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### Section 2-2 : Surface Area

2. Set up, but do not evaluate, an integral for the surface area of the object obtained by rotating $$y = \sin \left( {2x} \right)$$ , $$\displaystyle 0 \le x \le \frac{\pi }{8}$$ about the $$x$$-axis using,

1. $$\displaystyle ds = \sqrt {1 + {{\left[ {\frac{{dy}}{{dx}}} \right]}^2}} \,dx$$
2. $$\displaystyle ds = \sqrt {1 + {{\left[ {\frac{{dx}}{{dy}}} \right]}^2}} \,dy$$
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a $$\displaystyle ds = \sqrt {1 + {{\left[ {\frac{{dy}}{{dx}}} \right]}^2}} \,dx$$ Show All Steps Hide All Steps
Start Solution

We’ll need the derivative of the function first.

$\frac{{dy}}{{dx}} = 2\cos \left( {2x} \right)$ Show Step 2

Plugging this into the formula for $$ds$$ gives,

$ds = \sqrt {1 + {{\left[ {\frac{{dy}}{{dx}}} \right]}^2}} \,dx = \sqrt {1 + {{\left[ {2\cos \left( {2x} \right)} \right]}^2}} \,dx = \sqrt {1 + 4{{\cos }^2}\left( {2x} \right)} \,dx$ Show Step 3

Finally, all we need to do is set up the integral. Also note that we have a $$dx$$ in the formula for $$ds$$ and so we know that we need $$x$$ limits of integration which we’ve been given in the problem statement.

$SA = \int_{{}}^{{}}{{2\pi y\,ds}} = \int\limits_{0}^{{\frac{\pi }{8}}}{{2\pi y\,\sqrt {1 + 4{{\cos }^2}\left( {2x} \right)} \,dx}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\int\limits_{0}^{{\frac{\pi }{8}}}{{2\pi \sin \left( {2x} \right)\,\sqrt {1 + 4{{\cos }^2}\left( {2x} \right)} \,dx}}}}$

Be careful with the formula! Remember that the variable in the integral is always opposite the axis of rotation. In this case we rotated about the $$x$$-axis and so we needed a $$y$$ in the integral.

Note that with the $$ds$$ we were told to use for this part we had a $$dx$$ in the final integral and that means that all the variables in the integral need to be $$x$$’s. This means that the $$y$$ from the formula needs to be converted into $$x$$’s as well. Luckily this is easy enough to do since we were given the formula for $$y$$ in terms of $$x$$ in the problem statement.

As an aside, note that the $$ds$$ we chose to use here is technically immaterial. Realistically however, one $$ds$$ may be easier than the other to work with. Determining which might be easier comes with experience and in many cases simply trying both to see which is easier.

b $$\displaystyle ds = \sqrt {1 + {{\left[ {\frac{{dx}}{{dy}}} \right]}^2}} \,dy$$ Show All Steps Hide All Steps
Start Solution

In this case we first need to solve the function for $$x$$ so we can compute the derivative in the $$ds$$.

$y = \sin \left( {2x} \right)\hspace{0.5in} \to \hspace{0.5in}\,\,\,x = \frac{1}{2}{\sin ^{ - 1}}\left( y \right)$

The derivative of this is,

$\frac{{dx}}{{dy}} = \frac{1}{2}\frac{1}{{\sqrt {1 - {y^2}} }} = \frac{1}{{2\sqrt {1 - {y^2}} }}$ Show Step 2

Plugging this into the formula for $$ds$$ gives,

$ds = \sqrt {1 + {{\left[ {\frac{{dx}}{{dy}}} \right]}^2}} \,dy = \sqrt {1 + {{\left[ {\frac{1}{{2\sqrt {1 - {y^2}} }}} \right]}^2}} \,dy = \sqrt {1 + \frac{1}{{4\left( {1 - {y^2}} \right)}}} \,dy = \sqrt {\frac{{5 - 4{y^2}}}{{4\left( {1 - {y^2}} \right)}}} \,dy$ Show Step 3

Next, note that the $$ds$$ has a $$dy$$ in it and so we’ll need $$y$$ limits of integration.

We are only given $$x$$ limits in the problem statement. However, we can plug these into the function we were given in the problem statement to convert them to $$y$$ limits. Doing this gives,

$x = 0:y = \sin \left( 0 \right) = 0\hspace{0.25in}\hspace{0.25in}x = \frac{\pi }{8}:y = \sin \left( {\frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2}$

So, the corresponding $$y$$ limits are : $$0 \le y \le \frac{{\sqrt 2 }}{2}$$.

Show Step 4

Finally, all we need to do is set up the integral.

$SA = \int_{{}}^{{}}{{2\pi yds}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\int_{0}^{{\frac{{\sqrt 2 }}{2}}}{{2\pi y\sqrt {\frac{{5 - 4{y^2}}}{{4 - 4{y^2}}}} \,dy}}}}$

Be careful with the formula! Remember that the variable in the integral is always opposite the axis of rotation. In this case we rotated about the $$x$$-axis and so we needed an $$y$$ in the integral.

Also note that the $$ds$$ we chose to use is technically immaterial. Realistically one $$ds$$ may be easier than the other to work with but technically either could be used.