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Chapter 15: Problem 25

Use the Midpoint Rule for triple integrals (Exercise 24 to estimate the value of the integral. Divide \(B\) into eight sub-boxes of equal size. $$ \begin{array}{l}{\int \mathbb{N}_{B} \frac{1}{\ln (1+x+y+z)} d V, \text { where }} \\ {B=\\{(x, y, z) | 0 \leqslant x \leqslant 4,0 \leqslant y \leqslant 8,0 \leqslant z \leqslant 4\\}}\end{array} $$

Short Answer

Expert verified
Midpoint Rule estimate involves evaluating \( \frac{1}{\ln(1+x+y+z)} \) at 8 midpoints, multiplied by the volume element for each sub-box, and summing.

Step by step solution

01

Understand the Integration Region

The region \( B \) is defined as \( \{ (x, y, z) \ |\ 0 \leq x \leq 4, \ 0 \leq y \leq 8, \ 0 \leq z \leq 4 \} \). This region is a rectangular prism with a width of 4, height of 8, and depth of 4.
02

Determine the Dimensions of Sub-Boxes

Divide the region \( B \) into eight sub-boxes each of equal size. Since \( B \) has dimensions 4 (x-axis), 8 (y-axis), and 4 (z-axis), each sub-box should be half the dimension: \( \frac{4}{2} = 2 \) for x, \( \frac{8}{2} = 4 \) for y, and \( \frac{4}{2} = 2 \) for z.
03

Locate Midpoints of Sub-Boxes

For each sub-box, determine the center point or midpoint. Midpoints are calculated as follows: For the x-dimension, they are 1, 3; for the y-dimension, they are 2, 6; and for the z-dimension, they are 1, 3. These points represent the midpoints of the respective dimension ranges.
04

Calculate the Midpoint for Each Sub-Box

The midpoint for each of the 8 sub-boxes is a combination of the midpoints obtained from Step 3. For example, one sub-box midpoint is \( (1, 2, 1) \), another is \( (3, 2, 1) \), and so on, resulting in a total of 8 combinations.
05

Function Values at Midpoints

Calculate the function value \( \frac{1}{\ln(1+x+y+z)} \) at each midpoint. For example, at midpoint \( (1, 2, 1) \), evaluate the function: \( \frac{1}{\ln(1+1+2+1)} = \frac{1}{\ln(5)} \). Repeat this for all midpoints.
06

Compute the Volume Element (dV)

The volume of each sub-box is the product of the side lengths: \( 2 \times 4 \times 2 = 16 \). This is the volume element \( dV \) for each sub-box.
07

Estimate the Integral Using the Midpoint Rule

Sum the product of the function values at each midpoint and the volume element \( dV \) for all sub-boxes. This gives: \( \sum_{i=1}^{8} \frac{1}{\ln(1+x_i+y_i+z_i)} \times 16 \).
08

Calculation of the Final Result

Calculate the sum from Step 7, which provides the estimate for the integral using the Midpoint Rule. Each term is formed by multiplying the function value at each midpoint by \( 16 \), then adding these products together.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Volume of Rectangular Prism
Understanding the concept of a rectangular prism is crucial for estimating integrals over a three-dimensional region. Imagine a box or a brick鈥攖hat's essentially what a rectangular prism is. It has three dimensions: width, height, and depth.
For example, in our exercise, we are looking at a prism defined by:
  • Width along the x-axis: 4 units
  • Height along the y-axis: 8 units
  • Depth along the z-axis: 4 units
To find the volume of such a prism, you multiply these three dimensions together: \[ \text{Volume} = \text{Width} \times \text{Height} \times \text{Depth} \]Therefore, the volume of our prism is \[4 \times 8 \times 4 = 128 \text{ cubic units}.\]This understanding of volume is foundational when dividing the region into smaller parts for numerical integration.
Midpoint Calculation
The concept of a midpoint is often used in numerical integration methods like the Midpoint Rule. It helps in estimating averages over intervals or chunks of a region. Updating our focus to three dimensions, calculating the midpoint means determining the center of sub-regions within a rectangular prism.
For a sub-box in three dimensions:
  • The x-coordinate midpoint is the average of the two x-limits, e.g., between 0 and 2, it's 1.
  • The y-coordinate midpoint is similar, say between 0 and 4, it's 2.
  • For the z-coordinate, the midpoint between 0 and 2 is 1.
Here, you'd find a midpoint like (1, 2, 1) for one sub-box. Repeat this for each segment, and you'll have midpoints; key positions where the function's value is to be calculated in the upcoming steps of numerical integration.
Numerical Integration
Numerical integration allows us to approximate the value of integrals when finding an exact solution is difficult. It involves breaking down a region into smaller, manageable pieces and then estimating the area (or volume) by evaluating the function at specified points鈥攊n this case, the midpoints.
The Midpoint Rule involves these steps:
  • Subdivide the integration region into smaller boxes or segments.
  • Calculate the function's value at each midpoint.
  • Multiply each function value by the volume of its corresponding sub-box (16 in our exercise).
Sum up all these products to get an estimate for the integral. For our function, it involves computing \[\frac{1}{\ln(1+x+y+z)}\] at each midpoint, and then using this value to calculate a component of the integral's total estimate. This approach is practical and efficient for complex regions or functions.

