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Q: Find the volume of the solid of revolution obtained by revolving the region bounded by the line $y = 2$ and the curve $y = \sec^2x$, $-90 \lt x \lt 90$, around the x-axis.

Attempted solution: $(1/3) \tan x ( \sec^2x + 2)$

I'm sure I did the integral correctly, but now when I need to sub in for $90, \tan(90)$ does not exist? (The answer is $2(\pi)^2 - 8\pi/3$)

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up vote 1 down vote accepted

Presumably your arguments are degrees, not radians. The volume element of the solid of revolution formed by the rectangle of width $dx$ and extending from $y_1$ to $y_2$ is $\pi(y_2^2-y_1^2)dx$ You integrated $\sec^4(x)dx$ correctly, but that is not quite what you need to integrate.

$\sec^2(x)$ crosses 2-are you supposed to just consider the part of it below y=2? This will eliminate the problem of the tangent blowing up. If you are supposed to consider the area outboard of y=2 all the way to 90 degrees, then the volume is infinite, as your calculation shows.

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Ahhh I see what I did wrong, thanks. – khchan Feb 22 '11 at 0:10

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