Answer to Question #118199 in Calculus for Ferdie

Question

A triangular lamina in the Ty-plane such tha
its vertices are (0,0), (0,1) and (1,0). Suppose

that the density function of the lamina is
y) = 63ry g
defined by p(x.
centimetre. Use the information given to
answer the following questions (2 decimal
places):

The total mass of the lamina is
grams.
The moment about the x -axis of the lamina

The Moment about the y -axis of the lamina

The centre of gravity of the lamina ( x, y) is (

Answer 1

Area of the triangle having vertices (0,0), (1,0) and (0,1) is $\frac{1}{2}$ square unit.

The equation of hypotenuse of triangle is $y = -x+1$ .

Hence the region D is the triangle enclosed by the lines :

$x =0, y = 0, x+y = 1$ .

a) The total mass of the lamina is $m = \int\int_D \rho (x,y) dA$.

Given $\rho(x,y) = 63 y$.

So, $m = \int_0^1 \int_0^{-x+1} 63 y dydx = \int_0^1 [63 \frac{y^2}{2}]_0^{-x+1} dx = [\frac{63}{2} \frac{(1-x)^3}{-3}]_0^1 = \frac{63}{6}$.

b) The center of mass $(\bar{x},\bar{y})$ is given by

$\bar{x} = \frac{1}{m} \int\int_D x\rho(x,y)dA , \bar{y} = \frac{1}{m} \int\int_D y\rho(x,y)dA$.

Now,

$\int\int_D x\rho(x,y)dA = \int_0^1 \int_0^{-x+1} 63x y dydx = \int_0^1 [63x \frac{y^2}{2}]_0^{-x+1} dx \\ = \frac{63}{2} \int_0^1 x(1-x)^2 dx = \frac{63}{2} \int_0^1 (x^3-2x^2 +x) dx = \frac{63}{2} [\frac{x^4}{4} - \frac{2x^3}{3}+\frac{x^2}{2}]_0^1 = \frac{63}{24}$

and

$\int\int_D y\rho(x,y)dA = \int_0^1 \int_0^{-x+1} 63 y^2 dydx = \int_0^1 [63 \frac{y^3}{3}]_0^{-x+1} dx \\ = \frac{63}{3} \int_0^1 (1-x)^3 dx = \frac{63}{3} [\frac{(1-x)^4}{(-4)}]_0^1 = \frac{63}{12}$

So, $\bar{x} = \frac{6}{63} \times\frac{63}{24} = \frac{1}{4}, \ and \ \bar{y} = \frac{6}{63} \times\frac{63}{12} = \frac{1}{2}$ .