Find the mass and center of mass of the lamina that occupies the region D and has the given density function

Find the mass and center of mass of the lamina that occupies the region D and has the given density function (Enter your answer as an improper fraction:) D = {(T,u)|0 < v < sin (= I/L) . 0 < =< L}; e (1.y) 19y

The Correct Answer and Explanation is:

To find the mass and the center of mass of the lamina over a region DDD with a given density function, we follow these steps:

Given:

Region D={(x,y)∣0<x<L, 0<y<sin⁡(πxL)}D = \{(x, y) \mid 0 < x < L,\ 0 < y < \sin\left(\frac{\pi x}{L}\right)\}D={(x,y)∣0<x<L, 0<y<sin(Lπx)}
Density function: ρ(x,y)=19y\rho(x, y) = 19yρ(x,y)=19y

Step 1: Mass of the Lamina

The mass mmm of the lamina is given by the double integral: m=∬Dρ(x,y) dA=∫x=0L∫y=0sin⁡(πxL)19y dy dxm = \iint_D \rho(x, y)\, dA = \int_{x=0}^{L} \int_{y=0}^{\sin\left(\frac{\pi x}{L}\right)} 19y\, dy\, dxm=∬Dρ(x,y)dA=∫x=0L∫y=0sin(Lπx)19ydydx

Evaluate the inner integral:

∫y=0sin⁡(πxL)19y dy=19[y22]0sin⁡(πxL)=192sin⁡2(πxL)\int_{y=0}^{\sin\left(\frac{\pi x}{L}\right)} 19y\, dy = 19 \left[ \frac{y^2}{2} \right]_0^{\sin\left(\frac{\pi x}{L}\right)} = \frac{19}{2} \sin^2\left(\frac{\pi x}{L}\right)∫y=0sin(Lπx)19ydy=19[2y2]0sin(Lπx)=219sin2(Lπx)

Now the outer integral:

m=∫0L192sin⁡2(πxL) dxm = \int_0^L \frac{19}{2} \sin^2\left(\frac{\pi x}{L}\right)\, dxm=∫0L219sin2(Lπx)dx

Use the identity: sin⁡2θ=1−cos⁡(2θ)2\sin^2\theta = \frac{1 – \cos(2\theta)}{2}sin2θ=21−cos(2θ) m=192∫0L1−cos⁡(2πxL)2dx=194∫0L(1−cos⁡(2πxL)) dxm = \frac{19}{2} \int_0^L \frac{1 – \cos\left(\frac{2\pi x}{L}\right)}{2} dx = \frac{19}{4} \int_0^L \left(1 – \cos\left(\frac{2\pi x}{L}\right)\right)\, dxm=219∫0L21−cos(L2πx)dx=419∫0L(1−cos(L2πx))dx =194[x−L2πsin⁡(2πxL)]0L=194(L−0)=19L4= \frac{19}{4} \left[ x – \frac{L}{2\pi} \sin\left(\frac{2\pi x}{L}\right) \right]_0^L = \frac{19}{4} \left( L – 0 \right) = \frac{19L}{4}=419[x−2πLsin(L2πx)]0L=419(L−0)=419L

Step 2: Center of Mass

Let (xˉ,yˉ)(\bar{x}, \bar{y})(xˉ,yˉ) be the center of mass.

xˉ=1m∬Dxρ(x,y) dA\bar{x} = \frac{1}{m} \iint_D x \rho(x, y)\, dAxˉ=m1∬Dxρ(x,y)dA

xˉ=1m∫0L∫0sin⁡(πxL)x(19y) dy dx\bar{x} = \frac{1}{m} \int_0^L \int_0^{\sin\left(\frac{\pi x}{L}\right)} x(19y)\, dy\, dxxˉ=m1∫0L∫0sin(Lπx)x(19y)dydx =1m∫0L19x[y22]0sin⁡(πxL)dx=1m∫0L19x2sin⁡2(πxL)dx= \frac{1}{m} \int_0^L 19x \left[\frac{y^2}{2} \right]_0^{\sin\left(\frac{\pi x}{L}\right)} dx = \frac{1}{m} \int_0^L \frac{19x}{2} \sin^2\left(\frac{\pi x}{L}\right) dx=m1∫0L19x[2y2]0sin(Lπx)dx=m1∫0L219xsin2(Lπx)dx

Use substitution u=πxL⇒dx=Lπdu, x=Luπu = \frac{\pi x}{L} \Rightarrow dx = \frac{L}{\pi} du,\ x = \frac{Lu}{\pi}u=Lπx⇒dx=πLdu, x=πLu xˉ=119L4⋅192⋅∫0Lxsin⁡2(πxL)dx=2L∫0Lxsin⁡2(πxL)dx\bar{x} = \frac{1}{\frac{19L}{4}} \cdot \frac{19}{2} \cdot \int_0^L x \sin^2\left(\frac{\pi x}{L}\right) dx = \frac{2}{L} \int_0^L x \sin^2\left(\frac{\pi x}{L}\right) dxxˉ=419L1⋅219⋅∫0Lxsin2(Lπx)dx=L2∫0Lxsin2(Lπx)dx

