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.) $$ y=9-x^{2}, y=0, \rho=k y^{2} $$

Short answer

Once you have followed all the steps and performed the integrations and calculations, you will have mass M and position of the center of mass \(\bar{x}, \bar{y}\).

Step-by-step solution

Define the region

The region of the lamina can be found by observing that it is bound by the curves \(y=9-x^{2}\) and \(y=0\). It's clear that the region stretches from one end of the parabola to the other, from \(x=-3\) to \(x=3\).

Find the mass

The mass of the lamina is given by the double integral of its density over the region. Since the boundaries of the region were discussed in Step 1, the double integral can be written as \( M = \int_{-3}^{3} \int_{0}^{9-x^{2}} k y^{2} dy dx = k \int_{-3}^{3} (\frac{1}{3}(9-x^{2})^{3}) dx\). Calculate this integral to get the mass.

Find the moments

The moments \(M_x\) and \(M_y\) about the x- and y-axes respectively are given by double integrals of y times the density and x times the density over the region. Calculate these integrals: \(M_x = \int_{-3}^{3} \int_{0}^{9-x^{2}} y^{3} dy dx = \int_{-3}^{3} (\frac{1}{4}(9-x^{2})^{4}) dx, M_y = \int_{-3}^{3} \int_{0}^{9-x^{2}} y^{2} x dy dx = 0\). The \(M_y\) integral results to zero due to the symmetry of the region around the y-axis.

Compute the center of mass

The coordinates of the center of mass are calculated by \( \bar{x} = M_y/M, \bar{y} = M_x/M \). After substituting the known values from Step 2 and 3, calculate \(\bar{x}\) and \(\bar{y}\) to find the coordinates of the center of mass.