( 862, #9) Find the mass and center of mass of the lamina that occupies the region D , which is bounded by the parabola x = y² and the line y = x - 2, and has the density function rho(x,y) = 3.

Solution: By the graph, for each fixed y, x varies from y² to y +2:

The mass of the lamina is 27/2, and its center of mass is ( 8/5, 1/2):

mass = 27/2:

rho := (x,y)->3:
c := y -> y^2: d := y -> y+2:
Integrate_dxdy(rho, -1,2,c,d);

f(x,y) = 3, c(y) = y^2, d(y) = y+2

int(f(x,y), x) = 3*x

int(f(x,y), x = c(y) .. d(y)) = 3*y+6-3*y^2

int( int(f(x,y), x = c(y) ..d(y)), x) = 3/2*y^2+6*y-y^3

int(int(f(x,y), x = c(y) ..d(y)), y = a .. b) = 27/2

xbar calculation

xrho := (x,y) -> x*rho(x,y):
Integrate_dxdy(xrho, -1,2,c,d);

f(x,y) = 3*x, c(y) = y^2 , d(y) = y+2

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

int(f(x,y), x = c(y) .. d(y)) = 3/2*y^2+6*y+6-3/2*y^4

int( int(f(x,y), x = c(y) ..d(y)), x) = 1/2*y^3+3*y^2+6*y-3/10*y^5

int(int(f(x,y), x = c(y) ..d(y)), y = a .. b) = 108/5

xbar = (108/5)/(27/2) = 8/5

ybar calculation

yrho := (x,y) -> y*rho(x,y):
Integrate_dxdy(yrho, -1,2,c,d);

f(x,y) = 3*y, c(y) = y^2, d(y) = y+2

int(f(x,y), x) = 3*y*x

int(f(x,y), x = c(y) .. d(y)) = 3*y^2+6*y-3*y^3

int( int(f(x,y), x = c(y) ..d(y)), x) = y^3+3*y^2-3/4*y^4

int(int(f(x,y), x = c(y) ..d(y)), y = a .. b) = 27/4
ybar = 1/2

by: nl