Volume around $x$ axis using shell method

Question

I know the volume around $x$ axis of the region $$\mathcal{R}=\{(x,y)\in\mathbb{R}^{2}\mid 0\leq x\leq 1, y\leq e^{-x}, y\leq e^{x-1}, y\geq0\}$$ is (using the disk method) $$V=\pi\left\{\int_{0}^{1/2}(e^{x-1})^{2}dx+\int_{1/2}^{1}(e^{-x})^{2}dx\right\}$$ using the disk method.

However, I have to write it using the shell method. To the first integral: \begin{align*} y=e^{x-1} & \Rightarrow\ln(y)=\ln(e^{x-1}) \\ & \Rightarrow x=\ln(y)+1 \end{align*} and since, for $0\leq x\leq1/2\Rightarrow 0\leq y\leq e^{-1/2}$. To the second integral $$x=-\ln(y),\quad e^{-1}\leq y\leq e^{-1/2}$$ So, the volume would be $$V=2\pi\left\{\int_{0}^{e^{-1/2}}y(\ln(y)+1))dy+\int_{0}^{e^{-1/2}}y(-\ln(y))dy\right\}$$ Is it correct?

Answer 1

If you sketch the cross section, you will find that the minimum radius of the disk is $e^{-1}$ when $x = 0$ or $x = 1$. See the figure below.

While this does not create any issues when integrating for the volume using disks, it makes the calculation slightly more problematic when using cylindrical shells. Your setup will not work because when $y = 0$, $\log y$ is not defined. Instead, you would compute the $x$-height of the shells from $y = e^{-1}$ to $y = e^{-1/2}$, and then add in the volume of the cylinder with radius $e^{-1}$ and height $1$; e.g.,

$$V = \pi (e^{-1})^2 + \int_{y=e^{-1}}^{e^{-1/2}} 2\pi y \left( -\log y - (1 + \log y \right) \, dy.$$

Answer 2

There is a small mistake. Notice that for $y\le e^{-1}$ your $x$ limits are from $0$ to $1$. So you need to split the integral into two parts, one from $0$ to $e^{-1}$, and one from $e^{-1}$ to $e^{-1/2}$: $$V=2\pi\left\{\int_0^{e^{-1}}y(1-0)dy+\int_{e^{-1}}^{e^{-1/2}}y[-\ln y-(\ln y+1)]dy\right\}$$

Answer 3

y-axis ($x = 0$) intersects the curve $y = e^{x-1}$ at $y = 1/e$.

Similarly $x = 1$ intersects the curve $y = e^{-x}$ at $y = 1/e$

$i$) So for $0 \leq y \leq 1/e$, all you have is a cylinder of radius $1/e$ and height $1$ and the volume is $ V_1 = \displaystyle \frac{ \pi}{e^2}$ which the below integral using shell method -

$ \displaystyle \int_0^{1/e} 2 \pi y ~ dy$

Now at the intersection of both curves, $ y = e^{-x} = e^{x-1} \implies x = 1/2$ and $y = 1/ \sqrt e$

Also, for the curve on the left, $y = e^{x-1} \implies x = 1 + \ln y$

And for the curve on the right, $y = e^{-x} \implies x = - \ln y$

$ii$) So for $1/e \leq y \leq 1/\sqrt e$,

$V_2 = \displaystyle \int_{1/e}^{1/\sqrt e} 2 \pi y ~\left (- \ln y - \ln y - 1\right) ~ dy = \frac{\pi(e-2)}{e^2}$

Adding both, we get $V = V_1 + V_2 = \displaystyle \frac{\pi(e-1)}{e^2}$