Calculate the volume $V(a,b)$ of a barrel that is created by rotating the graph of $y=\ln x$ around $x$-axis. The barrel starts from $x=a>0$ and ends at $x=b>a$. Also calculate $\displaystyle\lim_{a\to 0^{+}} V(a,b)$.
So far I thought a good way is to calculate it by the disk method using the formula $$\int_a^b π(f(x))^{2}\, dx = π\int_a^b (\ln x)^{2}\, dx.$$ Is my thinking correct and what about $\lim_{a\to 0^{+}} V(a,b)$?
$$ V = \pi \int_a^b (f(x))^2 dx $$
Note $ y = f(x) = \ln(x) \Rightarrow x = \exp(y) \Rightarrow dx = \exp(y) dy $. Thus,
$$ V = \pi \int_{y=\ln a}^{\ln b} y^2 e^y dy $$
$$ V = \pi [( y^2 -2y + 2 ) e^y ]_{y=\ln a}^{\ln b} $$
$$ V = \pi [ (\ln^2 b -2 \ln b + 2)b - (\ln^2 a - 2 \ln a + 2) a ]. $$
Limit: as $a \to 0+$, we have $(a\ln^n a) \to 0 $ for all $n \ge 1$. Thus,
$$ \lim_{a \to 0+} V = \pi b(\ln^2 b -2 \ln b + 2). $$