Arc Length and Surface Area Revisited — Question 9

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Question 9

Problem

Design a consistency check for any computed surface area of revolution using dimensions and scaling.

See the diagram in the original worksheet below.

Original worksheet page 1: question and worked solution for 3-11-009
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Question 9 – Solution

See the diagram in the original worksheet below.

Solution

  1. Choose a representation and interval that trace the desired curve exactly once. The equivalent arc-length formulas are L=∫1+(dydx)2dx,L=∫(dxdt)2+(dydt)2dt,L=∫r2+(drdθ)2dθ.\begin{aligned} L&=\int\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx,\\ L&=\int\sqrt{\left(\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2}\,dt,\\ L&=\int\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}\,d\theta. \end{aligned}

  2. For a surface of revolution, multiply the appropriate arc-length element by 2π2\pi times the nonnegative distance to the axis. Check the tracing interval to prevent geometric double-counting.

  3. Area must have length-squared units.

  4. If every coordinate is scaled by k>0k>0, both radius and arc length scale by kk, so the answer must scale by k2k^2.

Original worksheet page 2: question and worked solution for 3-11-009

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