( 862, #3) Find the mass and center of mass of the lamina that occupies the region
D = {(x,y) | -1 <= x <= 1, 0 <= y <=1} and has the density function rho(x,y) = x².

The geometry:

rho := (x,y) ->x^2 :
c := x-> 0: d := x -> 1:
Graph_dydx(rho, -1,1,c,d,10);

Solution:

mass = 2/3

Integrate_dydx(rho,-1,1,c,d);

f(x,y) = x^2, c(x) = 0, d(x) = 1

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

int(f(x,y), y = c(x) .. d(x)) = x^2

int( int(f(x,y), y = c(x) ..d(x)), x) = 1/3*x^3

int(int(f(x,y), y = c(x) ..d(x)), x = a .. b) = 2/3

xbar calculation.

Since the mass is symmetric with respect to the x-axis we will havwexbar = 0. This
can be checked out:

xrho := (x,y) -> x*rho(x,y):

Integrate_dydx(xrho,-1,1,c,d);

f(x,y) = x^3, c(x) 0, d(x) = 1

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

int(f(x,y), y = c(x) .. d(x)) = x^3

int( int(f(x,y), y = c(x) ..d(x)), x) = 1/4*x^4

int(int(f(x,y), y = c(x) ..d(x)), x = a .. b) = 0

ybar calculation. From the graph, we should have ybar = 1/2. It is:

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

f(x,y) = y*x^2,c(x)= 0,d(x) = 1

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

int(f(x,y), y = c(x) .. d(x)) = 1/2*x^2

int( int(f(x,y), y = c(x) ..d(x)), x) = 1/6*x^3

,int(int(f(x,y), y = c(x) ..d(x)), x = a .. b) = 1/3

ybar = (1/3)/(2/3) = 1/2

by: nl