Question

Find the mass and center of mass of the lamina for each density. \(R:\) triangle with vertices \((0,0),(0, a),(a, 0)\) (a) \(\rho=k\) (b) \(\rho=x^{2}+y^{2}\)

Short answer

For density \(\rho = k\), the mass is \(M = \frac{1}{2}ka^2\) and the center of mass is \((\bar{x}, \bar{y}) = (\frac{a}{3}, \frac{a}{3})\). For density \(\rho = x^{2}+y^{2}\), the mass is \(M = \frac{1}{6}a^4\) and the center of mass is \((\bar{x}, \bar{y}) = (\frac{a}{4}, \frac{a}{4})\).

Step-by-step solution

find the mass with density \(\rho = k\)

The mass is equal to the integral over area R of density. We calculate the integral \(\int_0^a \int_0^{a-x} k dy dx\), where k is a constant. The mass (M) equals to \(M = \frac{1}{2}ka^2\).

find the center of mass with density \(\rho = k\)

We find the center of mass using the formula \((\bar{x}, \bar{y}) = (\frac{1}{M} \int_R x\rho(x,y) dA, \frac{1}{M} \int_R y\rho(x,y) dA)\). The center of mass equals to \((\bar{x}, \bar{y}) = (\frac{a}{3}, \frac{a}{3})\).

find the mass with density \(\rho = x^{2}+y^{2}\)

Now we calculate the mass using density function \(\rho = x^{2}+y^{2}\). We need to calculate the integral \(\int_0^a \int_0^{a-x} (x^{2}+y^{2}) dy dx\). This gives us \(M = \frac{1}{6}a^4\).

find the center of mass with density \(\rho = x^{2}+y^{2}\)

Using the formula for the center of mass like in step 2, we find that the center of mass equals to \((\bar{x}, \bar{y}) = (\frac{a}{4}, \frac{a}{4})\).