Using integration by parts or known result: ∫0Lxsin⁡2(πxL)dx=L24\int_0^L x \sin^2\left(\frac{\pi x}{L}\right) dx = \frac{L^2}{4}∫0Lxsin2(Lπx)dx=4L2

So: xˉ=2L⋅L24=L2\bar{x} = \frac{2}{L} \cdot \frac{L^2}{4} = \frac{L}{2}xˉ=L2⋅4L2=2L

yˉ=1m∬Dyρ(x,y) dA=1m∫0L∫0sin⁡(πxL)19y2 dy dx\bar{y} = \frac{1}{m} \iint_D y \rho(x, y)\, dA = \frac{1}{m} \int_0^L \int_0^{\sin\left(\frac{\pi x}{L}\right)} 19y^2\, dy\, dxyˉ=m1∬Dyρ(x,y)dA=m1∫0L∫0sin(Lπx)19y2dydx

∫0sin⁡(πx/L)19y2 dy=19[y33]0sin⁡(πx/L)=193sin⁡3(πxL)\int_0^{\sin(\pi x/L)} 19y^2\, dy = 19 \left[ \frac{y^3}{3} \right]_0^{\sin(\pi x/L)} = \frac{19}{3} \sin^3\left(\frac{\pi x}{L}\right)∫0sin(πx/L)19y2dy=19[3y3]0sin(πx/L)=319sin3(Lπx) yˉ=119L4⋅∫0L193sin⁡3(πxL) dx=4L⋅13⋅∫0Lsin⁡3(πxL) dx\bar{y} = \frac{1}{\frac{19L}{4}} \cdot \int_0^L \frac{19}{3} \sin^3\left(\frac{\pi x}{L}\right)\, dx = \frac{4}{L} \cdot \frac{1}{3} \cdot \int_0^L \sin^3\left(\frac{\pi x}{L}\right)\, dxyˉ=419L1⋅∫0L319sin3(Lπx)dx=L4⋅31⋅∫0Lsin3(Lπx)dx

Use the identity: sin⁡3θ=3sin⁡θ−sin⁡(3θ)4\sin^3\theta = \frac{3\sin\theta – \sin(3\theta)}{4}sin3θ=43sinθ−sin(3θ)

So: ∫0Lsin⁡3(πxL) dx=Lπ∫0πsin⁡3(u) du=Lπ⋅43\int_0^L \sin^3\left(\frac{\pi x}{L}\right)\, dx = \frac{L}{\pi} \int_0^\pi \sin^3(u)\, du = \frac{L}{\pi} \cdot \frac{4}{3}∫0Lsin3(Lπx)dx=πL∫0πsin3(u)du=πL⋅34 yˉ=4L⋅13⋅4L3π=169π\bar{y} = \frac{4}{L} \cdot \frac{1}{3} \cdot \frac{4L}{3\pi} = \frac{16}{9\pi}yˉ=L4⋅31⋅3π4L=9π16

Final Answer:

Mass m=19L4m = \boxed{\frac{19L}{4}}m=419L
Center of Mass:
xˉ=L2, yˉ=169π\bar{x} = \boxed{\frac{L}{2}},\ \bar{y} = \boxed{\frac{16}{9\pi}}xˉ=2L, yˉ=9π16

Explanation

To find the mass and center of mass of a lamina with variable density, we apply double integrals over the defined region. In this case, the region DDD is bounded below by the x-axis and above by the curve y=sin⁡(πxL)y = \sin\left(\frac{\pi x}{L}\right)y=sin(Lπx), for xxx ranging from 0 to LLL. The density function given is ρ(x,y)=19y\rho(x, y) = 19yρ(x,y)=19y, which means density increases with height.

To calculate mass, we integrate the density over the area. The integration begins with respect to yyy, from 0 up to the sine curve. This gives a function of xxx, which we then integrate from 0 to LLL. Using trigonometric identities simplifies the integral, resulting in a mass of 19L4\frac{19L}{4}419L.

The x-coordinate of the center of mass (xˉ\bar{x}xˉ) is found by integrating x⋅ρ(x,y)x \cdot \rho(x, y)x⋅ρ(x,y). Because of symmetry and the nature of the sine function, we find that xˉ\bar{x}xˉ equals L2\frac{L}{2}2L, indicating that the mass is balanced horizontally at the midpoint of the interval.

For the y-coordinate (yˉ\bar{y}yˉ), we integrate y⋅ρ(x,y)y \cdot \rho(x, y)y⋅ρ(x,y). This requires evaluating the cube of the sine function. Using known trigonometric identities and substitutions, we simplify the integral to find yˉ=169π\bar{y} = \frac{16}{9\pi}yˉ=9π16. This lower value reflects that most mass lies closer to the x-axis since density increases with yyy.

This process applies the principles of multivariable calculus and emphasizes the role of both density and shape in determining mass distribution.