Let $f$($x$)= cos($x$) and $g$($x$)= 1-${\frac{x^2}{2}} \\$. Let $D$ be the region bounded by the graph of $f$ and the $x$-axis between $x$= -${\frac{π}{2}} \\$ and $x$= ${\frac{π}{2}} \\$ and let $A$ be the region bounded by the graph $g$ and the $x$-axis. $$\\$$
a) Use the Shells Method to find the volume of the solids generated by rotating $D$ and $A$ around the $y$-axis.
b) Compare your 2 results from part a, and explain graphically which one must be larger.
c) Use this comparison to prove something you already knew about $π$. $$\\$$ This is what I have so far:
Region D:
$$2π\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} x(1-{\frac{x^2}{2}}+cosx)$$
By symmetry, we only need to consider the branch from $0$ to $\displaystyle \sqrt{2}$, so we have $\displaystyle 2π\int_0^{\sqrt{2}}x\left|1-\frac{x^2}{2}\right|\,dx=π=\boxed{3.141}$ for the A solid.
Similarly, we have $\displaystyle 2π \int_0^{\frac{π}{2}}x\left|\cos(x)\right|\,dx=π^2-2π=\boxed{3.5864}$ for the D solid.
This is consistent with what we see, because $\cos(x)$ is more than $\displaystyle 1-\frac{x^2}{2}$ on the interval.
So the cylindrical shells we obtain by rotating will be taller in the D solid than the A solid.
Not sure what this could prove about $π$.