Let's say that we are given the following image
which $r(x)=2\sqrt{x}$
and we want to know the volume of the solid whose cross section taken perpendicular to the y-axis is a rectangle whose height is 3 times the length of its base in region R. I want to find an integral expression that gives the volume of the solid
The question-
According to my class, I would naturally have $\int_{0}^{6}\frac{3}{16}{y^4}dy$
I understand to do this, but why do we not square this and why do we not multiply by $\pi$?
This is an image of the volume of interest, drawn to scale:
The cross sections comprising the intersection of planes perpendicular to the $y$-axis are rectangles, whose base is the horizontal distance between the $y$-axis and the curve $x = y^2/4$, and whose height is $3$ times the base.
In light of this diagram, the question becomes, why would $\pi$ be involved in such a volume calculation, any more than it would be involved in calculating the volume of a cube? There are no circular cross sections. Mathematics is about using deductive reasoning, logic, and rigor. It is not about arbitrary choices or blind adherence to formulas. Intuition can be motivation, but is never a sufficient foundation for proof.
To perform the requisite computation, it suffices to consider the differential volume of a representative rectangular cross section. Since we already know that for a cross section corresponding to a particular $y$-coordinate $y_0$ has base of width $y_0^2/4$ and height $3(y_0^2/4)$, it follows that such a cross section would have differential volume $$dV = \frac{3}{4} y^2 \cdot \frac{1}{4} y^2 \cdot dy = \frac{3}{16} y^4 \, dy,$$ as a function of a general $y \in [0, 6]$. Consequently, the total volume is now found by integrating with respect to $y$ on this interval: $$V = \int_{y=0}^6 \frac{3}{16} y^4 \, dy,$$ and the rest is straightforward.
Had the cross-sections been specified as, say, semicircles whose diameters are the horizontal distances from the $y$-axis to the curve $x = y^2/4$--that is to say, instead of rectangles, we used semicircular cross sections--then indeed we could see $\pi$ entering into the volume calculation, as the differential volume would then be $dV = \pi (y/2)^2/4 \, dy$. But no such cross-section was given.