Note: Its not a duplicate. The other one talks about triple integrals... while I am only doing a single one..!
So to get the area of a cube I can do the following integral,
$$\int_0^a a^2 \, dh = a^3$$
So I try to do the same thing for the square pyramid-- taking the area to be $a\cdot a$, I do the following integral.
$$\int_0^h b^2 \left(1 - \frac{h}{H}\right)dh$$
Thinking that the factor $(1 - h/H)$ should take care of the reducing cross section.
Evaluating this integral gives me $b^2H/2$ while the correct volume happens to be $b^2H/3$.
I looked at the solution of calculating the volume and I get what its doing (calculating the area as a function of height) But I can't still get around why my reasoning is wrong.
Thanks and regards, Dhruv Kapur.
Your mistake is to assume that as $h$ varies from $0$ to $H$, the area of the square at height $h$ decreases linearly. But it's the side length of the square that decreases linearly, so the area decreases quadratically. The side length of the square at height $h$ is $$ L(h) = b\left(1 - \tfrac{h}{H}\right) $$ so you have to integrate $L^2(h)$, i.e. compute $$ V = \int_0^H (L(h))^2 \,dh = \int_0^H b^2\left(1-\tfrac{h}{H}\right)^2 \,dt \text{.} $$
This yields $$ V = b^2\left(\frac{h^3}{3H^2} - 2\frac{h^2}{2H} + h\right)\Bigg|_{h=0}^{h=H} = b^2\left(\frac{H}{3} - H + H\right) = \frac{b^2H}{3} \text{,} $$ as it should.