Please use the method of cylindrical shells to find the volume generated by rotating the region bounded by
- $\displaystyle y=\begin{cases}\frac{\sin x}x&\text{ if }x>0\\1&\text{ if }x=0\end{cases}$
- $y=0$
- $x=0$
- $x=\pi$
about the $y$-axis.
I tried to do it using
\begin{align} V&=\int_0^\pi2\pi x\frac{\sin x}x\,\mathrm dx\\ &=2\pi\int_0^\pi\sin x\,\mathrm dx\\ &=4\pi. \end{align}
Is this correct? What about when $x=0$? Doesn't it should affect the integrand? so maybe something like
\begin{align}
V&=\int_{0^-}^{0^+} 2\pi x\mathrm dx\\
\end{align}
What should I do?
Yes, that is correct; the answer is $4\pi$ for the reasn that you gave. And, if $f$ is your function, then$$0\times f(0)=0\times1=0=\sin(0).$$ So, there is no problem with $x=0$.