Find the volume of the solid generated by rotating the region bounded by the curves I'm having a little problem with this question:
Find the volume of the solid generated by the curves $y=e^x$, $y=\cos (\frac{\pi x^2}{2})$ and $x=1$ about the $y$-axis.
Can this be done with the method of cylindrical shells? Would the integral then we use to find the volume look like this: $$2\pi\int_0^1 x(e^x - \cos (\frac{\pi x^2}{2}))   
dx= 2(\pi-1)?$$ Or is this complete nonsense??? 
 A: Note: I had a mistake in my earlier calculation and the answer didn't match yours. The cylindrical shells method also works, I found the volume by reasoning directly and subtracting different parts.

When you rotate a curve $y=f(x)$, $a\leqslant x\leqslant b$ about the $y$-axis, the volume is given by $V=\pi\int_{f^{-1}(a)}^{f^{-1}(b)} x^2\,dy$, where $x=f^{-1}(y)$.
Now the shape you are rotating is the following: 

You can find the volume first by rotating this: 

We can find this by subtracting the volume obtained when the part of the $e^x$ curve between $1\leqslant y\leqslant e$ is subtracted from the cylinder obtained by rotating the $1\times e$ rectangle about the $y$-axis. This has volume $\pi e$, and so the desired volume is 
$$V_1 = \pi e-\pi\int_1^e x^2\,dy = \pi e-\pi\int_1^e\ln^2 y\, dy = 2\pi.$$
Now we subtract what we get when we rotate this:

which has a volume of $$V_2 = \pi \int_0^1 x^2\,dy.$$
For $V_2$, rather than inverting $\cos\big(\frac{\pi x^2}2\big)$ we can leave things in $x$ and instead compute the differential $dy = f'(x)\, dx = -\pi x\sin\big(\frac{\pi x^2}2\big)\,dx$, and change the limits of the integral so that they are also in $x$ (just look at the graph): $y=0\Rightarrow x = 1$, $x=1\Rightarrow y=0$. 
Therefore by a substitution ($u = \frac{\pi x^2}2$, say), we have
$$V_2 = -\pi^2 \int_1^0 x^3\sin\big(\tfrac{\pi x^2}2\big)\,dx  = -2\int_{\frac\pi2}^0 u\sin u\,du = 2.$$
Therefore, the total volume is $V = V_1 - V_2 = \boxed{2(\pi-1)}$.