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Most popular questions from this chapter

Find the average value of \(f\) over region \(D .\) $$\begin{array}{l}{f(x, y)=x \sin y, \quad D \text { is enclosed by the curves } y=0} \\ {y=x^{2}, \text { and } x=1}\end{array}$$

\(\begin{array}{l}{\text { (a) Use cylindrical coordinates to show that the volume of }} \\ {\text { the solid bounded above by the sphere } r^{2}+z^{2}=a^{2} \text { and }} \\ {\text { below by the cone } z=r \cot \phi_{0}\left(\text { or } \phi=\phi_{0}\right), \text { where }} \\\ {0<\phi_{0}<\pi / 2, \text { is }}\end{array}\) $$V=\frac{2 \pi a^{3}}{3}\left(1-\cos \phi_{0}\right)$$ (b) Deduce that the volume of the spherical wedge given by \(\rho_{1} \leqslant \rho \leqslant \rho_{2}, \theta_{1} \leqslant \theta \leqslant \theta_{2}, \phi_{1} \leqslant \phi \leqslant \phi_{2}\) is $$\Delta V=\frac{\rho_{2}^{3}-\rho_{1}^{3}}{3}\left(\cos \phi_{1}-\cos \phi_{2}\right)\left(\theta_{2}-\theta_{1}\right)$$ (c) Use the Mean Value Theorem to show that the volume in part (b) can be written as $$\Delta V=\tilde{\rho}^{2} \sin \tilde{\phi} \Delta \rho \Delta \theta \Delta \phi$$ where \(\tilde{\rho}\) lies between \(\rho_{1}\) and \(\rho_{2}, \tilde{\phi}\) lies between \(\phi_{1}\) and \(\phi_{2}, \Delta \rho=\rho_{2}-\rho_{1}, \Delta \theta=\theta_{2}-\theta_{1},\) and \(\Delta \phi=\phi_{2}-\phi_{1}\)

Evaluate the integral by making an appropriate change of variables. \(\iint_{R}(x+y) e^{x^{2}-y^{2}} d A,\) where \(R\) is the rectangle enclosed by the lines \(x-y=0, x-y=2, x+y=0,\) and \(x+y=3\)

When studying the spread of an epidemic, we assume that th probability that an infected individual will spread the diseas an uninfected individual is a function of the distance betwee them. Consider a circular city of radius 10 mi in which the population is uniformly distributed. For an uninfected individual at a fixed point \(A\left(x_{0}, y_{0}\right),\) assume that the probability function is given by $$f(P)=\frac{1}{20}[20-d(P, A)]$$ where \(d(P, A)\) denotes the distance between \(P\) and \(A\) (a) Suppose the exposure of a person to the disease is the sum of the probabilities of catching the disease from all members of the population. Assume that the infected people are uniformly distributed throughout the city, with k infected individuals per square mile. Find a double integral that represents the exposure of a person residing at \(A .\) (b) Evaluate the integral for the case in which \(A\) is the center of the city and for the case in which \(A\) is located on the edge of the city. Where would you prefer to live?

\(29-32\) Evaluate the iterated integral by converting to polar coordinates. $$ \int_{0}^{2} \int_{0}^{\sqrt{2 x-x^{2}}} \sqrt{x^{2}+y^{2}} d y d x $$
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