mateomy
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Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line.
y= 1 + secx, y=3, about y=1
Set up:
1-1+secx= secx
1-3= -2
A(x)= \pi(secx)^2\,-\,\pi(-2)^2
=\pi(sec^2 x - 4)
<br /> \pi\int_{\frac{-\pi}{3}}^{\frac{\pi}{3}}\,(4-\sec^2x)\,dx<br />
All in all I come out with \pi(4\pi/3 - \sqrt{3})
The answer at the end of the book is telling me 2pi rather than pi, I can't figure out why. Can anyone explain this to me? THanks.
y= 1 + secx, y=3, about y=1
Set up:
1-1+secx= secx
1-3= -2
A(x)= \pi(secx)^2\,-\,\pi(-2)^2
=\pi(sec^2 x - 4)
<br /> \pi\int_{\frac{-\pi}{3}}^{\frac{\pi}{3}}\,(4-\sec^2x)\,dx<br />
All in all I come out with \pi(4\pi/3 - \sqrt{3})
The answer at the end of the book is telling me 2pi rather than pi, I can't figure out why. Can anyone explain this to me? THanks.