Question

In the Midpoint Rule for triple integrals we use a triple Riemann sum to approximate a triple integral over a box B, where f(x, y, z) is evaluated at the center (xi, yj, zk) of the box Bijk. Use the Midpoint Rule to estimate the value of the integral. Divide B into eight sub-boxes of equal size. (Round your answer to three decimal places.)triple integrals of cos(xyz) dV,where B = {(x, y, z) | 0 ≤ x ≤ 2, 0 ≤ y ≤ 2, 0 ≤ z ≤ 2}

Answer 1

To approximate the volume with 8 boxes, we have to split up the interval of integration for each variable into 2 subintervals, [0, 1] and [1, 2]. Each box will have midpoint $m_{i,j,k}$ that is one of all the possible 3-tuples with coordinates either 1/2 or 3/2. That is, we're sampling $f(x,y,z)=\cos(xyz)$ at the 8 points,

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

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

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

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

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

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

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

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

which are captured by the sequence

$m_{i,j,k}=\left(\dfrac{2i-1}2,\dfrac{2j-1}2,\dfrac{2k-1}2\right)$

with each of $i,j,k$ being either 1 or 2.

Then the integral of $f(x,y,z)$ over $B$ is approximated by the Riemann sum,

$\displaystyle\iiint_B\cos(xyz)\,\mathrm dV\approx\sum_{i=1}^2\sum_{j=1}^2\sum_{k=1}^2\cos m_{i,j,k}\left(\frac{2-0}2\right)^2$

$=\displaystyle\sum_{i=1}^2\sum_{j=1}^2\sum_{k=1}^2\cos\frac{(2i-1)(2j-1)(2k-1)}8$

$=\cos\dfrac18+3\cos\dfrac38+3\cos\dfrac98+\cos\dfrac{27}8\approx\boxed{4.104}$

(compare to the actual value of about 4.159)