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Chapter 5: Problem 23

For the following exercises, approximate the mass of the homogeneous lamina that has the shape of given surface \(S\). Round to four decimal places. Evaluate surface integral \(\iint_{S} g d S\), where \(g(x, y, z)=x z+2 x^{2}-3 x y\) and \(S\) is the portion of plane \(2 x-3 y+z=6\) that lies over unit square \(R: 0 \leq x \leq 1,0 \leq y \leq 1\).

Short Answer

Expert verified
The mass of the lamina is approximately 14.0260.

Step by step solution

01

Parameterize the Surface

Given the plane equation \(2x - 3y + z = 6\), we can solve for \(z\) in terms of \(x\) and \(y\). Thus, \(z = 6 - 2x + 3y\). The surface \(S\) is defined as this plane over the unit square region \(R: 0 \leq x \leq 1, 0 \leq y \leq 1\).
02

Compute the Surface Element (dS)

The surface element \(dS\) can be obtained from the gradient of \(F(x, y, z) = 2x - 3y + z - 6\), which gives \(abla F = (2, -3, 1)\). The magnitude \(|abla F| = \sqrt{2^2 + (-3)^2 + 1^2} = \sqrt{14}\). Thus, \(dS = \sqrt{14} \; dx\, dy\).
03

Set Up the Surface Integral

The integrand \(g(x, y, z) = xz + 2x^2 - 3xy\) needs \(z\) replaced with the parameterization from Step 1. Substitute \(z = 6 - 2x + 3y\) into \(g(x, y, z)\) to get \(x(6 - 2x + 3y) + 2x^2 - 3xy = 6x + x^2 + 3xy\).
04

Evaluate the Double Integral Over R

Integrate the expression \(6x + x^2 + 3xy\) over \(R\) using \(dS\). The integral becomes \[ \int_{0}^{1} \int_{0}^{1} (6x + x^2 + 3xy) \sqrt{14} \, dy \, dx \]. Integrate with respect to \(y\) first: \[ \int_{0}^{1} \left( 6xy + xy^2 + \frac{3}{2}xy^2 \right) \Bigg|_{y=0}^{y=1} \, dx = \int_{0}^{1} \left( 6x + x + \frac{3}{2}x \right) \, dx \].
05

Final Integration and Answer Round

This simplifies to \( \int_{0}^{1} 7.5x \, dx \), which evaluates to \(\frac{7.5}{2}x^2\Bigg|_{0}^{1} = 3.75\). Multiply by \(\sqrt{14}\) to obtain the final mass: \(3.75\sqrt{14} \, \approx 14.0260\). Thus, the mass of the lamina is approximately 14.0260.

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Key Concepts

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

Homogeneous Lamina
A homogeneous lamina is a flat, two-dimensional shape with uniform mass distribution across its entire area. In simpler terms, it is a surface that has the same mass per unit area throughout, allowing us to approach it mathematically without concerning variations in density. In this particular exercise, the lamina is defined by the portion of a plane over a specified region. This plane is represented by the equation \(2x - 3y + z = 6\), and it forms the shape of the lamina within the confines of a unit square, denoted as \(R: 0 \leq x \leq 1, 0 \leq y \leq 1\). Since the mass density is constant, calculating the total mass simplifies to finding an appropriate surface integral over this shape.
Parameterization
Parameterization is a method in mathematics used to express a surface or curve by a set of equations typically involving parameters like \(x\) and \(y\). In this exercise, we've parameterized the surface \(S\), which is part of the plane \(2x - 3y + z = 6\). By solving for \(z\) in terms of \(x\) and \(y\), we achieve \(z = 6 - 2x + 3y\). Thus, for any point on the surface \(S\), you can determine its position using \((x, y, 6-2x+3y)\).
  • This approach allows us to work with a 2D integral over the region \(R\), given that the surface lies above this rectangle in the \(xy\)-plane.
  • It simplifies evaluating the surface integral as a double integral over the parameterization, streamlining the process.
Double Integral
The double integral is central to calculating quantities like mass over two-dimensional regions, especially for shapes like laminae. It involves integrating a function of two variables over a certain domain. In our exercise, the double integral becomes \[ \int_{0}^{1} \int_{0}^{1} (6x + x^2 + 3xy) \sqrt{14} \, dy \, dx\].
This encapsulates:
  • A function to integrate: as derived, the expression \(6x + x^2 + 3xy\) represents mass contributions across the lamina.
  • Limits of integration: These guide the integration over \(x\) and \(y\), spanning from 0 to 1 for both variables, as given by the unit square.
  • Integration order: Start integrating in terms of \(y\), followed by \(x\). This sets the pathway to calculate the aggregate mass.
By performing these integrals in the specified order, we precisely capture the entire mass of the lamina.
Gradient of a Function
The gradient of a function is a vector that points in the direction of the greatest rate of increase of that function. For a surface defined implicitly by a function \(F(x, y, z)\), like \(2x - 3y + z - 6 = 0\) in our problem, the gradient \(abla F\) connects us to surface integrals. Here, \(abla F = (2, -3, 1)\), which helps compute the surface element \(dS\).
The magnitude \(|abla F| = \sqrt{2^2 + (-3)^2 + 1^2} = \sqrt{14}\), signifies the full weight of contributions from each coordinate.
The gradient not only informs us about directional change but effectively assists in transforming the surface integral into a more manageable form by defining \(dS\), simplifying it via \(dS = \sqrt{14} \, dx \, dy\). This invites ease in evaluating the mass of the homogeneous lamina as part of the double integral calculation.

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