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### Section 2-2 : Surface Area

4. Find the surface area of the object obtained by rotating $$y = 4 + 3{x^2}$$ , $$1 \le x \le 2$$ about the $$y$$-axis.

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Start Solution

The first step here is to decide on a $$ds$$ to use for the problem. We can use either one, however the function is set up for,

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

Using the other $$ds$$ will put fractional exponents into the function and make the $$ds$$ and integral potentially messier so we’ll stick with this $$ds$$.

Show Step 2

Let’s now set up the $$ds$$.

$\frac{{dy}}{{dx}} = 6x\hspace{0.5in} \Rightarrow \hspace{0.5in}ds = \sqrt {1 + {{\left[ {6x} \right]}^2}} \,dx = \sqrt {1 + 36{x^2}} \,dx$ Show Step 3

The integral for the surface area is,

$SA = \int_{{}}^{{}}{{2\pi x\,ds}} = \int_{1}^{2}{{2\pi x\,\sqrt {1 + 36{x^2}} \,dx}}$

Note that because we are rotating the function about the $$y$$-axis for this problem we need an $$x$$ in front of the root. Also note that because our choice of $$ds$$ puts a $$dx$$ in the integral we need $$x$$ limits of integration which we were given in the problem statement.

Show Step 4

Finally, all we need to do is evaluate the integral. That requires a quick Calc I substitution. We’ll leave most of the integration details to you to verify since you should be pretty good at Calc I substitutions by this point.

$SA = \int_{1}^{2}{{2\pi x\,\sqrt {1 + 36{x^2}} \,dx}} = \left. {\frac{\pi }{{54}}{{\left( {1 + 36{x^2}} \right)}^{\frac{3}{2}}}} \right|_1^2 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{\pi }{{54}}\left( {{{145}^{\frac{3}{2}}} - {{37}^{\frac{3}{2}}}} \right) = 88.4864}}$