I am having some trouble understanding why the integrals setup for finding surfaces consider the slant of the function, whereas while calculating volumes we don't bother with them.
Consider the setup in the link.
If I were to rotate the graph about the x-axis, and then try to write the equation for infinitesimal surface area, I would end up with
$$∆Surface = 2y * \sqrt(1 + (\frac{∆y}{∆x})^2) * ∆x$$
Whereas for Volume I would have,
$$∆Volume = y^2 * ∆x$$
In case of volume why don't we consider the slant, or why isn't the formula for infinitesimal volume
$$∆Volume = y^2 * ∆x + \frac{1}{2} * ∆y * ∆x * 2y$$ where the additional part would be the volume due to the enlarged triangle in the image rotated about the x - axis.
You are using the fundamental theorem of calculus, $$ V(b)=\int_a^b V'(x)\,dx, $$ to compute the volume. For both of your two expressions for $\Delta V$ the limit $$ \lim_{\Delta x\to0}\frac{\Delta V}{\Delta x}=V'(x) $$ is the same.