Calculus II - Surface Area

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Section 8.2 : Surface Area

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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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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 $ds$ .

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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.

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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.

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