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Chapter 11: Problem 3

Find the mass of the rectangular region \(0 \leq x \leq 2,0 \leq y \leq 1\) with density function \(\rho(x, y)=1-y\)

Short Answer

Expert verified
The mass of the rectangular region with the given density function \(\rho(x, y) = 1 - y\) and the limits \(0 \leq x \leq 2\) and \(0 \leq y \leq 1\) can be found by evaluating the double integral: \[\int_{0}^{2} \int_{0}^{1} (1-y)\, dy \, dx\] After performing the integration, we find that the mass of the rectangular region is 1 unit.

Step by step solution

01

Set up the limits of integration

We need to perform a double integral of the density function over the region 0 鈮� x 鈮� 2 and 0 鈮� y 鈮� 1. The limits of integration for x and y will be: 0 鈮� x 鈮� 2 and 0 鈮� y 鈮� 1
02

Express the density function as a double integral

Given that 蟻(x, y) = 1 - y, we set up a double integral as follows: \[\int_{0}^{2} \int_{0}^{1} (1-y)\, dy \, dx\]
03

Evaluate the inner integral with respect to y

We need to evaluate the integral of (1 - y) with respect to y in the given limits (0 to 1). \[ \int_{0}^{1} (1-y)\, dy = \left[y - \frac{1}{2}y^2\right]_{0}^{1}\] Substitute the limits and find the result: \[(1 - \frac{1}{2}(1)^2) - (0 - \frac{1}{2}(0)^2) = (1 - \frac{1}{2}) - 0 = \frac{1}{2}\]
04

Evaluate the outer integral with respect to x

We need to evaluate the integral of the result from step 3 with respect to x in the limits of integration (0 to 2). \[ \int_{0}^{2} \left(\frac{1}{2}\right)\, dx = (\frac{1}{2}x) \Big|_0^2\] Substitute the limits and find the result of the integral: \[ \frac{1}{2}(2) - \frac{1}{2}(0) = 1\]
05

Calculate the mass of the rectangular region

The mass of the rectangular region is given by the result of the double integral: Mass=1 The mass of the rectangular region with the given density function is 1 unit.

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Key Concepts

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

Double Integral
In multivariable calculus, a double integral is a way to integrate a function of two variables over a certain region in a plane. Imagine you have a flexible sheet representing your function and you want to find the total volume under that sheet above a rectangular region. To compute this using a double integral, you divide this sheet into little rectangles, sum up the volumes of the thin solid pieces on top of each rectangle, and take the limit as the rectangles become infinitesimally small.

Here's a simple analogy: if you were tiling a wavy floor, you'd count the number of tiles you need (which is like integration) and then adjust for the waves by cutting the tiles where necessary (which is like integrating a variable function).

The double integral is noted as \[\int\int\], which indicates that you perform integration twice, once for each variable. In the given exercise, the function \(\rho(x, y)=1-y\) is integrated over the rectangle \(0 \leq x \leq 2, 0 \leq y \leq 1\) to find the total mass.
Density Function
A density function in the context of physics represents how mass is distributed across a region. The concept can be transferred to points in a plane or space for mathematical analysis. In our example, the density function \(\rho(x, y)\) describes how the mass is spread out over the rectangular region. The function \(\rho(x, y)=1-y\) suggests that the density decreases linearly as the value of \(y\) increases.

With a density function, you can integrate over a certain area to find the total mass or charge or any other density-related quantity. In simpler terms, the density function tells you 'how much' of something is at any point, and integrating over an area or volume gives you the 'total amount' of that something.
Limits of Integration
The limits of integration determine the boundaries over which you are integrating. They can be thought of as the dimensions of the space you're calculating the total mass, charge, or other quantities for. In our rectangle from the exercise, the limitations \(0 \leq x \leq 2\) and \(0 \leq y \leq 1\) are like the coordinates of the corners of a piece of paper you're using to cover a specific portion of your function's graph.

The limits of integration are essential because they essentially 'clip out' the part of the function you're interested in, similar to using a cookie-cutter to get just the piece of dough in the shape you want. For the inner integral, the limits for \(y\) go from 0 to 1, while for the outer integral, the \(x\) limits go from 0 to 2.
Multivariable Integration
When we generalize integration to functions of more than one variable, we enter the realm of multivariable integration. It includes double, triple, and higher-order integrals. In this broader context, finding the mass of a region using a density function is a classic application of multivariable integration, often best approached by thinking about it in layers.

