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The region bounded by y = tan(x), y = 0, x=pi/4 is rotated about the x-axis. The volume generated equals?
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|dw:1401226796135:dw|
the sum of the areas of circles whose radius is equal to f(x) \[\int \pi~r^2\] \[\pi\int_{0}^{\pi/4} tan^2(x)~dx\]
\[\pi [(\sec^2(\pi/4)-1) -(\sec^2(0))] \] ?
tan^2 = sec^2 - 1 integrate gives us tan(x) - x, from 0 to pi/4 is just tan(pi/4) - pi/4; or simply 1 - pi/4, times pi of course
got it! thank you so much!
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