Problem 20 - Thomas Calculus

Question

In Exercises 19 and 20 , calculate the following. a. The amount of work done by (constant) gravity \(g\) in moving the liquid filling in the container to the \(x y\) -plane. (Hint: Partition the liquid into small volume elements \(\Delta V_{i}\) and find the work done (approximately) by gravity on each element. Summation and passage to the limit gives a triple integral to evaluate.) b. The work done by gravity in moving the center of mass down to the \(x y\) -plane. The container is in the shape of the region bounded by \(y=0, z=0, z=4-x^{2},\) and \(x=y^{2} .\) The density of the liquid filling the region is \(\delta(x, y, z)=k x y, \quad k\) a constant Exercise 14 ).

Step-by-Step Solution

Answer

Use triple integrals to find the total work done by integrating the density function weighted by gravity along the z-axis across given bounds. Evaluate the integrals to find the work. Calculate the center of mass and use it to find the work done on the center of mass.

Step 1: Understand the Problem

The problem requires calculating the work done by gravity in two different scenarios: moving the entire liquid to the xy-plane and moving the center of mass of the liquid. The container's region is bounded by the curves and surfaces: \(y=0\), \(z=0\), \(z=4-x^2\), and \(x=y^2\). The density of the liquid is given as \(\delta(x, y, z)=kxy\).

Step 2: Calculate Volumetric Work Elements

To find the work done by gravity, partition the liquid into small volume elements \(\Delta V_i\). The approximate work done by gravity on each element is given by \( dW = \delta(x, y, z) \cdot g \cdot z \cdot dV \) in the direction of gravity (z-direction), where \(g\) is the gravitational acceleration.

Step 3: Set Up the Triple Integral

Set up the triple integral to calculate the total work \(W\). The integral bounds for \(x\) are from 0 to 2, for \(y\) are from 0 to \(\sqrt{x}\), and for \(z\) are from 0 to \(4-x^2\). Thus, the total work is given by: \[ W = \int_{x=0}^{2} \int_{y=0}^{\sqrt{x}} \int_{z=0}^{4-x^2} k \cdot g \cdot x \cdot y \cdot z \cdot dz \cdot dy \cdot dx \]

Step 4: Evaluate the Triple Integral

Evaluate the integral step-by-step: first, integrate with respect to \(z\), then \(y\), and finally \(x\). - Integrating with respect to \(z\) yields: \[ \int_{0}^{4-x^2} z \cdot dz = \left[ \frac{1}{2}z^2 \right]_0^{4-x^2} = \frac{1}{2}(4-x^2)^2 \]- Next, integrate with respect to \(y\):\[ \int_{0}^{\sqrt{x}} y \cdot \frac{1}{2}(4-x^2)^2 \ dy = \frac{1}{4}(4-x^2)^2 \left[ y^2 \right]_0^{\sqrt{x}} = \frac{1}{4}(4-x^2)^2 \cdot x \]- Finally integrate with respect to \(x\):\[ W = k \cdot g \cdot \int_{0}^{2} \frac{x(4-x^2)^2}{4} \cdot dx \]

Step 5: Simplify and Compute the Remaining Integral

Expand \((4-x^2)^2\) and substitute into the integral. Solve the definite integral:\((4-x^2)^2 = 16 - 8x^2 + x^4\)Substitute in:\[ W = k \cdot g \cdot \int_{0}^{2} (16x - 8x^3 + x^5) \cdot \frac{1}{4} \cdot dx \]Evaluate:\[ W = k \cdot g \cdot \left[ \frac{16}{4} \cdot \frac{x^2}{2} - \frac{8}{4} \cdot \frac{x^4}{4} + \frac{x^6}{24} \right]_{0}^{2} \]Calculate values for each term at the bounds \(x=0\) and \(x=2\).

