# Proof- Volume of tetrahedron is abc/6

Prove that the volume of a tetrahedron with mutually perpendicular adjacent sides of lengths a, b, and c, is abc/6.

Keep in mind: - Finding the volume of a solid bounded by the 3 coordinate planes and a given plane - 3 noncollinear points make a plane

I was looking at this two ways...

one, I know that I can place the three sides that are mutually perpendicular on the xyz plane so that the point where they intersect is the origin. I labeled the side along the z-axis c, with a,b along the x and y planes respectively. I can see that the the three sides end on noncollinear points, making a plane. But how do I get the equation of that plane??

The second way was through a similar question posted (but instead of a,b and c there were real numbers)... same visulaization but looking for the area of a cross section perpendicular to the xy plane, with height z. The cross section would be a similar triangle and this is where I get lost... why would this cross section be with sides scaled at (c-z)/c? and why is its area (1/2)(a)(b)(c-z/c)^2? I understand that I have to take the integral from z=0 to z=c of (1/2)(a)(b)(c-z/c)^2, but I don't understand how to get to that point. Any help is appreciated. Thank you!

Your plane has to pass through $(a,0,0), (0,b,0)$ and $(0,0,c)$. In the 2D case the line would be $x/a + y/b = 0$ (or equivalently $y = -bx/a$ and it is not hard to see the analogy to the plane being $x/a+y/b+z/c=0$. Hence, your volume is $$\int_{x=0}^{x=a} \int_{y=0}^{y=-bx/a} \int_{z=0}^{z=-cx/a-cy/b} dzdydx.$$ Can you finish this?
• @mays yes, you are integrating $1$. Can you do double integrals instead? – gt6989b Nov 21 '17 at 16:24