Question

Question: (a) In the Midpoint Rule for triple integrals we use a triple Riemann sum to approximate a triple integral over a box \({\rm{B}}\), where \({\rm{f(x,y,z)}}\) is evaluated at the center \({\rm{(}}\overline {{{\rm{x}}_{\rm{i}}}} {\rm{,}}\overline {{{\rm{y}}_{\rm{j}}}} {\rm{,}}\overline {{{\rm{z}}_{\rm{k}}}} {\rm{)}}\) of the box \({{\rm{B}}_{{\rm{ijk}}}}\) . Use the Midpoint Rule to estimate, where \({\rm{B}}\) is the cube defined by \({\rm{0}} \le {\rm{x}} \le {\rm{4,0}} \le {\rm{y}} \le {\rm{4,0}} \le {\rm{z}} \le {\rm{4}}\). Divide \({\rm{B}}\) into eight cubes of equal size.

(b) Use a computer algebra system to approximate the integral in part (a) correct to the nearest integer. Compare with the answer to part (a).

Short answer

1. The value of
2. The value is \({\rm{246}}\).

Step-by-step solution

Define integral

In mathematics, an integral assigns numbers to functions to express displacement, area, volume, and other concepts arising from linking infinitesimal data.

Explanation

a. Use \({\rm{l = m = n = 2}}\) since we require \({\rm{8}}\)sub-boxes, and middle points will be,

\(\begin{array}{l}{\rm{A(1,1,1)}}\\{\rm{B(1,1,3)}}\\{\rm{C(1,3,1)}}\\{\rm{D(1,3,3)}}\\{\rm{E(3,1,1)}}\\{\rm{F(3,1,3)}}\\{\rm{G(3,3,1)}}\\{\rm{H(3,3,3)}}\end{array}\)

Let's compute\({\rm{\Delta V}}\)now. Because we used\({\rm{l = m = n = 2}}\), all of our intervals were halved, giving us the following result,

\(\begin{array}{c}{\rm{\Delta V = 2 \times 2 \times 2}}\\{\rm{ = 8}}\end{array}\)

Let's write down the Midpoint Rule and apply it,

As a result, we arrive to the following conclusion,

\(\begin{array}{l}{\rm{ = 8}}\left( {\sqrt {\rm{3}} {\rm{ + }}\sqrt {{\rm{11}}} {\rm{ + }}\sqrt {{\rm{11}}} {\rm{ + }}\sqrt {{\rm{19}}} {\rm{ + }}\sqrt {{\rm{11}}} {\rm{ + }}\sqrt {{\rm{19}}} {\rm{ + }}\sqrt {{\rm{19}}} {\rm{ + }}\sqrt {{\rm{27}}} } \right)\\ \approx {\rm{239}}{\rm{.64}}\end{array}\)

Therefore,

Explanation

b. It is said that \({\rm{(x,y,z)}} \in {\rm{D}}\) such that \({\rm{0}} \le {\rm{x}} \le {\rm{4,0}} \le {\rm{y}} \le {\rm{4,0}} \le {\rm{z}} \le {\rm{4}}\),

Therefore,

As stated in the problem, use a computational tool to approximate this Integral.

Part(a) yielded an approximation of\({\rm{240}}\), which is near to the correct answer of\({\rm{246}}\).

Therefore, the value is \({\rm{246}}\).