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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

Find the center of mass of a lamina in the shape of an isosceles right triangle with equal sides of length a if the density at any point is proportional to the square of the distance from the vertex opposite the hypotenuse.

To determine

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

Explanation

Given:

The region D is the isosceles triangle with equal sides of length a.

The density function is proportional to the square of the distance from the vertex opposite to the hypotenuse, that is ρ(x,y)=k(x2+y2) .

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

Calculation:

From the given conditions it is observed that x varies from 0 to a and y varies from 0 to ax . Then, the total mass of the lamina is,

m=Dρ(x,y)dA=0a0axk(x2+y2)dydx

Integrate with respect to y and apply the corresponding limit.

m=k0a(x2y+y33)0axdx=k0a[(x2(ax)+(ax)33)(x2(0)+(0)33)]dx=k0a[ax2x3+a333a2x3+3ax23x3300]dx=k0a[ax2x3+a33a2x+ax2x33]dx

    = k0a(2ax2a2x+a334x33)dx

Integrate with respect to x using the technique of integration by parts.

Let u=x .

Then, dv=e2xdx .

Thus, the total mass of the lamina is obtained as follows.

m=k[2ax33a2x22+a3x34x43(4)]0a=k[(2a(a)33a2(a)22+a3(a)3(a)43)(2a(0)33a2(0)22+a3(0)3(0)43)]=k[(2a43a42+a43a43)(00+00)]=k(2a43a42)

Apply the limit of x.

m=k(4a43a46)=ka46=ka46

In order to get the coordinates of the center of mass, find x¯ and y¯ .

x¯=1mDxρ(x,y)dA=1(ka46)0a0axk(x2+y2)(x)dydx=6kka40a0ax(x3+xy2)dydx=6a40a0ax(x3+xy2)dydx

Integrate with respect to y and apply the corresponding limit.

x¯=6a40a(x3y+xy33)0axdx=6a40a[(x3(ax)+x(ax)33)(x3(0)+x(0)33)]dx=6a40a[(ax3x4+x(a33a2x+3ax2x3)3)(0+0)]dx=6a40a[ax3x4+a3x33a2x23+3ax33x43]dx

Further simplify the terms as shown below

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