Volume of a pyramid whose vertices are $(0,0,0) , (0,2,1),(2,0,0),(1,0,2)$. 
Find the volume of a pyramid whose vertices are $(0,0,0) , (0,2,1),(2,0,0),(1,0,2)$.


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*I have found out that it should be $\dfrac43$ by using the elementary approach (finding the base area and multiply by the altitude over three).

*However, I ought to find it using triple integral.

*So, we have $3$ points located on $y=0$ plane and the other on $y=2$ plane. Therefore, $y \in [0,2]$.

*Also, looking at the coordinates, I thought $x\in [0,2]$ too. But I'm not sure.

*About $z$, however, I barely have any ideas; maybe splitting the figure into some pieces. But that doesn't seem to be correct and optimal approach.

*Together with this, I would really appreciate the general case ideas on the problem of the similar kind (possibly a proof of the formula here using triple integral)

 A: To express this as a triple integral, we'll want to know equations for the planes containing the four faces. Let's call $O=(0,0,0), A=(0,2,1), B=(2,0,0), C=(1,0,2)$. One way to find a plane equation is to calculate the cross product of two vectors to find a vector perpendicular to the plane. Some of these are fairly easy just to guess and verify. In any case, the planes are:
$$ \begin{align*}
OAB&: & y-2z &= 0 \\
OAC&: & 4x+y-2z &= 0 \\
OBC&: & y &= 0 \\
ABC&: & 4x+3y+2z &= 8
 \end{align*} $$
Now to decide the order of integration.
If we integrate over $x$ first, the projections of the points onto the $yz$-plane are
$$O_{yz} = (0,0), A_{yz} = (2,1), B_{yz} = (0,0), C_{yz} = (0,2)$$
Since we have $O_{yz}=B_{yz}$, the edge $OB$ projects to that single point, the face $\triangle OAB$ projects to the line segment $O_{yz}A_{yz}$, and the face $\triangle OBC$ projects to the line segment $O_{yz}C_{yz}$. The projection is a single triangle $\triangle O_{yz}A_{yz}C_{yz}$, and won't need to split into pieces.
If we integrate over $y$ first, the projections on the $xz$-plane are
$$O_{xz} = (0,0), A_{xz} = (0,1), B_{xz} = (2,0), C_{xz} = (1,2)$$
This forms a convex quadrilateral with no parallel sides and just one right angle. Each triangular face of the tetrahedron projects onto a triangle sharing two sides with the quadrilateral and one of its diagonals as the third side. So equations for the lower or upper limits for $y$ will change when crossing either diagonal, and we would need to split into four pieces to write the triple integrals.
If we integrate over $z$ first, the projections of the points onto the $xy$-plane are
$$ O_{xy} = (0,0), A_{xy} = (0,2), B_{xy} = (2,0), C_{xy} = (1,0) $$
Point $C_{xy}$ is on line segment $O_{xy}B_{xy}$, so the face $\triangle OBC$ projects to the same line segment $O_{xy}B_{xy}$, and the tetrahedron projects to $\triangle O_{xy}A_{xy}B_{xy}$. Equations for the upper limit of $z$ will change when crossing the line segment $A_{xy}C_{xy}$, and we would need to split into two pieces to write the triple integrals.
Obviously integrating over $x$ first will be easiest. Since planes $OAB$ and $OBC$ projected to lines on the $yz$-plane, we just need to find $x$ on planes $OAC$ and $ABC$.
$$ \begin{align*}
OAC&: & x &= -\frac{y}{4} + \frac{z}{2} \\
ABC&: & x &= 2 -\frac{3y}{4} - \frac{z}{2} 
\end{align*} $$
Again, the projected region of integration on the $yz$-plane after integrating with respect to $x$ is $\triangle O_{yz}A_{yz}C_{yz}$. To find the lines containing the triangle's sides, first note $O_{yz}A_{yz}$ has the same equation as plane $OAB$, and $O_{yz}C_{yz}$ has the same equation as plane $OBC$, since these planes projected to just lines. $A_{yz}C_{yz}$ is the intersection of plane $OAC$ and plane $ABC$, so take those two equations as simultaneously true, eliminate $x$, and simplify. These lines are:
$$ \begin{align*}
O_{yz}A_{yz}&: & y - 2z &= 0 \\
O_{yz}C_{yz}&: & y &= 0 \\
A_{yz}C_{yz}&: & y + 2z &= 4
\end{align*} $$
