삼중적분

First we have to get some idea as to what $E$ looks like. Here is a sketch.

We are going to need the equation of the plane that contains the points $(-1,0,0)$, $(0,-2,0)$ and $(0,0,3)$. This plane does not contain the origin and so has an equation of the form $ax+by+cz=1$.

• $(-1,0,0)$ lies on the plane $ax+by+cz=1$ if and only if $a(-1)+b(0)+c(0)=1$. So $a=-1$.
• $(0,-2,0)$ lies on the plane $ax+by+cz=1$ if and only if $a(0)+b(-2)+c(0)=1$. So $b=-\frac{1}{2}$.
• $(0,0,3)$ lies on the plane $ax+by+cz=1$ if and only if $a(0)+b(0)+c(3)=1$. So $c=\frac{1}{3}$.

So the plane that contains the points $(-1,0,0)$, $(0,-2,0)$ and $(0,0,3)$ is $-x-\frac{y}{2}+\frac{z}{3}=1$.

We can now get a detailed mathematical description of $E$. A point $(x,y,z)$ is in $E$ if and only if

• $(x,y,z)$ lies above the $xy$–plane, i.e. $z\ge 0$, and
• $(x,y,z)$ lies to the left of the $xz$–plane, i.e. $y\le 0$, and
• $(x,y,z)$ lies behind the $yz$–plane, i.e. $x\le 0$, and
• $(x,y,z)$ lies on the same side of the plane $-x-\frac{y}{2}+\frac{z}{3}=1$ as the origin. That is $-x-\frac{y}{2}+\frac{z}{3}\le 1$. (Go ahead and check that $(0,0,0)$ obeys this inequality.)

So

$$E=\big\{ \ (x,y,z)\ \big|\ x\le 0,\ y\le 0,\ z\ge 0,\ -x-\tfrac{y}{2}+\tfrac{z}{3}\le 1\ \big\}$$

(a) Note that we want the outside integral to be the $x$–integral. On $E$

• $x$ runs from $-1$ to $0$ and
• for each fixed $x$ in that range $(y,z)$ runs over

$$\begin{aligned} E_x=\big\{ \ (y,z)\ \big|\ y\le 0,\ z\ge 0,\ -\tfrac{y}{2}+\tfrac{z}{3}\le 1+x\ \big\} \end{aligned}$$

Here is a sketch of $E_x$.

• On $E_x$, $y$ runs from $-2(1+x)$ to $0$ and
• for each fixed such $y$, $z$ runs from $0$ to $3(1+x+y/2)$

So

$$\begin{aligned} I = \int_{x=-1}^{x=0}\int_{y=-2(1+x)}^{y=0}\int_{z=0}^{z=3(1+x+y/2)} f(x,y,z)\ \mathrm{d}z\,\mathrm{d}y\,\mathrm{d}x \end{aligned}$$

(b) This time we want the outside integral to be the $z$–integral. Looking back at the sketch of $E$, we see that, on $E$,

• $z$ runs from $0$ to $3$ and
• for each fixed $z$ in that range $(x,y)$ runs over

$$\begin{aligned} E_z=\big\{ \ (x,y)\ \big|\ x\le 0,\ y\le 0,\ -x-\tfrac{y}{2}\le 1-\tfrac{z}{3}\ \big\} \end{aligned}$$

Here is a sketch of $E_z$.

• On $E_z$, $x$ runs from $-(1-z/3)$ to $0$ and
• for each fixed such $x$, $y$ runs from $-2(1+x-z/3)$ to $0$

So

$$\begin{aligned} I = \int_{z=0}^{z=3}\int_{x=-(1-z/3)}^{x=0}\int_{y=-2(1+x-z/3)}^{y=0} f(x,y,z)\ \mathrm{d}y\,\mathrm{d}x\,\mathrm{d}z \end{aligned}$$