Volume: The Disk Method AP Calculus AB

Question

Finding the Volume of a Solid - Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the $x$-axis

$$y = e^{-x}, \space y = 0, \space x = 0, \space y = 1$$

Why would the upper limit of integration be $1$?

Answer 1

The region is unbounded, thus the solid of revolution will be unbounded as well.

It would be best to take a 'slice' parallel to the axis of revolution, so one wants

\begin{equation} V=\int_0^12\pi rh\,dy \end{equation}

with $r=y$ and $h=x=-\ln(y)$

Therefore

\begin{equation} V=-2\pi\int_0^1y\ln(y)\,dy \end{equation}

Which is an improper integral.

\begin{equation} V=-2\pi\left[ \dfrac{y^2}{2}\ln(y)-\dfrac{y^2}{4}\right]_0^1 \end{equation}

Since \begin{equation} \lim_{x\to0^+}y^2\ln(y)=0 \end{equation}

Then

\begin{equation} V=\dfrac{\pi}{2} \end{equation}

Answer 2

It looks as if there is a typo involved. I think the last bounding line is meant to be $x=1$, not $y=1$.

At least this is consistent with the given answer, since $$\int_0^1 \pi e^{-2x}\,dx=\frac{\pi}{2}(1-e^{-2})\approx 1.358.$$

Answer 3

Like André Nicolas said, it should be $x = 1$, not $y = 1$.

If so, then

Shell Method: $$\int_{\frac{1}{e}}^{1}-2\pi ylnydy+ \int_{0}^{\frac{1}{e}}2\pi y(1)dy$$

Washer Method: $$\int_{0}^{1}\pi(e^{-x})^{2}dx$$