Side $O_{yz}C_{yz}$ is on the $z$-axis. Integrating with respect to $y$ before integrating with respect to $z$, the upper bound for $y$ would be on segment $O_{yz}A_{yz}$ if $z<1$ or on segment $A_{yz}C_{yz}$ if $z>1$, so we would need to split into two pieces to write the triple integrals.
Or integrating with respect to $z$ before integrating with respect to $y$, the lower bound for $z$ is always on $O_{yz}A_{yz}$ and the upper bound is always on $A_{yz}C_{yz}$, so splitting the region is not needed. This will be easiest, so solve for $z$:
$$ \begin{align*}
O_{yz}A_{yz}&: & z &= \frac{y}{2} \\
A_{yz}C_{yz}&: & z &= 2 - \frac{y}{2}
\end{align*} $$
The overall minimum for $y$ is $y=0$, anywhere on the line segment $O_{yz}C_{yz}$. (In 3D, $y=0$ anywhere on face $\triangle OBC$.) The overall maximum for $y$ is $y=2$, at point $A_{yz}$. (In 3D, at point $A$ only.) We could also conclude this from looking at the original four vertices $O,A,B,C$.
Now we can work back to make sure we get the limits in lesser-to-greater order. $0 \leq y \leq 2$, so $y/2 \leq 2-y/2$ and the $z$ limits given $y$ are
$$\frac{y}{2} \leq z \leq 2-\frac{y}{2}$$
Then
$$ \left( 2 - \frac{3y}{4} - \frac{z}{2} \right) - 
\left( -\frac{y}{4} + \frac{z}{2} \right) = 2 - \frac{y}{2} - z \geq 0 $$
so the $x$ limits given $y$ and $z$ are
$$ -\frac{y}{4} + \frac{z}{2} \leq x \leq 2 - \frac{3y}{4} - \frac{z}{2} $$
Finally,
$$ V = \int_0^2 \int_{y/2}^{2-y/2} \int_{-y/4+z/2}^{2-3y/4-z/2} 1\, dx\, dz\, dy $$
A: I will leave out some details for you to work out.
Coordinates of vertices - $A(0, 0, 0), B(0, 2, 1), C(2, 0, 0), D(1, 0, 2)$
Step$1$: find equations of planes. For example, to obtain equation of plane in which points $B, C$ and $D$ lie,
$\vec {CB} = (-2, 2, 1), \vec {CD} = (-1, 0, 2)$. Take the cross product to get the normal vector to the plane which is $(4,3,2)$. So equation of plane is,
$4 (x-2) + 3 (y-0) + 2 (z-0) = 0 \implies 4x + 3y + 2z = 8$
At the end of step $1$, we have (leave details for you to work out),
Plane thru $ABC: y - 2z = 0$
Plane thru $ABD: 4x + y - 2z = 0$
Plane thru $BCD: 4x+3y+2z=8$
Plane thru $ACD: y = 0$
Step $2$: Try and see which plane as base and which order of integral is easier to compute and if it is possible to do this in one integral vs. split.
I decide to go with plane $ABC$ as base and order $dz \ dx \ dy$.
Step $3$: Once we have made the decision, we find the bounds for integral(s).
$x$ is bound between line segments $AB$ and $BC$ in the plane. So using its intersection with $ABD$ and $BCD$, we get $0 \leq x \leq 2 - y, 0 \leq y \leq 2$.
Using intersection of $ABD$ and $BCD$, $x = \dfrac{2-y}{2}$ so for $0 \leq x \leq \dfrac{2-y}{2}$, $z$ is bound above by plane $ABD$ and for  $\dfrac{2-y}{2} \lt x \leq 2 - y$, $z$ is bound above by plane $BCD$.
So final bounds are (again I leave certain details for you to fill in):
$\dfrac{y}{2} \leq z \leq \dfrac{4x+y}{2}, 0 \leq x \leq \dfrac{2-y}{2}, 0 \leq y \leq 2$
$\dfrac{y}{2} \leq z \leq \dfrac{8-4x-3y}{2}, \dfrac{2-y}{2} \leq x \leq {2 - y}, 0 \leq y \leq 2$
Step $4$: Set up the integral and compute. Validate against result obtained using any elementary method to make sure we have not made any computation errors.
A: Once you decide to have $y$ range from $0$ to $2$ that is your outer integral.  If the $x$ integral is to be the next one in you need to find the range of $x$ as a function of $y$.  At $y=0$ it ranges from $0$ to $2$, but at $y=2$ it is just $x=0$.  As things are linear, $x$ ranges from $0$ to $2-y$.  Your innermost integral is over $z$.  You need to find the range of $z$ as a function of $x$ and $y$.  That gives the triple integral over the volume of the pyramid.  Integrate $1$ over that volume and you have the answer.
A: The edges of a tetrahedron are given by the vectors $2\vec j+\vec k, \vec i,\vec i+2\vec k$, the volume of it is $1/6$ times their box product.
So $$V=\left|\frac{1}{6}\left[2\vec j+\vec k, \vec i,\vec i+2\vec k\right]\right|=2/3.$$
