0
$\begingroup$

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!

$\endgroup$
0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ for this problem it's assumed we haven't learned about or done triple integrals yet. Triple integrals are in the next section, so I'm assuming we're supposed to look for an answer without using triple integrals. I see what you are saying though... but I'm not sure how I would finish the integral, is there supposed to be an expression I'm integrating? $\endgroup$ – mays Nov 20 '17 at 22:36
  • $\begingroup$ @mays yes, you are integrating $1$. Can you do double integrals instead? $\endgroup$ – gt6989b Nov 21 '17 at 16:24
  • $\begingroup$ gt6989b Thank you. I think I can finish that triple integral now but I'm still unsure of how to solve this problem using only double integrals, which is what I think we're supposed to do. $\endgroup$ – mays Nov 25 '17 at 1:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.