Triple Integrals

$0 \le z \le 3$.

The shadow is the rectangle $0 \le x \le 2$, $0 \le y \le 1$.

$0 \le y \le 1$, $0 \le x \le 2$: all numbers. And $xy^2z = x \cdot y^2 \cdot z$, so the integral splits into three.

$\displaystyle\int_0^2 x\,dx \cdot \int_0^1 y^2\,dy \cdot \int_0^3 z\,dz = 2 \cdot \tfrac{1}{3} \cdot \tfrac{9}{2} = 3$.

$0 \le z \le 2 - 2x - y$.

Set $z = 0$: the shadow is the triangle under $2x + y = 2$ in the first quadrant.

$0 \le y \le 2 - 2x$, $0 \le x \le 1$.

$\displaystyle\int_0^1\int_0^{2 - 2x} y(2 - 2x - y)\,dy\,dx = \int_0^1 \frac{(2 - 2x)^3}{6}\,dx = \frac{1}{3}$.

$0 \le z \le x$: the plane $z = x$ is the top, $z = 0$ the bottom.

The shadow lies under $y = 4 - x^2$ in the first quadrant of the $xy$-plane.

$0 \le y \le 4 - x^2$, $0 \le x \le 2$.

$\displaystyle\int_0^2\int_0^{4 - x^2} xy\,dy\,dx = \int_0^2 \frac{x(4 - x^2)^2}{2}\,dx = \frac{16}{3}$.

$0 \le x \le 1$, $0 \le y \le 1 - x$, $0 \le z \le 2 - 2x - 2y$.

The tetrahedron $x, y, z \ge 0$, $2x + 2y + z \le 2$, with corners $(0, 0, 0)$, $(1, 0, 0)$, $(0, 1, 0)$, $(0, 0, 2)$.

$0 \le z \le 2$.

At height $z$: the triangle $x, y \ge 0$, $x + y \le 1 - \tfrac{z}{2}$, so $0 \le x \le 1 - \tfrac{z}{2}$.

$0 \le y \le 1 - x - \tfrac{z}{2}$.

$0 \le y \le 2$, $0 \le x \le y$, $0 \le z \le 2 - y$.

Bounded by $x = 0$, $z = 0$, the plane $y = x$ and the plane $y + z = 2$. Corners: $(0, 0, 0)$, $(0, 2, 0)$, $(2, 2, 0)$, $(0, 0, 2)$. So $y$ runs from $y = x$ to $y = 2 - z$.

Remove $y$: $x \le 2 - z$ with $z \ge 0$ gives $0 \le x \le 2$.

For fixed $x$: $0 \le z \le 2 - x$.

$x \le y \le 2 - z$.

$0 \le x \le 1$, $0 \le z \le 1$, $0 \le y \le 2 - x - z$.

Bounded by $x = 0$, $x = 1$, $y = 0$, $z = 0$, $z = 1$ and the plane $x + y + z = 2$.

$0 \le x \le 1$.

For fixed $x$, $y$ runs from $0$ to $2 - x$. The top for $z$ is $z = 1$ while $y \le 1 - x$, then the plane. So split at $y = 1 - x$.

$0 \le z \le 1$ on the first piece; $0 \le z \le 2 - x - y$ on the second.

$0 \le x \le 1$, $0 \le y \le 1$, $0 \le z \le 2$.

The box and $\rho$ are symmetric about $y = \tfrac{1}{2}$: $\bar y = \tfrac{1}{2}$.

$\iiint x\,dV = \tfrac{1}{2} \cdot 1 \cdot 2 = 1$ and $\iiint z\,dV = 1 \cdot 1 \cdot 2 = 2$, so $M = 3$.

$\iiint x(x + z)\,dV = \tfrac{2}{3} + 1 = \tfrac{5}{3}$ and $\iiint z(x + z)\,dV = 1 + \tfrac{8}{3} = \tfrac{11}{3}$. So $\bar x = \tfrac{5}{9}$, $\bar z = \tfrac{11}{9}$.

$0 \le z \le 2\left(1 - \tfrac{x}{3} - y\right)$, $0 \le y \le 1 - \tfrac{x}{3}$, $0 \le x \le 3$.

No symmetry: the three edges on the axes have different lengths.

Constant density, so use the volume: $V = \tfrac{1}{6}(3)(1)(2) = 1$.

$\iiint x\,dV = \tfrac{3}{4}$, $\iiint y\,dV = \tfrac{1}{4}$, $\iiint z\,dV = \tfrac{1}{2}$; divide each by $V = 1$.

$0 \le z \le 1 - x^2$, $-1 \le x \le 1$, $0 \le y \le 2$.

Symmetric about the plane $x = 0$: $\bar x = 0$. Symmetric about the plane $y = 1$: $\bar y = 1$.

$V = \displaystyle\int_0^2\int_{-1}^{1}(1 - x^2)\,dx\,dy = 2 \cdot \tfrac{4}{3} = \tfrac{8}{3}$.

$\displaystyle\iiint z\,dV = \int_0^2\int_{-1}^{1}\tfrac{(1 - x^2)^2}{2}\,dx\,dy = \tfrac{16}{15}$, so $\bar z = \dfrac{16/15}{8/3} = \dfrac{2}{5}$.