( 862, #7) Find the mass and center of mass of the lamina that occupies the region D , which is the region in the first quadrant bounded by the parabola y = x² and the line y = 1, and has the density function rho(x,y) = x y..

Solution: The region D is sketched below. Note that for fixed x, y varies from x² to 1

The mass of the lamina is 1/6 and its center of mass is ( 4/7, 3 /4):

mass:

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

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

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

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

int( int(f(x,y), y = c(x) ..d(x)), x) = 1/4*x^2-1/12*x^6

int(int(f(x,y), y = c(x) ..d(x)), x = a .. b) = 1/6

xbar calculation:

xrho := (x,y) -> x*rho(x,y):
Integrate_dydx(xrho, 0,1,c,d);
f(x,y) = x^2,*y c(x) = x^2 d(x), = 1

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

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

int( int(f(x,y), y = c(x) ..d(x)), x) = 1/6*x^3-1/14*x^7

int(int(f(x,y), y = c(x) ..d(x)), x = a .. b) = 2/21

xbar = (2/21)/(1/6) = 4/7

ybar calculation:
yrho := (x,y) -> y*rho(x,y):
Integrate_dydx(yrho, 0,1,c,d);

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

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

int(f(x,y), y = c(x) .. d(x)) = 1/3*x-1/3*x^7

int( int(f(x,y), y = c(x) ..d(x)), x) = 1/6*x^2-1/24*x^8

int(int(f(x,y), y = c(x) ..d(x)), x = a .. b) = 1/8
ybar = (1/8)/(1/6) = 3/4


by: nl