Question

Question: Use the Midpoint Rule for triple integrals (Exercise \({\rm{22}}\)) to estimate the value of the integral. Divide into eight sub-boxes of equal size. where \({\rm{B = \{ (x,y,z)|0}} \le {\rm{x}} \le {\rm{1,0}} \le {\rm{y}} \le {\rm{1,0}} \le {\rm{z}} \le {\rm{1\} }}\).

Short answer

The value of the integral is \( \approx {\rm{0}}{\rm{.985}}\).

Step-by-step solution

Define triple integral

A triple integral is a three-variable iterated integral over a three-dimensional area. The definition of triple integrals is similar to that of double integrals, except that this time we're integrating over a volume rather than an area.

Evaluating integral

Let's start by figuring out \({\rm{\Delta V}}\). We'll put our points at the following locations since we'll need \({\rm{8}}\)sub-boxes.

\(\left( {\frac{{\rm{1}}}{{\rm{2}}}{\rm{,0,0}}} \right)\left( {{\rm{0,}}\frac{{\rm{1}}}{{\rm{2}}}{\rm{,0}}} \right)\left( {{\rm{0,0,}}\frac{{\rm{1}}}{{\rm{2}}}} \right)\)

As a result, we arrive to the following conclusions,\({\rm{l = m = n = 2}}\)and

\(\begin{array}{c}{\rm{\Delta V = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ \times }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ \times }}\frac{{\rm{1}}}{{\rm{2}}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{8}}}\end{array}\)

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

As a result, we arrive to the following conclusion,

\(\begin{array}{c}{\rm{ = }}\frac{{\rm{1}}}{{\rm{8}}}\left( {{\rm{cos}}\frac{{\rm{1}}}{{{\rm{64}}}}{\rm{ + cos}}\frac{{\rm{3}}}{{{\rm{64}}}}{\rm{ + cos}}\frac{{\rm{3}}}{{{\rm{64}}}}{\rm{ + cos}}\frac{{\rm{9}}}{{{\rm{64}}}}{\rm{ + cos}}\frac{{\rm{3}}}{{{\rm{64}}}}{\rm{ + cos}}\frac{{\rm{9}}}{{{\rm{64}}}}{\rm{ + cos}}\frac{{\rm{9}}}{{{\rm{64}}}}{\rm{ + cos}}\frac{{{\rm{27}}}}{{{\rm{64}}}}} \right)\\ \approx {\rm{0}}{\rm{.985}}\end{array}\)

Therefore, the value of the integral is \( \approx {\rm{0}}{\rm{.985}}\).