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.) cos(xyz) dv, where b = {(x, y, z) | 0 ≤ x ≤ 2, 0 ≤ y ≤ 2, 0 ≤ z ≤ 2} b

Answer 1

The sub-boxes will have dimensions $\frac{2-0}{2} \times \frac{2-0}{2} \times \frac{2-0}{2} =1\times1\times1=1 \ cubic \ units$

x sub-intervals are 0 to 1 and 1 to 2. Midpoints are at $x= \frac{1}{2}$ and $x= \frac{3}{4}$
y sub-intervals are 0 to 1 and 1 to 2. Midpoints are at $y= \frac{1}{2}$ and $y= \frac{3}{4}$
z sub-intervals are 0 to 1 and 1 to 2. Midpoints are at $z= \frac{1}{2}$ and $z= \frac{3}{4}$

Let $f(x,y,z)=\cos{(xyz)}$

$\int\limits \int\limits \int\limits {f(x,y,z)} \, dV \approx f\left( \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right)+f\left( \frac{1}{2} , \frac{1}{2} , \frac{3}{4} \right)+f\left( \frac{1}{2} , \frac{3}{4} , \frac{1}{2} \right)+f\left( \frac{1}{2} , \frac{3}{4} , \frac{3}{4} \right)$
$+f\left( \frac{3}{4} , \frac{1}{2} , \frac{1}{2} \right)+f\left( \frac{3}{4} , \frac{1}{2} , \frac{3}{4} \right)+f\left( \frac{3}{4} , \frac{3}{4} , \frac{1}{2} \right)+f\left( \frac{3}{4} , \frac{3}{4} , \frac{3}{4} \right) \\ \\ \approx\cos{ \frac{1}{8} }+\cos{ \frac{3}{16} }+\cos{ \frac{3}{16} }+\cos{ \frac{9}{32} }+\cos{ \frac{3}{16} }+\cos{ \frac{9}{32} }+\cos{ \frac{9}{32} }+\cos{ \frac{27}{64} } \\ \\ \approx0.9922+0.9825+0.9825+0.9607+0.9825+0.9607+0.9607 \\ +0.9123 \\ \\ \approx\bold{7.734}$