Step 6: Calculate Work on Center of Mass (Part b)

To find the center of mass, calculate its coordinate \(z_{cm}\) using similar integration methods. The center of mass \(z_{cm}\) along the z-direction is given by:\[ z_{cm} = \frac{1}{M} \int_{0}^{2} \int_{0}^{\sqrt{x}} \int_{0}^{4-x^2} z \cdot \delta(x, y, z) \cdot dz \cdot dy \cdot dx \] \( M \) is the total mass of the liquid obtained from:\[ M = \int_{0}^{2} \int_{0}^{\sqrt{x}} \int_{0}^{4-x^2} \delta(x, y, z) \cdot dz \cdot dy \cdot dx \]Substitute \(z_{cm}\) into \[ W_{cm} = M \cdot g \cdot z_{cm} \].

Key Concepts

Triple Integral

A triple integral is a powerful mathematical tool used to calculate volumes, masses, or other quantities that are distributed throughout a three-dimensional space. In this particular exercise, we use the triple integral to find the work done by gravity as it moves a liquid filling a container to the xy-plane. This method involves slicing the region into tiny sub-volumes, each of which has a small volume element denoted as \( dV \). This volume element is then integrated over all dimensions of the region.Here are the key steps involved in setting up a triple integral:

• Identify the region of integration bounded by the given surfaces, in this case: \( y = 0 \), \( z = 0 \), \( z = 4 - x^2 \), and \( x = y^2 \).
• Determine the limits of integration for each variable sequentially: \( x \) from 0 to 2, \( y \) from 0 to \( \sqrt{x} \), and \( z \) from 0 to \( 4-x^2 \).
• Set up the triple integral by multiplying the integrand, in this case, the product of density function \( \delta(x, y, z) = kxy \), gravitational constant \( g \), and the displacement \( z \).

The resulting integral allows for the calculation of total work done as it accounts for contributions from each volume element within the predefined bounds.

Center of Mass

The center of mass (CM) is a critical concept when evaluating the work done by gravity. It represents the average position of all the mass in a body or system. For the exercise, calculating the center of mass helps us determine the work involved in moving the entire mass to the xy-plane.Finding the center of mass involves:

• Calculating the total mass \( M \) of the liquid within the container. This is done using a triple integral of the density function over the volume of the region.
• Determining the coordinates of the center of mass, particularly the \( z_{cm} \) for this scenario. This is achieved through calculating \( z_{cm} = \frac{1}{M} \int_{0}^{2} \int_{0}^{\sqrt{x}} \int_{0}^{4-x^2} z \cdot \delta(x, y, z) \cdot dz \cdot dy \cdot dx \).

By knowing the center of mass, we can ascertain not just the position where all mass could theoretically be concentrated, but also compute the necessary work \( W_{cm} = M \cdot g \cdot z_{cm} \) to move this concentrated mass to the desired plane.

Density Function

In physics and engineering, a density function helps describe how mass or other quantities are distributed in space. Here, the liquid within the container is characterized by a density function \( \delta(x, y, z) = kxy \), where \( k \) is a constant.Understanding the density function involves:

• Identifying that the function \( \delta(x, y, z) = kxy \) suggests that the density varies with both \( x \) and \( y \) coordinates.
• Recognizing that higher densities occur in regions with greater \( x \) and \( y \) product values. This impacts the mass distribution within the container.
• Integrating this function over the specified region to determine total parameters such as mass or work when considering the entire container.

The density function's role is crucial in forming the basis for calculating integrals related to mass, center of mass, and the work done by gravitational forces on each volume element.

Volume Element Partitioning

Volume element partitioning is an essential technique used in calculus for breaking down a three-dimensional region into smaller, manageable pieces, known as volume elements \( \Delta V_i \). This is especially beneficial when calculating integrals pertaining to complex shapes.The process of partitioning involves:

• Dividing the entire region defined by the container into infinitesimally small elements \( dV = dx \cdot dy \cdot dz \) suitable for integration.
• Calculating approximate contributions such as work done or mass for these small elements individually using the density function and gravitational factors.
• Summing up these contributions and transitioning to a limit as these elements become infinitely small, resulting in the triple integral over the entire volume.

This approach allows for accurate evaluation of complicated integrals by leveraging the principle that the integration over volumes can be conducted by integrating over very small constituent parts.