# Finding Surface Area using Integrals

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

## Find the surface area of the object obtained by revolving $$f(x)=\frac{1}{14}(e^{7x}+e^{-7x})$$ on $$[-2,2]$$ around the $$x$$-axis.

We begin by drawing the function over the interval $$[-2,2]$$.

We can now revolve this graph around the $$x$$-axis and we will get the following picture.

Since we want to find the surface area, we will cut cross sections of the face of the object. The cross sections will look like the following.

Now that we have our cross sections, we note that they will be rectangle if we would cut them and stretch them out. In this case, the length of the rectangle will be the circumference of the circle whose radius is the function value at the given $$x$$. The width of the rectangle will then be given as the length of the function over the given interval. That is, the width is the arc-length of the function. We therefore find that
\begin{align*}
A_{cs}&=2 \pi r \sqrt{1+f'(x)}\Delta x \\
&=2 \pi (y)\sqrt{1+f'(x)}\Delta x.
\end{align*}

We can now note that $$f'(x)=\frac{1}{14}(e^{7x}+e^{-7x}) =\frac{1}{14}(7e^{x}-7e^{-7x})$$, so the area of the cross section will be
\begin{align*}
A_{cs}&=2 \pi r \sqrt{1+f'(x)}\Delta x \\
&=2 \pi (y)\sqrt{1+f'(x)}\Delta x \\
&=2 \pi \frac{1}{14}(e^{7x}+e^{-7x}) \sqrt{1+(\frac{1}{14}(7e^{7x}-7e^{-7x}))^{2}}\Delta x \\
&=2 \pi \frac{1}{14}(e^{7x}+e^{-7x})\sqrt{1+\frac{49}{196}(e^{7x}-e^{-7x})^{2}} \Delta x.
\end{align*}

We can then find that the total area will be
\begin{align*}
A_{T}&=\int_{a}^{b}2 \pi \frac{1}{14}(e^{7x}+e^{-7x})\sqrt{1+\frac{49}{196}(e^{7x}-e^{-7x})^{2}}dx.
\end{align*}
Here, $$a$$ is given as $$-2$$ and $$b$$ is given in the problem as $$2$$. We will then find the anti-derivative of the integrand before find the definite integral.

For this integral, we will let $$i=2(e^{7x}-e^{7x})$$, then $$di=14(e^{7x}+e^{-7x})dx$$ and $$dx=\frac{di}{14(e^{7x}+e^{-7x})}$$. We therefore have that
\begin{align*}
&\int 2 \pi \frac{1}{14}(e^{7x}+e^{-7x})\sqrt{1+\frac{49}{14}(e^{7x}-e^{-7x})^{2}}dx \\
&=\int \frac{\pi}{7}(e^{7x}+e^{-7x})\sqrt{1+i^{2}}\frac{di}{14(e^{7x}+e^{-7x})} \\
&=\int \frac{\pi}{2}\sqrt{1+i^{2}}di
\end{align*}

To see how to find $$\int \sqrt{1+i^{2}}di$$ see our previous post How long is that mountain path?. Following this process, we get the resulting surface area to be
\begin{align*}
A_{T}\approx 4.6 *10^{10}.
\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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