   Chapter 15.4, Problem 11E

Chapter
Section
Textbook Problem

A lamina occupies the part of the disk x2 + y2 ≤ 1 in the first quadrant. Find its center of mass if the density at any point is proportional to its distance from the x-axis.

To determine

To find: The center of mass of the lamina occupied by the given disk D.

Explanation

Given:

The region D is the disk x2+y21 in the first quadrant.

The density function is proportional to x-axis, that is ρ(x,y)=ky .

Formula used:

The total mass of the lamina is, m=limk,li=1kj=1lρ(xij*,yij*)ΔA=Dρ(x,y)dA .

Here, the density function is given by ρ(x,y) and D  is the region that is occupied by the lamina.

The center of mass of the lamina that occupies the given region D is (x¯,y¯) .

Here, x¯=Mym=1mDxρ(x,y)dA and y¯=Mxm=1mDyρ(x,y)dA

If f is a polar rectangle R given by 0arb,αθβ, where 0βα2π , then, Rf(x,y)dA=αβabf(rcosθ,rsinθ)rdrdθ (1)

If g(x) is the function of x and h(y) is the function of y then,

abcdg(x)h(y)dydx=abg(x)dxcdh(y)dy (2)

Calculation:

Convert into polar coordinates to make the problem easier. So, from the given conditions it is observed that r varies from 0 to 1 and θ varies from 0 to π2 . Then, by (1) the total mass of the lamina is,

m=Dρ(x,y)dA=0π201kydydx=0π201krsinθ(r)drdθ=k0π201r2sinθdrdθ

Integrate with respect to r and θ by using the property (2).

m=k0π2sinθdθ01r2dr=k[cosθ]0π2[r33]01=k(cos(π2)+cos(0))(133033)=k(0+1)(130)

= k3

In order to get the coordinates of the center of mass, find x¯ and y¯ . Therefore, by the equation (1),

x¯=1mDxρ(x,y)dA=1(k3)0π201ky(x)dydx=3kk0π201(rcosθ)(rsinθ)(r)drdθ=30π201r3cosθsinθdrdθ

Integrate with respect to r and θ by using the property (2)

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