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### Section 2-2 : Surface Area

1. Set up, but do not evaluate, an integral for the surface area of the object obtained by rotating $$x = \sqrt {y + 5}$$ , $$\sqrt 5 \le x \le 3$$ about the $$y$$-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

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

$x = \sqrt {y + 5} \hspace{0.5in} \to \hspace{0.5in}\,\,\,y = {x^2} - 5$

The derivative of this is,

$\frac{{dy}}{{dx}} = 2x$ Show Step 2

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

$ds = \sqrt {1 + {{\left[ {\frac{{dy}}{{dx}}} \right]}^2}} \,dx = \sqrt {1 + {{\left[ {2x} \right]}^2}} \,dx = \sqrt {1 + 4{x^2}} \,dx$ 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.

$SA = \int_{{}}^{{}}{{2\pi x\,ds}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\int_{{\sqrt 5 }}^{3}{{2\pi x\,\sqrt {1 + 4{x^2}} \,dx}}}}$

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 $$y$$-axis and so we needed an $$x$$ in the integral.

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

We’ll need the derivative of the function first.

$\frac{{dx}}{{dy}} = \frac{1}{2}{\left( {y + 5} \right)^{ - \,\frac{1}{2}}} = \frac{1}{{2{{\left( {y + 5} \right)}^{\,\frac{1}{2}}}}}$ Show Step 2

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

$ds = \sqrt {1 + {{\left[ {\frac{{dx}}{{dy}}} \right]}^2}} \,dy = \sqrt {1 + {{\left[ {\frac{1}{{2{{\left( {y + 5} \right)}^{\,\frac{1}{2}}}}}} \right]}^2}} \,dy = \sqrt {1 + \frac{1}{{4\left( {y + 5} \right)}}} \,dy = \sqrt {\frac{{4y + 21}}{{4\left( {y + 5} \right)}}} \,dy$ 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 derived in Step 1 of the first part to convert them to $$y$$ limits. Doing this gives,

$x = \sqrt 5 :y = 0\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}x = 3:y = 4$

So, the corresponding $$y$$ limits are : $$0 \le y \le 4$$.

Show Step 4

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

\begin{align*}SA & = \int_{{}}^{{}}{{2\pi xds}} = \int_{0}^{4}{{2\pi x\sqrt {\frac{{4y + 21}}{{4\left( {y + 5} \right)}}} \,dy}} = \int_{0}^{4}{{2\pi \sqrt {y + 5} \sqrt {\frac{{4y + 21}}{{4\left( {y + 5} \right)}}} \,dy}}\\ & = \int_{0}^{4}{{2\pi \sqrt {y + 5} \frac{{\sqrt {4y + 21} }}{{2\sqrt {y + 5} }}\,dy}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\int_{0}^{4}{{\pi \sqrt {4y + 21} \,dy}}}}\end{align*}

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 $$y$$-axis and so we needed an $$x$$ in the integral.

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

Finally, make sure you simplify these as much as possible as we did here. Had we not taken the square root of the numerator and denominator of the rational expression we would not have seen the cancelation that can happen there. Without that cancelation the integral would be much more difficult to do!

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.