Question 612717
Most of these proofs require a bit of calculus. If we integrate along the height (h), we can say that the volume V is 


*[tex \LARGE V = \int_0^{h} b(\frac{x}{h})^2 dx] 


Here, x is a real number which varies between 0 and h. The (x/h)^2 term "scales" the area of each cross-section of b. As you can see, if x = 0, the cross-sectional area is 0, if x = h, the cross-sectional area is b.


Integrating, we obtain


*[tex \LARGE V = b\frac{x^3}{3h^2}]. We evaluate at x = h, subtract the value obtained at x = 0:


*[tex \LARGE V = b\frac{h^3}{3h^2} - b\frac{0^3}{3h^2}]


Simplifying everything, we obtain


*[tex \LARGE V = \frac{1}{3}bh]