Exercises: Surface Area

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Various exercises related to the computation of areas of surfaces of revolution.

Find the surface area of the solid formed by revolving $y=2x$ on $[0,1]$ about the $x$ -axis. $A = \answer {2\pi \sqrt {5}}.$

Find the surface area of the solid formed by revolving $y=x^2$ on $[0,1]$ about the $y$ -axis.

To compute the integral, you will need to make a substitution like $u = 1 + 4x^2$ or something similar.

$A = \answer {\frac {(5 \sqrt {5} - 1) \pi }{6}}.$

Find the surface area of the solid formed by revolving $y=x^3$ on $[0,1]$ about the $x$ -axis.

To compute the integral, you will need to make a substitution like $u = 1 + 9 x^4$ or something similar.

$L = \answer {\frac {(10 \sqrt {10} - 1) \pi }{27}}.$

Sample Exam Questions

Give an integral formula for the area of the surface generated by revolving the curve $y = \ln x$ between $x=1$ and $x=2$ about the $y$ -axis. Explain your answer. You do not need to evaluate the integral.

$\displaystyle \int _1^2 2 \pi \sqrt {x^2+1} ~ dx$ $\displaystyle \int _1^2 2 \pi (\ln x) \frac {\sqrt {x^2+1}}{x} ~ dx$ $\displaystyle \int _1^2 \frac {2 \pi }{x} \sqrt {1 + (\ln x)^2} ~ dx$ $\displaystyle \int _1^2 \frac {1}{2 \pi \sqrt {x^2+1}} ~ dx$ $\displaystyle \int _1^2 2 \pi (\ln x) \frac {x}{\sqrt {x^2+1}} ~ dx$ $\displaystyle \int _1^2 \frac {2 \pi x}{\sqrt {1 + (\ln x)^2}} ~ dx$

The curve $y = \frac {x^2}{8}$ between $x=0$ and $x = 3$ is revolved around the $y$ -axis. Compute the surface area of the resulting surface.

$\displaystyle \frac {31 \pi }{6}$ $\displaystyle \frac {41 \pi }{6}$ $\displaystyle \frac {61 \pi }{6}$ $\displaystyle \frac {71 \pi }{6}$ $\displaystyle \frac {91 \pi }{6}$ none of the above