.5 = 7.5
  14. f(3.5,2.5)=6(3.5)9(2.5)=2122.5=1.5f(3.5, 2.5) = 6(3.5) - 9(2.5) = 21 - 22.5 = -1.5
  15. f(3.5,3.5)=6(3.5)9(3.5)=2131.5=10.5f(3.5, 3.5) = 6(3.5) - 9(3.5) = 21 - 31.5 = -10.5
  16. f(3.5,4.5)=6(3.5)9(4.5)=2140.5=19.5f(3.5, 4.5) = 6(3.5) - 9(4.5) = 21 - 40.5 = -19.5

Step 4: Multiply by the Area of Each Subrectangle

The area of each subrectangle is ΔA=Δx×Δy=1×1=1\Delta A = \Delta x \times \Delta y = 1 \times 1 = 1.

Step 5: Calculate the Riemann Sum

The Riemann sum is the sum of the function values at the center of each subrectangle, multiplied by the area ΔA\Delta A. Since ΔA=1\Delta A = 1, we just sum the function values:

Riemann sum=(10.5)+(19.5)+(28.5)+(37.5)+(4.5)+(13.5)+(22.5)+(31.5)\text{Riemann sum} = (-10.5) + (-19.5) + (-28.5) + (-37.5) + (-4.5) + (-13.5) + (-22.5) + (-31.5) +1.5+(7.5)+(16.5)+(25.5)+7.5+(1.5)+(10.5)+(19.5)+ 1.5 + (-7.5) + (-16.5) + (-25.5) + 7.5 + (-1.5) + (-10.5) + (-19.5)

Summing these values:

Riemann sum=240\text{Riemann sum} = -240

Final Answer:

The Riemann sum approximation of the double integral is 240-240.


Would you like further details or have any questions?

Here are 5 related questions:

  1. How does a Riemann sum work for single integrals?
  2. What is the effect of increasing the number of subrectangles in the Riemann sum?
  3. How does the choice of sample points (center, corners, etc.) affect the Riemann sum?
  4. What is the geometric interpretation of a double integral over a rectangular region?
  5. Can we extend this method to non-rectangular regions?

Tip: Increasing the number of subrectangles (i.e., using a finer partition) generally leads to a more accurate approximation of the integral.

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Math Problem Analysis

Mathematical Concepts

Riemann Sum
Double Integral
Partitioning
Multivariable Calculus

Formulas

Riemann Sum for Double Integral: ∑ f(x_i, y_j) ΔA
Δx = (b - a) / n
Δy = (d - c) / m

Theorems

Riemann Sum Approximation Theorem
Double Integration over Rectangular Regions

Suitable Grade Level

College Level Calculus (Calculus III or Multivariable Calculus)