Question

Find the mass of the lamina described by the inequalities, given that its density is \(\rho(x, y)=x y .\) (Hint: Some of the integrals are simpler in polar coordinates.) $$ x \geq 0,0 \leq y \leq 9-x^{2} $$

Short answer

The mass of the lamina is found by evaluating the given double integral. This value represents the mass of the lamina over the region covered by it.

Step-by-step solution

Define Region

First, define the region of integration. It is given by the inequalities: \(x \geq 0\), and \(0 \leq y \leq 9-x^{2}\).

Convert to polar coordinates

The conversion to polar coordinates will involve replacing \(x\) with \(rcos(\theta)\) and \(y\) with \(rsin(\theta)\). The inequality \(x \geq 0\) becomes \(rcos(\theta) \geq 0\), which implies that \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\). As \(0 \leq y \leq 9-x^{2}\), which after conversion becomes \(0 \leq rsin(\theta) \leq 9-r^{2}cos^{2}(\theta)\). As \(rsin(\theta)\) can't be negative, equating \(rsin(\theta) = 9-r^{2}cos^{2}(\theta)\) will give, \( r = 3cos(\theta)\). Hence, the bounds of \(r\) are \(0 \leq r \leq 3cos(\theta)\).

Compute integral

Replace the density function in polar coordinates, getting: \(\rho(r, \theta) = r^2 sin(\theta) cos(\theta)\). Then compute the double integral over the region expressed in polar coordinates: \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{3cos(\theta)} r^2 sin(\theta) cos(\theta) * r dr d\theta\). The \(r\) outside the integral represents the Jacobian determinant for the change of variables to polar coordinates.

Evaluating the integral

After evaluating the inner integral, followed by the outer integral one will compute the final value.