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+\frac{i(1-0)}2=\frac 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+\frac{j(1-0)}3=\frac 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)\frac{1-0}m\frac{1-0}n=\frac16\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)\frac{1-0}m\frac{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+\frac{(i-1)(2-(-2))}8=\frac{i-5}2`

`r_i=-2+\frac{i(2-(-2))}8=\frac{i-4}2`

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

`\ell_j=0+\frac{(j-1)(2-0)}2=j-1`

`r_j=0+\frac{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.)