( 862, #5) Find the mass and center of mass of the lamina that occupies the region D , which is the triangular region with vertices ( 0, 0), (2, 1), (0, 3), and has the density function rho(x, y) = x + y.

Solution: The equations governing the domain are given below. Note that for fixed x, y varies from x/2 to -x + 3.

The mass of the lamina is 6, and its center of mass is ( 3/4, 3/2):

mass = 6:

rho := (x,y)->x+y:
c := x -> x/2: d := x -> 3-x:
Integrate_dydx(rho, 0,2,c,d);

f(x,y) = x+y, c(x) = 1/2*x, d(x) = 3-x

int(f(x,y), y) = x*y+1/2*y^2

int(f(x,y), y = c(x) .. d(x)) = -9/8*x^2+9/2

int( int(f(x,y), y = c(x) ..d(x)), x) = -3/8*x^3+9/2*x

int(int(f(x,y), y = c(x) ..d(x)), x = a .. b) = 6

xbar calculation
xrho := (x,y) -> x*rho(x,y):
Integrate_dydx(xrho, 0,2,c,d);

f(x,y) = x*(x+y), c(x) = 1/2*x, d(x) = 3-x

int(f(x,y), y) = x*(x*y+1/2*y^2)

int(f(x,y), y = c(x) .. d(x)) = -9/8*x^3+9/2*x

int( int(f(x,y), y = c(x) ..d(x)), x) = -9/32*x^4+9/4*x^2

int(int(f(x,y), y = c(x) ..d(x)), x = a .. b) = 9/2

xbar = (9/2)/6 = 3/4

ybar calculation

yrho := (x,y) -> y*rho(x,y):
Integrate_dydx(yrho, 0,2,c,d);

f(x,y) = y*(x+y), c(x) = 1/2*x, d(x) = 3-x

int(f(x,y), y) = 1/3*y^3+1/2*x*y^2

int(f(x,y), y = c(x) .. d(x)) = 9-9/2*x

int( int(f(x,y), y = c(x) ..d(x)), x) = 9*x-9/4*x^2

int(int(f(x,y), y = c(x) ..d(x)), x = a .. b) = 9

ybar = 9/6 =3/2


by: nl