LogFAQs > #888569349

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Topic	Board 8 math people, got a question for you (calc 2)
azuarc 10/16/17 9:28:20 PM #12:	What you need to recall about surface area of a rotated figure is that you're basically adding up circumferences over and over. So if sqrt(4-x^2) is the radius, the circumference is 2pisqrt(4-x^2). You want to be integrating the "slices" of that from -1 to 1. --- http://imgur.com/ekqlssN <cite>azuarc posted [p:888569349] in Board 8 ma... [t:75886889]...</cite> <quote>What you need to recall about surface area of a rotated figure is that you're basically adding up circumferences over and over. So if sqrt(4-x^2) is the radius, the circumference is 2pisqrt(4-x^2). You want to be integrating the "slices" of that from -1 to 1. --- http://imgur.com/ekqlssN</quote> ... Copied to Clipboard!
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