Question

Estimate the volume of the solid that lies below the surface z = ex+y and above the rectangle

Answer 1

1. The volume under the surface $f(x,y)=e^{x+y}$ is given by the double integral,

$\displaystyle\int_0^1\int_0^1e^{x+y}\,\mathrm dx\,\mathrm dy$

We split up the integration region into a 2x3 grid of rectangles whose upper right corners are determined by the right endpoints of the partition along either axis. That is, we split up the $x$ interval [0, 1] into 2 subintervals,

[0, 1/2], [1/2, 1]

with right endpoints given by the arithmetic sequence,

$r_i=0+\dfrac{i(1-0)}2=\dfrac i2$

for $i\in\{1,2\}$, and the $y$ interval [0, 1] into 3 subintervals,

[0, 1/3], [1/3, 2/3], [2/3, 1]

with right endpoints

$r_j=0+\dfrac{j(1-0)}3=\dfrac j3$

for $j\in\{1,2,3\}$.

Then the upper right corners of the 6 rectangles are the points

(1/2, 1/3), (1/2, 2/3), (1/2, 1), (1, 1/3), (1, 2/3), (1, 1)

generated by the sequence $(r_i,r_j)$.

The integral is thus approximated by the sum

$\displaystyle\sum_{j=1}^3\sum_{i=1}^2f(r_i,r_j)\dfrac{1-0}m\dfrac{1-0}n=\dfrac16\sum_{j=1}^3\sum_{i=1}^2f(r_i,r_j)=\frac{e^{5/6}+e^{7/6}+e^{4/3}+e^{5/3}}6$

or approximately 2.4334. (Compare to the actual value of the integral, which is close to 2.952.)

For the midpoint rule estimate, we replace the sampling points $(r_i,r_j)$ with $(m_i,m_j)$, i.e. the midpoints of each subinterval, so the set of sampling points is

(1/4, 1/6), (3/4, 1/6), (1/4, 1/2), (3/4, 1/2), (1/4, 5/6), (3/4, 5/6)

and the integral is approximately

$\displaystyle\sum_{j=1}^3\sum_{i=1}^2f(m_i,m_j)\dfrac{1-0}m\dfrac{1-0}n=\frac{e^{5/12}+e^{3/4}+e^{11/12}+e^{13/12}+e^{5/4}+e^{19/12}}6$

or about 2.908.

2. We approach the second integral the same way. Split up the $x$ interval into 8 subintervals with left and right endpoints given respectively by

$\ell_i=-2+\dfrac{(i-1)(2-(-2))}8=\dfrac{i-5}2$

$r_i=-2+\dfrac{i(2-(-2))}8=\dfrac{i-4}2$

for $i\in\{1,2,\ldots,8\}$, and the $y$ interval into 2 subintervals with

$\ell_j=0+\dfrac{(j-1)(2-0)}2=j-1$

$r_j=0+\dfrac{j(2-0)}2=j$

for $j\in\{1,2\}$.

The upper left corners of the rectangles in this grid are given by the sequence $(\ell_i,r_j)$. So the integral is approximately

$\displaystyle\sum_{j=1}^2\sum_{i=1}^8f(\ell_i,r_j)\frac{2-(-2)}m\frac{2-0}n=51$

(Compare to the actual value, 32.)