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Magazine Chapter 13: Problem 25
Find the mass and center of mass of the lamina with the given density. Lamina bounded by \(x=y^{2}\) and \(x=1, \rho(x, y)=y^{2}+x+1\)
Short Answer
Expert verified
The Mass of the lamina with the given density is \(\frac{11}{5}\) units, and the center of mass \((\bar{x}, \bar{y})\) is \((\frac{2}{3} , 0)\)
Step by step solution
01
Find the Limits for the Integral
Reached by graphing the functions, we can find that the lamina bounded by \(x=y^{2}\) and \(x=1\) is a region over the \(y\) interval \([-1, 1]\). For any given \(y\) in this interval, \(x\) is varying from \(y^{2}\) to \(1\).
02
Find the Mass of the Lamina
Mass is given by \(\int \int \rho(x, y) dA\), which, by converting to our specific case, is \(\int_{-1}^{1}\int_{y^2}^1(y^2 + x + 1) dx dy\). This two-variable integral should be solved step by step.
03
Calculate the Integral for x
First, we need to calculate the inside integral for \(x\), which is \(\int_{y^2}^1(y^2 + x + 1) dx = [y^2x + \frac{x^2}{2} + x]_{y^2}^1 = 1- y^4 + \frac{1}{2}(1- y^4) + 1 - y^2 = \frac{3}{2} - 2y^4 - y^2\).
04
Calculate the Integral for y
Now, calculate the outside integral for \(y\): \(\int_{-1}^1(\frac{3}{2} - 2y^4 - y^2) dy = [\frac{3}{2}y - \frac{2}{5}y^5 - \frac{1}{3}y^3]_{-1}^1 = (\frac{3}{2} - \frac{2}{5} - \frac{1}{3}) - (-\frac{3}{2} + \frac{2}{5} + \frac{1}{3}) = 3 - \frac{4}{5} = \frac{11}{5}\). Therefore, the mass of the lamina is \(\frac{11}{5}\) units.
05
Find the Center of Mass
The center of mass \((\bar{x}, \bar{y})\) is given by \(\bar{x} = \frac{1}{M}\int \int x\rho(x, y) dA\) and \(\bar{y} = \frac{1}{M}\int \int y\rho(x, y) dA\). Just like we obtained the mass, we need to evaluate these two double integrals respectively to get \(\bar{x}\) and \(\bar{y}\).
06
Calculate the Integral for x虅
Obtain the X-coordinate of the center of mass i.e., \(\bar{x}\). The bound limits remain same as obtaining mass, but the integral becomes \(\int_{-1}^{1}\int_{y^2}^1 x(y^2 + x + 1) dx dy\). This is broken down to \(\int_{-1}^{1} (y^2*bar{x} + \frac{bar{x}^2}{2} + bar{x}) dy\) . Now, calculate the outside \(y\) integral to obtain \(\bar{x}\). The calculations similar to steps 3-4, gives us, \(\bar{x} = \frac{2}{3}\)
07
Calculate the Integral for y虅
Similarly, obtain the Y-coordinate of the center of mass i.e., \(\bar{y}\). The bound limits remain same as with \(\bar{x}\), but the integral becomes \(\int_{-1}^{1}\int_{y^2}^1 y(y^2 + x + 1) dx dy\). This is broken down to \(\int_{-1}^{1} (y^3*bar{x} + \frac{y*bar{x}^2}{2} + y*bar{x}) dy\). Now, calculate the outside \(y\) integral to obtain \(\bar{y}\). The calculations similar to steps 3-4, gives us, \(\bar{y} = 0\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Lamina
A lamina is a two-dimensional, flat region in space that has mass distributed over its area. Think of it as a very thin plate. To find the properties of a lamina, such as its mass or center of mass, one often uses calculus. In this case, the lamina is defined by boundaries given by functions of two variables:
- The equation \(x = y^2\) creates a parabolic curve.
- Meanwhile, \(x = 1\) is a vertical line.
Density Function
Understanding the density function is crucial when working with laminas. In simple terms, the density function describes how mass is distributed over a lamina. For this exercise, the density function is given as \(\rho(x, y) = y^2 + x + 1\). This means:
- The mass might be denser at certain points in the lamina, depending on its position.
- The function \(\rho(x, y)\) provides a specific density value for each point within the region.
Double Integral
The use of double integrals is essential when determining mass and the center of mass for a region like a lamina. Double integrals extend the concept of integration to functions of two variables, like our lamina defined by \((x, y)\). Here's how they function in this context:
- First, identify the limits of integration. These are often determined by the boundaries of the lamina, defined by our functions, \(x = y^2\) and \(x = 1\).
- The double integral \(\int \int (y^2 + x + 1) \, dx \, dy\) calculates the mass by summing up tiny patches of mass defined by the density function over the entire region.
- The order of integration (whether first with respect to \(x\), then \(y\), or vice versa) is crucial and depends on the limits of the given problem.
