Help with work problems

**khuezy** · June 5th 2008, 06:58 PM

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.

**o_O** · June 5th 2008, 07:16 PM

Note that: $\sin \left(\frac{\pi x}{2}\right) \geq x^{2} \quad 0 \leq x \leq 1$

A graph of this would be helpful in visualizing what's going on.

By method of shells: $V = 2\pi \int_{0}^{1} x \left(\sin\left(\frac{\pi}{2}x\right) - x^{2}\right)dx$

x = the radius of the shell; $\sin\left(\frac{\pi}{2}x\right) - x^{2}$ = represents the height of te shell

**khuezy** · June 5th 2008, 07:29 PM

how would i solve that integral? it looks complicated

**o_O** · June 5th 2008, 07:40 PM

It's not too bad:

$V = 2\pi \int_{0}^{1} x \left(\sin\left(\frac{\pi}{2}x\right) - x^{2}\right)dx$
$V = 2\pi \int_{0}^{1} \left[ x\sin \left(\frac{\pi}{2} x\right) - x^{3}\right] dx$

You can integrate term by term (since: $\int \left(f(x) + g(x)\right) dx = \int f(x) dx + \int g(x) dx$ ) in which case the 2nd term should be rather easy.

The first one, i.e. $\int_{0}^{1} x\sin \left(\frac{\pi}{2}x\right) dx$ requires integration by parts:
$\begin{array}{lll} u = x & \quad & dv = \sin \left(\frac{\pi}{2}x\right) \\ du = dx & \quad & v = -\frac{2}{\pi} \cos \left(\frac{\pi}{2}x\right) \end{array}$

Sub it into the integration by parts formula and you should be on your way.

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Help with work problems

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