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.) $$ y=\sqrt{x}, y=0, x=4, \rho=k x y $$

Short answer

The mass of lamina is 8k and the center of mass is located at the point (8/15, 4/15).

Step-by-step solution

Set up the integral for the mass

First, make use of the given density function \(\rho=k x y\) and the formula for mass \(M=\int\int_R ρ(x,y) dA\), where R represents the region in the xy-plane. The region R is bounded by y=0, y= \(\sqrt{x}\) , and x=4, which is a region in the first quadrant from x=0 to x=4. So, the mass M becomes \(M= \int_{0}^{4}\int_{0}^{\sqrt{x}} kxy \, dy \, dx\).

Evaluate the integral for mass

Now, evaluate the double integral through simple integration. The inner integral, with respect to y, is 0.5*k*x*y² evaluated from y=0 to y= \(\sqrt{x}\). Substituting these limits and simplifying followed by the outer integral calculation, we obtain the mass \(M=8k\).

Set up the integrals for the center of mass

To find the center of mass, we need to find the x-coordinate (\(x̄\)) and the y-coordinate (\(ȳ\)) of the center of mass. The formulas are \(x̄ = (1/M) * \int \int_R x*ρ(x,y)dA \) and \(ȳ = (1/M) *\int \int_R y*ρ(x,y) dA\). Following the same steps as in step 1, we set up the double integrals.

Evaluate the integrals for the center of mass

Evaluating the double integrals for \(x̄\) and \(ȳ\), we can find the coordinates of the center of mass. After substituting the limits and simplifying, we find \(x̄ = 64k/15M\) and \(ȳ = 32k/15M\).

Normalizing the obtained center of mass coordinates

As a final step, taking M=8k (found in Step 2), substituting it into each coordinate and simplifying, we find that the center of mass of the plate region is located at (\(x̄\), \(ȳ\)) = (8/15, 4/15).