R is bounded below by the $\displaystyle x-axis$ and above by the curve$\displaystyle y=2cosx,0 \leq x \leq \frac{\pi}{2}$. Find the volume of the solid generated by revolving R around the $\displaystyle y-axis $.by the method of cylindrical shells.
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R is bounded below by the $\displaystyle x-axis$ and above by the curve$\displaystyle y=2cosx,0 \leq x \leq \frac{\pi}{2}$. Find the volume of the solid generated by revolving R around the $\displaystyle y-axis $.by the method of cylindrical shells.
Hello, Yogi!
Quote:
$\displaystyle R$ is bounded below by the $\displaystyle x$-axis and above by the curve $\displaystyle y = 2\cos x,\;0 \leq x \leq \frac{\pi}{2}$.
Find the volume of the solid generated by revolving $\displaystyle R$ around the $\displaystyle y$-axis
by the method of cylindrical shells.
The shells formula is: .$\displaystyle V \:=\:2\pi\int^b_axy\,dx$
We have: .$\displaystyle V \;= \;2\pi\int^{\frac{\pi}{2}}_0x\cdot2\cos x\,dx \;=\;4\pi\int^{\frac{\pi}{2}}_0x\cos x\,dx$
Integrate by parts:
. . $\displaystyle \begin{array}{cc}u = x & dv = \cos x\,dx \\ du = dx & v = \sin x\end{array}$
We have: .$\displaystyle V \:=\:4\pi\left[x\sin x - \int\sin x\,dx\right] \:=\:4\pi\bigg[x\sin x + \cos x \bigg]^{\frac{\pi}{2}}_0$
. . $\displaystyle = \;4\pi\left[\left(\frac{\pi}{2}\!\cdot\!\sin\frac{\pi}{2} + \cos\frac{\pi}{2}\right) - \left(0\!\cdot\!\sin0 + \cos0\right)\right] $
. . $\displaystyle = \;4\pi\left[\left(\frac{\pi}{2} + 0\right) - \left(0 + 1\right)\right] \;=\;4\pi\left(\frac{\pi}{2} - 1\right)\;=\;2\pi(\pi - 2)$
We can also check by doing it the 'other' way....washers.
$\displaystyle {\pi}\int_{0}^{2}(cos^{-1}(\frac{y}{2}))^{2}dy=2{\pi}({\pi}-2)$
