# Volume using Washer Method

As part of my Calculus 2 class, I have created a practice exam to help my students study.  In doing so, I felt that providing the questions and the solutions in a blog series would be helpful not just to my students, but also to anyone that may be taking, or teaching, calculus 2.  I hope you find the following solutions helpful in your studies.

You can also view me presenting the solution on YouTube at

## Let $$R$$ be the region bounded by $$f(x)=\ln(x^{2})$$, $$y=0$$ and $$x=e$$. Find the volume of the object obtained by revolving this region about the $$y$$-axis.

Here, we again begin by graphing $$f(x)=\ln(x^{2})$$, $$y=0$$ and $$x=e$$. We can now revolve this region around the $$y$$-axis in order to arrive at a three dimensional object. This will look like ### Washers

We are trying to find the total volume, so we will cut cross sections out and find the volume of these cross sections. In this case, we will cut out the washers.

We now note that the washers will have a volume given by the area of the face multiplied by the depth. Therefore,
\begin{align*}
V_{cs}=(\pi R^{2}-\pi r^{2})d.
\end{align*}
We now note that the larger radius is $$x$$ when $$x=e$$ and the smaller radius is $$x$$ when $$y=\ln(x^{2})$$. Therefore $$R=e$$ and
\begin{align*}
y&=\ln(r^{2}) \\
e^{y}&=r^{2} \\
e^{\frac{y}{2}}&=r.
\end{align*}
Note we choose the positive square root because we are finding a radius.

Therefore, the volume of the cross sections will be
\begin{align*}
V_{cs}&=\pi(e^{2})-\pi (e^{\frac{y}{2}})^{2}) \Delta y \\
&=\pi(e^{2}-e^{y})\Delta y.
\end{align*}

The total volume of the object will then be
\begin{align*}
V_{T}=\int_{a}^{b}\pi(e^{2}-e^{y})dy.
\end{align*}

The smallest $$y$$ we need for the integral is given as the intersection of $$y=0$$. The largest value of $$y$$ is given as the intersection of $$y=\ln(x^{2})$$ and $$x=e$$. Therefore, $$y=\ln(e^{2})=2$$. We will now have that
\begin{align*}
V_{T}&=\int_{0}^{2}\pi(e^{2}-e^{y})dy \\
&=\pi(e^{2}y-e^{y}|_{0}^{2} \\
&=\pi(2e^{2}-e^{2}-(0-e^{0})) \\
&=\pi(e^{2}+1).
\end{align*}

### Shells

Here, we again begin by graphing $$f(x)=\ln(x^{2})$$, $$x=1$$ and $$x=e$$.

We can now revolve this region around the $$y$$-axis in order to arrive at a three dimensional object. This will look like

We are trying to find the total volume, so we will cut cross sections out and find the volume of these cross sections. In this case, we will cut out the shells.

We now note that the shells will have a volume given by the area of the face multiplied by the depth. Therefore,
\begin{align*}
V_{cs}=2\pi *r*d.
\end{align*}

We now note that the larger radius is $$x$$. The height is then given by $$f(x)=\ln(x^{2})$$. Therefore, the volume of the cross sections will be
\begin{align*}
V_{cs}&=2 \pi x \ln(x^{2}) \Delta x. \\
\end{align*}

The total volume of the object will then be
\begin{align*}
V_{T}=\int_{a}^{b}2 \pi x \ln(x^{2}) dx.
\end{align*}

The smallest and largest $$x$$ we need for the integral are given in the problem as the intersection of $$y=0$$ with $$\ln(x^{2})$$, that is $$x=1$$ and $$x=e$$, respectively. We will now have that
\begin{align*}
V_{T}&=\int_{1}^{e}2 \pi x \ln(x^{2}) dx
\end{align*}

In order to find this integral, we will first need an anti-derivative. In order to find this we will factor out the $$2 \pi$$ and let our inside function be $$i=x^{2}$$. Therefore, $$di=2xdx$$ so $$dx=\frac{di}{2x}$$. We now can find that
\begin{align*}
\int x \ln(x^{2})dx &=\int x \ln(i)\frac{di}{2x} \\
&=\frac{1}{2} \int \ln(i)di.
\end{align*}

We now have to find the anti-derivative of $$\ln(i)$$. In order to do this, we will do integration by parts. Here, we have that $$F=\ln(i)$$ and $$dS=1di$$. Therefore,
\begin{align*}
\int \ln(i)di&=i \ln(i)-\int i\frac{1}{i}di \\
&=i \ln (i) -i+c.
\end{align*}

Substituting back in $$i=x^{2}$$, we find that
\begin{align*}
\int x \ln(x^{2})dx&=x^{2} \ln(x^{2})-x^{2}+c.
\end{align*}

We can now find the total volume as
\begin{align*}
V_{T}&=\int_{1}^{e}2 \pi x \ln(x^{2}) dx \\
&=2\pi(\frac{1}{2}(x^{2} \ln(x^{2})-x^{2}))|_{1}^{e} \\
&=\pi(e^{2}\ln(e^{2})-e^{2}-(1\ln(1)-1)) \\
&=\pi (e^{2}+1).
\end{align*}

## Other Resources

I have it linked above, but you can find the practice exam here.  You can find the other solutions to the practice exam in the other posts available here.

If you found the post, or the YouTube video helpful, please like the post or the video.  Also remember to follow the blog and subscribe to the YouTube channel.  By doing so you will not only be able to find future content, you will also make it easier for other students to find the content when they are studying for Calculus.  Thank you.

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