삼중적분

The domain of integration is

$$V=\big\{ \ (x,y,z)\ \big|\ x,y,z\ge 0,\ \tfrac{x}{a}+\tfrac{y}{b}+\tfrac{z}{c}\le 1\ \big\}$$

• In $V$, $\tfrac{z}{c}\le 1-\tfrac{x}{a}-\tfrac{y}{b}$ and $x,y\ge0$, so the biggest value of $z$ in $V$ is achieved when $x=y=0$ and is $c$. Thus, in $V$, $z$ runs from $0$ to $c$.
• For each fixed $0\le z\le c$, $(x,y)$ takes all values in

$$D_z=\big\{ \ (x,y)\ \big|\ x,y\ge 0,\ \tfrac{x}{a}+\tfrac{y}{b}\le 1-\tfrac{z}{c}\ \big\}$$

The biggest value of $y$ on $D_z$ is achieved when $x=0$ and is $b\left(1-\tfrac{z}{c}\right)$. Thus, on $D_z$, $y$ runs from $0$ to $b\left(1-\tfrac{z}{c}\right)$.

• For each fixed $0\le z\le c$ and $0\le y \le b\left(1-\tfrac{z}{c}\right)$, $x$ runs over

$$D_{y,z}=\big\{ \ x\ \big|\ 0\le x \le a\left(1-\tfrac{y}{b}-\tfrac{z}{c}\right)\ \big\}$$

This is pictured in the figure on the right below.

So the specified integral is

$$\begin{aligned} \iiint _R x\ \mathrm{d}V &=\int_0^c \mathrm{d}z \iint _{D_z} \mathrm{d}x\,\mathrm{d}y\ x = \int_0^c \mathrm{d}z\int_0^{b(1-{z\over c})}\mathrm{d}y\int_{D_{y,z}}\mathrm{d}x\ x\\ &=\int_0^c \mathrm{d}z\int_0^{b(1-{z\over c})}\mathrm{d}y \int_0^{a(1-{y\over b}-{z\over c})} \mathrm{d}x\ x =\int_0^c \mathrm{d}z\int_0^{b(1-{z\over c})} \mathrm{d}y\ \frac{a^2}{2} \left(1-\frac{y}{b}-\frac{z}{c}\right)^2 \\ &=\int_0^c \mathrm{d}z\ \left[-\frac{a^2b}{6} \left(1-\frac{y}{b}-\frac{z}{c}\right)^3 \right]_0^{b(1-{z\over c})} =\int_0^c \mathrm{d}z\ \frac{a^2b}{6}\left(1-\frac{z}{c}\right)^3 \\ &=\left[-\frac{a^2bc}{24}\left(1-\frac{z}{c}\right)^4 \right]_0^c =\frac{a^2bc}{24} \end{aligned}$$