How would I find the volume of a pyramid

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    Pyramid Volume
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How would I find the volume of a pyramid with an equilateral triangle as it's base? I know how to do a pyramid with a square base (the integral from 0 to it's height of 2*(equation solved for y of a side)^2 dy)). Would this be the same method?
 
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Didn't your text tell you that the area of a pyramid is 1/3 the product of the area of its base and its height?
 
You can make an equation for the area of a given slice of the pyradmid with respect to the height (since the height will determine the side length, which will of course influence the area since it is equilateral..I believe it's ((a)^(3/2))/4 but I forget.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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