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Paul
February 18, 2026
Section 8.2 : 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 SolutionThe 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 2Let’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 3The 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 4Finally, 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.
\[SA = \int_{1}^{2}{{2\pi x\,\sqrt {1 + 36{x^2}} \,dx}} = \left. {\frac{\pi }{{54}}{{\left( {1 + 36{x^2}} \right)}^{\frac{3}{2}}}} \right|_1^2 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{\pi }{{54}}\left( {{{145}^{\frac{3}{2}}} - {{37}^{\frac{3}{2}}}} \right) = 88.4864}}\]