Cylindrical Shells Help

**Yogi_Bear_79** · November 7th 2006, 02:36 PM

R is bounded below by the $x-axis$ and above by the curve $y=2cosx,0 \leq x \leq \frac{\pi}{2}$ . Find the volume of the solid generated by revolving R around the $y-axis$ .by the method of cylindrical shells.

**Soroban** · November 7th 2006, 04:40 PM

Hello, Yogi!

$R$ is bounded below by the $x$ -axis and above by the curve $y = 2\cos x,\;0 \leq x \leq \frac{\pi}{2}$ .
Find the volume of the solid generated by revolving $R$ around the $y$ -axis
by the method of cylindrical shells.

The shells formula is: . $V \:=\:2\pi\int^b_axy\,dx$

We have: . $V \;= \;2\pi\int^{\frac{\pi}{2}}_0x\cdot2\cos x\,dx \;=\;4\pi\int^{\frac{\pi}{2}}_0x\cos x\,dx$

Integrate by parts:
. . $\begin{array}{cc}u = x & dv = \cos x\,dx \\ du = dx & v = \sin x\end{array}$

We have: . $V \:=\:4\pi\left[x\sin x - \int\sin x\,dx\right] \:=\:4\pi\bigg[x\sin x + \cos x \bigg]^{\frac{\pi}{2}}_0$

. . $= \;4\pi\left[\left(\frac{\pi}{2}\!\cdot\!\sin\frac{\pi}{2} + \cos\frac{\pi}{2}\right) - \left(0\!\cdot\!\sin0 + \cos0\right)\right]$

. . $= \;4\pi\left[\left(\frac{\pi}{2} + 0\right) - \left(0 + 1\right)\right] \;=\;4\pi\left(\frac{\pi}{2} - 1\right)\;=\;2\pi(\pi - 2)$

**galactus** · November 8th 2006, 01:04 PM

We can also check by doing it the 'other' way....washers.

${\pi}\int_{0}^{2}(cos^{-1}(\frac{y}{2}))^{2}dy=2{\pi}({\pi}-2)$

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