Problem 27 - Advanced Engineering Mathematics

Question

Find the mass of the given surface with the indicated density function. \(S\) that portion of the plane \(x+y+z=1\) in the first octant; density at a point \(P\) directly proportional to the square of the distance from the \(y z\) -plane

Step-by-Step Solution

Answer

The mass is \(\frac{k \sqrt{3}}{12}\).

Step 1: Understand the Problem

We need to find the mass of the portion of the plane \(x+y+z=1\) that lies in the first octant and is defined by the density function. The density at any point \(P\) is proportional to the square of its distance from the \(yz\)-plane.

Step 2: Define the Region

The plane \(x+y+z=1\) in the first octant implies \(x, y, z \geq 0\). We must determine the limits for the surface within the first octant, which are \(0 \leq x \leq 1\), \(0 \leq y \leq 1-x\), and \(0 \leq z \leq 1-x-y\).

Step 3: Define the Density Function

The density function \(\rho(x, y, z)\) is proportional to \(x^2\), since the distance from the \(yz\)-plane is \(x\) for a point \((x, y, z)\). Thus, \(\rho(x, y, z) = k x^2\), where \(k\) is the proportional constant.

Step 4: Set Up the Surface Integral

The plane is defined by \(z = 1 - x - y\). The differential surface area \(dS\) can be found by the formula for the area of a plane: \(dS = \sqrt{1 + (\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2} \, dx \, dy\). Calculate the partial derivatives: \(\frac{\partial z}{\partial x} = -1\) and \(\frac{\partial z}{\partial y} = -1\). Substituting gives \(dS = \sqrt{1 + 1 + 1} \, dx \, dy = \sqrt{3} \, dx \, dy\).

Step 5: Integrate to Find the Mass

The mass \(M\) is found by integrating the density function over the surface: \[ M = \int_0^1 \int_0^{1-x} k x^2 \sqrt{3} \, dy \, dx \]Simplify the integrals: \[ M = k \sqrt{3} \int_0^1 x^2 \int_0^{1-x} 1 \, dy \, dx \]\[ M = k \sqrt{3} \int_0^1 x^2 (1-x) \, dx \]Evaluate the integral: \[ M = k \sqrt{3} \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_0^1 \]\[ M = k \sqrt{3} \left( \frac{1}{3} - \frac{1}{4} \right) = k \sqrt{3} \left( \frac{1}{12} \right) = \frac{k \sqrt{3}}{12} \]

Step 6: Final Calculation

The mass of the surface is \(\frac{k \sqrt{3}}{12}\). Since we are not given a specific constant \(k\), the mass is expressed in terms of \(k\).

Key Concepts

Density Function

In this exercise, the density function plays a crucial role in finding the mass of a surface. The density function tells us how mass is distributed across a region. For this problem, the density at any point \( P(x, y, z) \) is determined by its distance from the \( yz \)-plane.
This implies that the density increases with distance from this plane.To be specific, the density is directly proportional to \( x^2 \), so we define the density function as \( \rho(x, y, z) = k x^2 \). Here, \( k \) is a proportional constant that will factor into our calculations. Understanding how this function affects our integral is crucial for determining the mass of the surface.

Surface Integral

The surface integral plays a key role in calculating the mass of a surface based on its density. Essentially, it aggregates contributions from the entire surface, factoring in how the density varies at different points.To compute the surface integral for our problem, we need to evaluate the integral of the density function \( \rho(x, y, z) \) over the surface we are considering. The integral formula used is:\[ M = \int \int_S \rho(x, y, z) \cdot dS \]Where \( dS \) is the differential surface area element. Integrating this across the specified region (the part of our plane in the first octant) allows us to account for variations in density and ultimately compute the total mass.

Plane Equation

The equation for the plane we are dealing with is \( x + y + z = 1 \). This equation outlines all the points that satisfy this relationship, forming a flat surface in three dimensions.For this exercise, focusing on the first octant, this plane is bounded by the positive axes. The equation also gives us \( z \) in terms of \( x \) and \( y \):\[ z = 1 - x - y \]This formula is important because it allows us to express \( z \) in our integrals, ensuring we are evaluating across the right region. Knowing this plane equation and how to manipulate it is essential when working with surface integrals and defining the bounds for integration.

First Octant

The concept of the first octant refers to the section of three-dimensional space where all coordinates are positive: \( x, y, z \geq 0 \).In the context of planes, like the one in this exercise, we must ensure our integration limits respect these constraints, keeping \( x, y, \) and \( z \) non-negative.Thus, for our plane equation \( x + y + z = 1 \), the first octant limits for \( x, y, \) and \( z \) become:

• \( 0 \leq x \leq 1 \)
• \( 0 \leq y \leq 1 - x \)
• \( 0 \leq z \leq 1 - x - y \)

These boundaries delineate the portion of the plane we will integrate over for our mass calculation, ensuring we stay within the first octant.