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M408M Learning Module Pages
Main page Chapter 10: Parametric Equations and Polar CoordinatesLearning module LM 10.1: Parametrized Curves:Learning module LM 10.2: Calculus with Parametrized Curves:Slope and areaArc length and surface area Summary and simplification Higher Derivatives Learning module LM 10.3: Polar Coordinates:Learning module LM 10.4: Areas and Lengths of Polar Curves:Learning module LM 10.5: Conic Sections:Learning module LM 10.6: Conic Sections in Polar Coordinates:Chapter 12: Vectors and the Geometry of SpaceChapter 13: Vector FunctionsChapter 14: Partial DerivativesChapter 15: Multiple Integrals |
Arc length and surface areaNext we want to figure out the length of a parametrized curve. As with all integrals, we break it into pieces, estimate each piece, add the pieces together, and take a limit. A short segment has length about $$\Delta L = \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{\left ( \frac{\Delta x}{\Delta t}\right )^2 + \left ( \frac{\Delta y}{\Delta y} \right )^2} \Delta t.$$ Adding these up and taking a limit gives length $$L = \int_{t_1}^{t_2} \sqrt{\left ( \frac{dx}{dt}\right )^2 + \left ( \frac{dy}{dt}\right )^2} \;dt,$$ where $t_1$ and $t_2$ are the starting and ending times. If we rotate a parametrized curve around the $x$ axis we get a surface, called a surface of revolution. The area of this surface is $$\int_{t_1}^{t_2} 2 \pi y \; \sqrt{\left ( \frac{dx}{dt}\right )^2 + \left ( \frac{dy}{dt}\right )^2} \;dt.$$ |