Volume of triangular pyramid given 4 vertices I want to calculate the colume of the polyhedra with vertices $(a_1,0,0),(0,a_2,0),(0,0,a_3),(a_1,a_2,a_3)$. My solution is:
Using translation and a reflection I am considering the polyhedra  $(a_1,a_2,0),(0,a_2,a_3),(a_1,0,a_3),(0,0,0)$. I calculated the area of the triangle with vertices $(a_1,a_2,0),(0,a_2,a_3),(a_1,0,a_3)$ and found $A=|a_1a_2|/2$ and the distance from $(0,0,0)$ to the plane passing from the other 3 vertices and found $H=\frac{2|a_1a_2a_3|}{\sqrt{a_1^2a_2^2+a_2^2a_3^2+a_1^2a_3^2}}$. Finally the volume of the pyramid is $V=1/3 A*H=\frac{|a_1a_2||a_1a_2a_3|}{3\sqrt{a_1^2a_2^2+a_2^2a_3^2+a_1^2a_3^2}}$. 

Is this correct and also is there a more simple solution using integration maybe?
A perhaps simpler approach is to start with the unit cube. Let 
$$A=(1,0,0),\ B=(0,1,0), C=(0,0,1),\ D=(1,1,1)$$
and note that if the volume of this tetrahedron is found, then your coordinate version can be found by applying three successive transforms, one carrying $(x,y,z)$ to $(a_1x,y,z),$ the next taking $(x,y,z)$ to $(x,a_2,z),$ and the third carrying $(x,y,z)$ to $(x,y,a_3z).$ These three multiply volumes respectively by $a_1,a_2,a_3$ and so your desired volume is the unit cube tetrahedron's volume multiplied by the product $a_1a_2a_3.$

For the unit cube version, note that the plane through $A,B,C$ has equation $x+y+z-1=0.$ Thus the vector $(1,1,1)$ is normal to that plane, and we can find where that normal hits the plane by putting $x=y=z=t$ into $x+y+z-1=0,$ giving $t=1/3$ so that the normalfrom $D$ to the plane of the triangle $ABC$ hits the triangle at the point $(1/3,1/3,1/3)$ and the length of that normal is $\sqrt{(2/3)^2+(2/3)^2+(2/3)^2}=2/\sqrt{3}.$ The area of the triangle $ABC$ is, since it is equilateral of side length $\sqrt{2},$ conveniently $\sqrt{3}/2,$ so applying the cone volume formula we get
$$\frac{1}{3}\cdot \frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2}=\frac{1}{3}$$
for the volume of the tetrahedron $ABCD$ inside the unit cube. 

Thus in your coordinate choice the volume is $a_1a_2a_3/3$ after stretching this as outlined above.

A simpler proof of unit cube tetrahedron volume. The complement of the tetrahedron $ABCD$ consists of four smaller tetrahedra, each the hull of one of the omitted vertices of the unit cube and the three of $A,B,C,D$ which are adjacent to it. These four tetrahedra are all congruent, and each may be viewed as a cone of altitude one over a triangle of area $1/2$, so has volume $(1/3)\cdot(1/2) \cdot 1=1/6.$ Thus the complement of tetrahedron $ABCD$ has volume $4\cdot(1/6)=2/3,$ leaving $1/3$ for the volume of $ABCD.$
You can also find the volume of the pyramid, or any continuous shape for that matter, by setting up a triple integral. It is best to first draw a picture of the figure to help you set up your bounds of integration. Your method is much simpler to compute for this example; however, the integral method is much more general.