Answer to Question #118104 in Calculus for Benjamin

Question

A triangular lamina in the xy -plane such that its vertices are (0,0), (0,1) and (1,0). Suppose that the density function of the lamina is defined by p(x,y)=120xy. What is the total mass of the lamina and the center of gravity?

Answer 1

We assume all units are in S.I unit.

Let's draw the rough sketch of the given triangular lamina.

Given, density of the triangular lamina

$$\rho(x,y)=120xy$$

Since, by definition of density,

$$\frac{dm}{dA}=\rho(x,y)\\ \implies m=\int_{0}^{1}\int_{0}^{1}\rho(x,y)dA=\int_{0}^{1}\int_{0}^{1}\rho(x,y)dxdy\\ \implies m=\int_{0}^{1}\int_{0}^{1}120xy \:dxdy\\\implies m=120\int_{0}^{1}\bigg (y\int_{0}^{1}xdx\bigg)dy=30Kg$$

Let's denote the center of gravity by $(X_G,Y_G)$ .

Now,

$$X_G=\frac{1}{m}\int_{0}^{1}\int_{0}^{1}x\rho(x,y)dxdy\\ \implies X_G=\frac{1}{m}\int_{0}^{1}\int_{0}^{1}x(120xy)dxdy\\ \implies X_G=\frac{120}{m}\int_{0}^{1}\int_{0}^{1}x^2ydxdy\\ \implies X_G=\frac{120}{m}\int_{0}^{1}x^2\bigg(\int_{0}^{1}ydy\bigg)dx\\ \implies X_G=\frac{60}{m}\int_{0}^{1}x^2dx\\ \implies X_G=\frac{20}{m}=\frac{2}{3}m$$

Similarly,

$$Y_G=\frac{1}{m}\int_{0}^{1}\int_{0}^{1}y\rho(x,y)dxdy$$

Clearly, if we replace $x\leftrightarrow y$ and vice versa we get exactly similar expression as of $X_G$ ,thus we conclude that

$$Y_G=\frac{2}{3}m$$

Therefore center of gravity of the given triangular lamina is

$$(X_G,Y_G)=\bigg(\frac{2}{3},\frac{2}{3}\bigg)$$