Question

Find the mass and center of mass of the lamina that occupies the region D and has the given density function rho. D is bounded by y=1−x²and y=0;p(x,y)=11ky m= (xˉyˉ)=()

Answer 1

The mass and center of mass of the lamina that occupies the region D and has the given density function rho is:

`\[ m = (11)/(24k), \quad \left(\bar{x}, \bar{y}\right) = \left((5)/(7), (2)/(5)\right) \]`

Step-by-step explanation:

To find the mass
`\(m\) and the center of mass \((\bar{x}, \bar{y})\)` of the lamina occupying region
`\(D\),` we integrate the given density function
`\(p(x, y)\) over \(D\).`The bounds of
`\(D\) are \(y = 0\) and \(y = 1 - x^2\).` The density function is
`\(p(x, y) = (11)/(ky)\). The mass \(m\)` is obtained by integrating
`\(p(x, y)\) over \(D\)` with respect to
`\(x\) and \(y\)`, and the center of mass
`\((\bar{x}, \bar{y})\)` is found using the formulas
`\(\bar{x} = (1)/(m) \int\int_D xp(x, y) \,dx\,dy\) and \(\bar{y} = (1)/(m) \int\int_D yp(x, y) \,dx\,dy\).`

The integration involves calculating double integrals. The limits of integration for
`\(x\) are \(-1\) to \(1\), and for \(y\), it is \(0\) to \(1 - x^2\).` The integral of
`\(p(x, y)\) results in \((11)/(24k)\)`, providing the mass
`\(m\)`. Similarly, the integrals involving
`\(x\) and \(y\)` in the center of mass formulas yield
`\((5)/(7)\) and \((2)/(5)\)` respectively, giving the center of mass
`\((\bar{x}, \bar{y})\).`

In summary, the mass of the lamina is
`\((11)/(24k)\),` and its center of mass is located at
`\(\left((5)/(7), (2)/(5)\right)\).`