Let R be the region bounded by the curves y=x^2, y=sin(xpi/2), x=0, and x=1.
Find the volume of the solid obtained by rotating R around the y-axis.
Note that: $\displaystyle \sin \left(\frac{\pi x}{2}\right) \geq x^{2} \quad 0 \leq x \leq 1$
A graph of this would be helpful in visualizing what's going on.
By method of shells: $\displaystyle V = 2\pi \int_{0}^{1} x \left(\sin\left(\frac{\pi}{2}x\right) - x^{2}\right)dx$
x = the radius of the shell; $\displaystyle \sin\left(\frac{\pi}{2}x\right) - x^{2}$ = represents the height of te shell
It's not too bad:
$\displaystyle V = 2\pi \int_{0}^{1} x \left(\sin\left(\frac{\pi}{2}x\right) - x^{2}\right)dx$
$\displaystyle V = 2\pi \int_{0}^{1} \left[ x\sin \left(\frac{\pi}{2} x\right) - x^{3}\right] dx$
You can integrate term by term (since: $\displaystyle \int \left(f(x) + g(x)\right) dx = \int f(x) dx + \int g(x) dx$) in which case the 2nd term should be rather easy.
The first one, i.e. $\displaystyle \int_{0}^{1} x\sin \left(\frac{\pi}{2}x\right) dx$ requires integration by parts:
$\displaystyle \begin{array}{lll} u = x & \quad & dv = \sin \left(\frac{\pi}{2}x\right) \\ du = dx & \quad & v = -\frac{2}{\pi} \cos \left(\frac{\pi}{2}x\right) \end{array}$
Sub it into the integration by parts formula and you should be on your way.