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

5. D is the triangular region with vertices (0, 0), (2, 1), (0, 3); ρ(x, y) = x + y

To determine

To find: The total mass and the center of mass of the lamina.

Explanation

Given:

The region D is the triangle with vertices (0,0),(2,1),(0,3).

The density function is ρ(x,y)=x+y.

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 information, it is clear that one of the sides of the triangle is y=0. Equations of other two lines ( Say l1=(0,0) to (2,1) and l2=(2,1) to (0,3)) are found as follows.

The slope of the line l1 is,

m1=y2y1x2x1=1020=12

The equation of the line l1 is,

yy1=m(xx1)y0=12(x0)y=x2

Similarly, the slope of the line l2 is,

m1=y2y1x2x1=3102=22=1

The equation of the line l1 is,

yy1=m(xx1)y3=(1)(x0)y=3x

Thus, the equations of the sides of the triangle are, x2 and 3x. Then, the total mass of the lamina is,

m=Dρ(x,y)dA=02x23x(x+y)dydx

Integrate with respect to y and apply the corresponding limit.

m=02(xy+y22)x23xdx=02[(x(3x)+(3x)22)(x(x2)+(x2)22)]dx=02[3xx2+926x2+x22x22x22(4)]dx=02(92x28x2)dx

Integrate with respect to x and apply the corresponding limit.

m=[9x29x38(3)]02=(9(2)29(2)324)(9(0)29(0)324)=(97224)(00)=93=6

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

x¯=1mDxρ(x,y)dA=1602x23xx(x+y)dydx=1602x23x(x2+xy)dydx

Integrate with respect to y and apply the corresponding limit

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