Chapter 5: Problem 35

A triangular lamina is bounded by the coordinate axes and the line \(x+y=6\). Find its mass if its density at each point \(P\) is proportional to the square of the distance from the origin to \(P\).

Short Answer

Expert verified

The mass is \(432k\).

Step by step solution

- Understand the Geometry of the Triangle

The triangle is bounded by the coordinate axes (x = 0 and y = 0) and the line equation given by \(x + y = 6\). The vertices of the triangle are at \((0,0)\), \((6,0)\), and \((0,6)\).

- Set Up the Integral for Mass

To find the mass of a lamina with a density function, set up a double integral. The density at a point \((x, y)\) is proportional to the square of the distance from the origin, hence, the density function is \(k(x^2 + y^2)\), where \( k \) is a constant of proportionality.

- Express the Limits of Integration

For the triangular region, \(y\) ranges from 0 to \(6 - x\) and \(x\) ranges from 0 to 6. The mass can be computed as a double integral: \(M = \int_{0}^{6} \int_{0}^{6-x} k(x^2 + y^2) dy \, dx\)

- Evaluate the Inner Integral

Evaluate the inner integral with respect to \(y\): \[\int_{0}^{6-x} k(x^2 + y^2) dy = k \left[ x^2y + \frac{y^3}{3} \right]_{0}^{6-x} = k \left[ x^2(6 - x) + \frac{(6-x)^3}{3} \right]\].

- Simplify the Inner Integral

Simplify the expression obtained: \[k \left[ 6x^2 - x^3 + \frac{(216 - 108x + 18x^2 - x^3)}{3} \right]\] which simplifies to \[k \left[ 6x^2 - x^3 + 72 - 36x + 6x^2 - \frac{x^3}{3} \right]\].

- Combine Like Terms

Combine like terms: \[k \left[ 12x^2 - x^3 - \frac{x^3}{3} + 72 - 36x \right], which simplifies to k \left[ 12x^2 - \frac{4x^3}{3} + 72 - 36x \right].\]

- Evaluate the Outer Integral

Evaluate the outer integral: \[M = \int_{0}^{6} k \left[ 12x^2 - \frac{4x^3}{3} + 72 - 36x \right] dx.\] Solve term by term: \[\int_{0}^{6} 12kx^2 \, dx + \int_{0}^{6} - \frac{4}{3}kx^3 \, dx + \int_{0}^{6} 72k \, dx - \int_{0}^{6} 36kx \, dx \].

- Compute Each Term

Evaluate each integral: \[12k \int_{0}^{6} x^2 \, dx = 12k \left[ \frac{x^3}{3} \right]_{0}^{6} = 12k \cdot 72 = 864k, \frac{-4k}{3} \int_{0}^{6} x^3 \, dx = \frac{-4k}{3}\left[ \frac{x^4}{4} \right]_{0}^{6} = -4k \cdot 54 = -216k, 72k \int_{0}^{6} dx = 72k \cdot 6 = 432k, and -36k \int_{0}^{6} x \, dx = -36k \left[ \frac{x^2}{2} \right]_{0}^{6} = -36k \cdot 18 = -648k.\]

- Sum the Results

Sum the results of each term: \[864k - 216k + 432k - 648k = 432k\]. Thus, the mass of the triangular lamina is \(432k\).

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

double integral

A double integral is a way to integrate over a two-dimensional area. It’s like zooming in twice: first, focus on integrating a function over one variable (let's say y), and then integrate the result over another variable (x). To understand this better, think of it as calculating volume beneath a surface above a region in the xy-plane.

When calculating the mass of a lamina, the double integral aggregates infinitesimally small masses over the entire region.
For our triangular region bounded by the axes and the line \(x + y = 6 \), the mass \(M\) can be expressed as:

Set up the limits of integration based on the geometry of the region. Here, \(x\) ranges from 0 to 6, and for each \(x\), \(y\) ranges from 0 to \(6 - x\).
Use the density function in the integrand, which in this exercise is a function of both \(x\) and \(y\): \(k(x^2 + y^2)\).
The double integral is represented as: \(M = \int_{0}^{6}\int_{0}^{6-x} k(x^2 + y^2) \ dy \ dx \).

density function

A density function tells you how much mass is concentrated at each point within a given region. In this exercise, the density is proportional to the square of the distance from the origin.

For a point \((x,y)\) in our triangular lamina, the distance from the origin is given by \(\rho = \sqrt{x^2 + y^2}\).

Therefore, if the density is proportional to the square of this distance, it means:

The density function \(D(x,y)\) can be expressed as \(k(x^2 + y^2)\), where \(k\) is the proportionality constant.
The term \(x^2 + y^2\) inside the density function represents the squared distance from the origin.

Understanding the density function is crucial because it allows us to set up the integrand in the double integral and ensures we're calculating the mass correctly. This concept is vital in physics and engineering, especially when dealing with objects of variable density.
By integrating this density function over the area of our lamina, we find the total mass distributed over that area.

triangular region

Understanding the geometry of a region is essential when setting up double integrals. In this case, we are working with a triangular lamina bounded by the coordinate axes and the line \(x + y = 6\).

This triangle has the following vertices:

The origin \((0,0)\)
Point \((6,0)\) on the x-axis
Point \((0,6)\) on the y-axis

The equations for the boundaries of the region are:

The x-axis \(y = 0\)
The y-axis \(x = 0\)
The line \(x + y = 6\)

To find its mass using double integrals, you need to:

Visualize the region and correctly identify the limits of integration.
For this region, \(x\) ranges from 0 to 6, and for each \(x\), \(y\) ranges from 0 to \(6 - x\).

With these limits and our density function as the integrand, we can set up the double integral for mass calculation. Understanding the triangular region helps in correctly setting up and evaluating the integral for accurate results.