1
$\begingroup$

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 $2(\pi)^2 - 8\pi/3$)

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Ahhh I see what I did wrong, thanks. $\endgroup$ – khchan Feb 22 '11 at 0:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.