Question

Estimate the volume of the solid that lies below the surface z=xy and above the rectangle R. Use a Riemann sum with m=3, n=2, and take the sample point to be the upper right corner of each square.
R = {(x,y): 0<=x<=6, 0<=y<=4}

Answer 1

For The whole question here is the answer.

Δx = (6 - 0)/3 = 2 and Δy = (12 - 8)/2 = 2.

So, our region R is partitioned as follows:
(0, 12) (2, 12) (4, 12) (6, 12) ,

(0, 10) (2, 10) (4, 10) (6, 10) ,

(0, 8) (2, 8) (4, 8) (6, 8)
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(a) Using upper right endpoints for each of the little 'grids' yields (with f(x,y) = xy)
∫∫R xy dA
≈ 2 * 2 [f(2, 10) + f(4, 10) + f(6, 10) + f(2, 12) + f(4, 12) + f(6, 12)]
= 4 [20 + 40 + 60 + 24 + 48 + 72]
= 1056.

(b) Using the midpoint for each of the little 'grids' yields
∫∫R xy dA
≈ 2 * 2 [f(1, 9) + f(3, 9) + f(5, 9) + f(1, 11) + f(3, 11) + f(5, 11)]
= 4 [9 + 27 + 45 + 11 + 33 + 55]
= 720.

Answer 2

@nathan917 thanks for the reply! Would you care to help me through this step by step?

Answer 3

Firstly, the [\Delta\] x and y are the increments in which our square will be partitioned along the x and y axes?

Answer 4

I'll just answer my own question then, lol.

@nathan917 your answer was actually incorrect, however, it helped me get on course with what is asked.

Okay so the m and n is the intervals.

So:
\[\Delta x=\frac{6-0}{3}=2\]\[\Delta y=\frac{4-0}{2}\]

Thus, the /real/ sample points will be:
(2,2) (2,4) (4,2) (4,4) (6,2) (6,4)


Now
\[M(m,n)=M(3,2)=\Delta x \Delta y \sum_{i=1}^{3} \sum_{j=1}^{2}f(x_{i},y_{j})\]\[=4[f(2,2)+f(2,4)+f(4,2)+f(4,4)+f(6,2)+f(6,4)]=288\]