Question

Mass In Exercises 11 and \(12,\) find the mass of the surface lamina \(S\) of density \(\rho .\) S: 2 x+3 y+6 z=12, \text { first octant, } \rho(x, y, z)=x^{2}+y^{2}

Short answer

The short answer will be the numerical value obtained after solving the mass integral.

Step-by-step solution

Convert the surface equation into the desired form

The surface S is given by 2x + 3y + 6z = 12, let's isolate z by making remaining terms one side: \(z = 2 - \frac{1}{3}x - \frac{1}{2}y\). So, the surface can be rewritten as \(S: z = 2 - \frac{1}{3}x - \frac{1}{2}y, x,y \geq 0, z \geq 0\).

Define the integral bounds

If z >= 0 and, based on the surface equation, z = 2 - \frac{1}{3}x - \frac{1}{2}y, then x varies from 0 to 6 and y varies from 0 to 4 - 2x/3.

Set up the surface integral for mass

The mass m of the lamina can be found by integrating the density function over the surface. We have \(m = \int \int_S \rho(x, y, z) dS\), where \( \rho(x, y, z) = x^{2} + y^{2} \) is the density of the lamina. In terms of x and y, the mass integral becomes \(m = \int_{0}^{6} \int_{0}^{4 - 2x/3} (x^{2} + y^{2}) \sqrt{1+(dx/dz)^2+(dy/dz)^2} dy dx \), where sqrt indicates the square root, dx/dz and dy/dz are the partial derivatives of z with respect to x and y respectively. This requires calculating the partial derivatives and substituting back in the equation.

Solve the integral

After calculating the derivatives, substituting them and simplifying, solve the integral for m to find the mass of the lamina.