Estimate the volume of the solid that lies below the surface z = xy and above the rectangle R = {(x, y)| 0 <= x <= 6, 0 <= y <= 4}

Solution:

We have to find the volume of the solid that lies below the surface z = xy and above the rectangle R.

a) Using upper right endpoints for each of the little ‘grids’ yields (with f(x, y) = xy)

\(\int \int R\, xy\, dA\)

≈ 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]

= 4[264]

= 1056

b) Using the midpoint for each of the little ‘grids’ yields

\(\int \int R\, xy\, dA\)

≈ 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]

= 4[180]

= 720

Therefore, the volume of the solid is 1056 and 720 cubic units.

---

Estimate the volume of the solid that lies below the surface z = xy and above the rectangle R = {(x, y)| 0 <= x <= 6, 0 <= y <= 4}

Summary:

The volume of the solid that lies below the surface z = xy and above the rectangle R. Using a Riemann sum with m = 3, n = 2, and the sample point to be the upper right corner of each square. R = {(x, y)| 0 <= x <= 6, 0 <= y <= 4} is 1056 and 720 cubic units.