Question

Find the mass and center of mass of the lamina bounded by the graphs of the equations for the given density or densities. (Hint: Some of the integrals are simpler in polar coordinates.) $$ y=\sin \frac{\pi x}{L}, y=0, x=0, x=L, \rho=k y $$

Short answer

The mass of the lamina is zero, hence its center of mass is not defined.

Step-by-step solution

Calculate the Mass of the Lamina

The mass is given by \( M = \int_A \rho \, dA \), where A is the area of the lamina, and \( \rho \) is the density function. For this problem, \(\ rho = k*y \), and the area is given by the integral over the strip from 0 to L (x direction). Hence we have: \[ M = \int_0^L \int_0^{\sin \frac{\pi x}{L}} k*y \, dy \, dx = k \int_0^L x \sin \frac{\pi x}{L} \, dx \]

Compute the Integral

Applying techniques of integration we obtain: \[ M = [ - k L cos (\pi x / L) / \pi]_0^L = 0 \]

Calculate the Center of Mass of the Lamina

The center of mass is given by the coordinates \((x^\prime , y^\prime) \), which are defined by \(x^\prime = \int_A x \rho \, dA / M \) and \(y^\prime = \int_A y \rho \, dA / M \). Since the mass M is zero, the center of mass is not defined.