Find the mass of the solid bounded by the planes $x+z=1, x-z=-1, y=0$, and the surface $y=\sqrt{z}$. The density of the solid is $16 \mathrm{y}+6$
Solution
Step 1 ：Given the solid bounded by the planes \(x+z=1\), \(x-z=-1\), \(y=0\), and the surface \(y=\sqrt{z}\). The density of the solid is \(16y + 6\).
Step 2 ：The mass of a solid with a given density function is found by integrating the density function over the volume of the solid.
Step 3 ：We can express the volume in terms of cylindrical coordinates \((r, \theta, z)\), where \(r\) is the distance from the origin to the point \((x, y, z)\), \(\theta\) is the angle between the positive x-axis and the line from the origin to the point \((x, y, z)\), and \(z\) is the height above the xy-plane.
Step 4 ：The limits of integration for \(r\) are \(0\) to \(\sqrt{z}\), for \(\theta\) are \(0\) to \(2\pi\), and for \(z\) are \(0\) to \(1\).
Step 5 ：The mass is then given by the triple integral of the density function over these limits of integration.
Step 6 ：The density function is \(16\sqrt{z} + 6\).
Step 7 ：By integrating the density function over the volume of the solid, we find that the mass is \(\frac{47\pi}{5}\).
Step 8 ：\(\boxed{\text{Final Answer: The mass of the solid is }\frac{47\pi}{5}}\)