Imagine you're building a layered cake. Each layer might have a different flavor or thickness, which could represent a different cross-section of your 3D volume or area. Multivariable integration, in the case of a double integral, involves integrating over these layers twice, once for each variable, summing up the contributions from all layers, to find the total volume or mass, as with the rectangular region and density function in the exercise.

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Most popular questions from this chapter

Evaluate the integral by reversing the order of integration. \(\int_{0}^{1} \int_{3 y}^{3} e^{x^{2}} d x d y=\) _______________________.

Decide if the points given in polar coordinates are the same. If they are the same, enter \(T\). If they are different, enter \(F\). $$ \begin{array}{l} \text { a.) }\left(5, \frac{\pi}{3}\right),\left(-5, \frac{-\pi}{3}\right) \\ \text { b.) }\left(2, \frac{35 \pi}{4}\right),\left(2,-\frac{35 \pi}{4}\right) \\\ \text { c.) }(0,5 \pi),\left(0, \frac{4 \pi}{4}\right) \\ \text { d.) }\left(1, \frac{141 \pi}{4}\right),\left(-1, \frac{\pi}{4}\right) \\\ \text { e.) }\left(2, \frac{92 \pi}{3}\right),\left(-2, \frac{-\pi}{3}\right) \\\ \text { f. })(5,15 \pi),(-5,15 \pi) \end{array} $$

A pile of earth standing on flat ground has height 36 meters. The ground is the xy-plane. The origin is directly below the top of the pile and the Z-axis is upward. The cross-section at height \(z\) is given by \(x^{2}+y^{2}=36-z\) for \(0 \leq z \leq 36,\) with \(x, y,\) and \(z\) in meters. (a) What equation gives the edge of the base of the pile? $$ \begin{array}{l} x^{2}+y^{2}=36 \\ x+y=36 \\ x^{2}+y^{2}=6 \\ x+y=6 \end{array} $$ None of the above (b) What is the area of the base of the pile? (c) What equation gives the cross-section of the pile with the plane \(z=5 ?\) $$ \begin{aligned} x^{2}+y^{2} &=25 \\ x^{2}+y^{2} &=\sqrt{31} \\ x^{2}+y^{2} &=31 \\ x^{2}+y^{2} &=5 \end{aligned} $$ None of the above (d) What is the area of the cross-section \(z=5\) of the pile? (e) What is \(A(z)\), the area of a horizontal cross-section at height \(z\) ? \(A(z)=\) _________________________ square meters (f) Use your answer in part (e) to find the volume of the pile. Volume \(=\) _____________________cubic meters

The region \(W\) lies below the surface \(f(x, y)=8 e^{-(x-3)^{2}-y^{2}}\) and above the disk \(x^{2}+y^{2} \leq 16\) in the \(x y\) -plane. (a) Think about what the contours of \(f\) look like. You may want to using \(f(x, y)=1\) as an example. Sketch a rough contour diagram on a separate sheet of paper. (b) Write an integral giving the area of the cross-section of \(W\) in the plane \(x=3\). Area \(=\int_{a}^{b}\) _________________________ \(d\) where \(a=\) _________________ and \(b=\) _________________ (c) Use your work from (b) to write an iterated double integral giving the volume of \(W\), using the work from (b) to inform the construction of the inside integral. \(\text { Volume }=\int_{a}^{b} \int_{c}^{d}\) __________________ \(d\) ____________ \(d\) ____________, where \(a\)= __________, \(b\)=___________\(c\)= ______________and \(d=\)

The temperature at any point on a metal plate in the \(x y\) plane is given by \(T(x, y)=100-4 x^{2}-y^{2},\) where \(x\) and \(y\) are measured in inches and \(T\) in degrees Celsius. Consider the portion of the plate that lies on the rectangular region \(R=[1,5] \times[3,6]\) a. Write an iterated integral whose value represents the volume under the surface \(T\) over the rectangle \(R\). b. Evaluate the iterated integral you determined in (a). c. Find the area of the rectangle, \(R\). d. Determine the exact average temperature, \(T_{\mathrm{AVG}(R)},\) over the region \(R\)
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