I'm not sure how to start with this problem so any help is appreciated. I'm not sure how to set-up the integrals and which integrals I should be using.
2 Answers
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This post does not contain a direct answer because the OP's question may be homework.
This octahedron is made up of two square-based pyramids, which means that we only need to find the volume of one and multiply that by $2$ to obtain the required result. First, find the height ($h$) of the pyramid and picture the figure being set up like this, with one of its vertices positioned at $(0,0)$, just to simplify our calculations:
(one face depicted)
Now, the volume of any figure is simply the integral of its cross-sectional area.
$$ V = \int_{0}^{h}A(x)dx$$
If you pick an arbitrary point $(x,y)$ on side $\mathbf{c}$, the height of the cross-section ($s_c$) at that point is twice the value of the $y$-coordinate. Since $A(x)$ = $s_c^2$:
$$ V = \int_{0}^{h}s_c^2dx$$ $$ s_c = 2y$$ $$\therefore V = \int_{0}^{h}(2y)^2dx = \int_{0}^{h}4y^2dx$$
Now, try relating $y$ with $x$ by figuring out the equation of the side $\mathbf{c}$. After plugging in wherever needed, calculate the integral to obtain the equation for the volume of this pyramid and multiply that by $2$ for the octahedron, which, if you have done correctly, will match the one given in the question.
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$\begingroup$ I got it now. Thank you for the steps! $\endgroup$– bblCommented Apr 25, 2022 at 4:41
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Hint: If you make a horizontal cut, the cross-section of an octahedron is a square. Octahedrons are also symmetric so you can reduce this to a square pyramid.
Now you need to find the side length for the tip and base of the pyramid and the differential volume of some slice through the octahedron. Then you can integrate this differential volume $V = \int dV$ (your volume will be an integral defined function which you can reduce to this result).
