Find the area of the shape between f(x) = x and g(x) = x^2 when revolved around the line x=5.
[spoiler]this is actually really easy but eh [/spoiler]
English
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Edited by Acronocos: 3/27/2016 3:20:32 AMDammit I forgot how to do this shit
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Pi * integral [(5-f(y))^2 - (5-g(y))^2]dy Lower value: 0 Upper value: 1
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That's the cylinder equation right?
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It's the washer method since it revolves around a line it doesn't touch
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[quote]It's the washer method since it revolves around a line it doesn't touch[/quote] *brain explodes*
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3[b][/b]
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I have no idea lmao, I just made up the question and didn't bother doing it. That sounds about correct though.
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I just went with the first number that I thought of lol
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I'll do this one later when I have motivation to work it out