Mass of solid in 3 dimensions

Question

Find the mass of the right pyramid that has a square base in the $xy$-plane. $-1<x<1, -1<y<1$, the vertex at $(0,0,8)$ and density function $f(x,y) = x^2$

Answer 1

For $z \in [0,8]$, the cross-section is the square $[-k,k] \times [-k,k] \times \{z\}$, where $k$ decreases linearly from $1$ at $z = 0$ to $0$ at $z = 8$. We deduce from this that $k = 1-\frac{1}{8}z$, so the mass is $$\newcommand{\dif}{\mathop{}\!\mathrm{d}} \begin{align*}&\int_{z=0}^{z=8}\int_{y = -\left(1-\frac{1}{8}z\right)}^{y = 1-\frac{1}{8}z}\int_{x = -\left(1-\frac{1}{8}z\right)}^{x = 1-\frac{1}{8}z}x^2 \dif x \dif y \dif z \\ &= \frac{2}{3}\int_{z=0}^{z=8}\int_{y = -\left(1-\frac{1}{8}z\right)}^{y = 1-\frac{1}{8}z}\left(1-\frac{1}{8}z\right)^3 \dif y \dif z \\ &= \frac{4}{3}\int_{z=0}^{z=8}\left(1-\frac{1}{8}z\right)^4 \dif z \\ &= \frac{32}{15}.\end{align*}$$