Calculating Volume of a Pyramid Frustum with Square Base and Top

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Find the volume of frustum of a pyramid with square base of side b, square top of side a, and height h.



Usually when I do these volume problems, I treat them as an equation rotating around an axis, but this object has flat sides so I don't know how to begin.

http://upload.wikimedia.org/wikipedia/en/f/f8/Pyramid_frustum_for_Moscow_papyrus_14.jpg

The solid looks like that.

I know I need to make an integral from 0 to h of the area of the square but while I usually replace the radius in the formula with an equation, I cannot do so here.

Thanks in advance.
 
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Lamoid said:
I know I need to make an integral from 0 to h of the area of the square but while I usually replace the radius in the formula with an equation, I cannot do so here.

Yes you can: Let s(y) be the length of the side of the square at height y. You know that s(0) = b and s(h) = a. You also know that s(y) has to a linear function of y. (Why?) Use these facts to find s(y).
 
So S(y) is (a - b)x / h ?
 
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I think you meant (a - b)y / h. You know this is not it because when y = h, it yields a - b. You want it to yield just a.
 
Oh whoop, I should have seen my linear equation needed a "+ b" on the end based on the 0,b point. So the equation should look like s(y) = (a-b)y/h + b ?
 
You got it.
 
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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