Question

a) Estimate the volume of the solid that lies below the surface z = 7x + 5y2 and above the rectangle R = [0, 2]⨯[0, 4]. Use a Riemann sum with m = n = 2 and choose the sample points to be lower right corners.

Answer 1

In the $x$ direction we consider the $m=2$ subintervals [0, 1] and [1, 2] (each with length 1), while in the $y$ direction we consider the $n=2$ subintervals [0, 2] and [2, 4] (with length 2). Then the lower right corners of the cells in the partition of $R$ are (1, 0), (2, 0), (1, 2), (2, 2).

Let $f(x,y)=7x+5y^2$. The volume of the solid is approximately

$\displaystyle\iint_Rf(x,y)\,\mathrm dx\,\mathrm dy\approx f(1,0)\cdot1\cdot2+f(2,0)\cdot1\cdot2+f(1,2)\cdot1\cdot2+f(2,2)\cdot1\cdot2=\boxed{164}$

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More generally, the lower-right-corner Riemann sum over $m=\mu$ and $n=\nu$ subintervals would be

$\displaystyle\sum_{m=1}^\mu\sum_{n=1}^\nu\left(7\frac{2m}\mu+5\left(\frac{4n-4}\nu\right)^2\right)\frac{2-0}\mu\frac{4-0}\nu=\frac83\left(101+\frac{21}\mu+\frac{40}{\nu^2}-\frac{120}\nu\right)$

Then taking the limits as $\mu\to\infty$ and $\nu\to\infty$ leaves us with an exact volume of $\dfrac{808}3$.