Find the volume of triangular pyramid

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I have a plane, $x+2y+3z=12$ with $x$-intercept of $12$, $y=6$, and $z=4$. So between origin and these points, I have a pyramid. $(0,0,0)$, $(12,0,0)$, $(0,6,0)$, $(0,0,4)$.

I need to work out the volume of this pyramid. I did it a few times and understand mostly how to do it, but I'm not getting the answer on the answer sheet, so maybe somebody could work it out and tell me what you get :)

Thanks a lot...

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3 Answers

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Well, that pyramid's base (in fact, prism's area) is a right-angled triangle with known lengths of its legs, so its area $\;A\;$ is very easy to calculate, and since this is also a straight pyramid also its height $\;h\;$ is easy to calculate.

Now just use the volume formula

$$V=\frac13Ah$$

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If you don't know, or can't use, the formula for volume of a pyramid, this is not too hard to do as an integral. In general, the volume of a solid $E$ is given by the integral $\iiint\limits_E dV$.

To do this you first need to figure out what the base triangle as. As you noted, the vertices in the $xy$-plane have coordinates $(12, 0)$ and $(0, 6)$. (I'm assuming this pyramid lives in the first octant / is bounded by the $xy$-plane below.) The equation of the line from $(12, 0)$ to $(0, 6)$ is $y = 6 - \frac{x}{2}$. So the base of the pyramid is the set

$$ \left\{ (x, y) \, \big| \, 0 \leq x \leq 12, \, 0 \leq y \leq 6 - \frac{x}{2} \right\} $$

Solving for $z$ in your equation of the plane tells you the $z$-values are between $z = 0$ and $z = \frac{1}{3}(12 - x- 2y)$.

Putting all of this together, the pyramid is the following.

$ \text{Volume}(E) = \iiint\limits_E dV \\ = \int_0^{12} \int_0^{6 - x/2} \int_0^{(12 - x - 2y)/3} \, dz \, dy \, dx \\ = ... \quad \text{(I cheated and did this in WolframAlpha)} \\ = 48 $

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The volume of a pyramid is the $\frac{1}{6}$th of the volume of respective parallelepiped. So, we have $$V=\frac{1}{6} \cdot 12 \cdot 6 \cdot 4=48.$$

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