Answer to Question #161475 in Calculus for Mano

Question

Find the mass of a triangular lamina whose vertices are (0,0), (1, 1), (0.1). if the density of the lamina is p(x, y) = sin (y^2).

Answer 1

Consider the triangular lamina as shown in the figure below:

The region $R$ in cartesian coordinates is given by,

$R=\{(x,y)|0\eqslantless y \eqslantless1, 0 \eqslantless x \eqslantless y \}$

The mass of the lamina with density $\rho(x,y)=sin(y^2)$ is evaluated as,

Mass$(m)=\iint_{R}\rho(x,y)dA$

$=\int_{y=0}^{1}\int_{x=0}^{y}sin(y^2)dxdy$

$=\int_{y=0}^{1}sin(y^2)[x]_{x=0}^{y}dy$

$=\int_{y=0}^{1}ysin(y^2)dy$

Let $u=y^2$ , then $\frac{1}{2}du=ydy$ , $0\eqslantless u \eqslantless 1$ , so the integral is,

$=\int_{u=0}^{1}\frac{1}{2}sin(u)du$

$=\frac{1}{2}[-cos(u)]_{=0}^{1}$

$=\frac{1}{2}(1-cos(1)) \approx 0.22984$

Therefore, the mass of the lamina is $m=\frac{1}{2}(1-cos(1)) \approx 0.22984$ .