Calc Volume of Disk - Rotate Region Bounded by y=1+secx, y=3 About y=1

[mateomy]

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)= [itex]\pi(secx)^2\,-\,\pi(-2)^2[/itex]
=[itex]\pi(sec^2 x - 4)[/itex]

[tex]
\pi\int_{\frac{-\pi}{3}}^{\frac{\pi}{3}}\,(4-\sec^2x)\,dx
[/tex]

All in all I come out with [itex]\pi(4\pi/3 - \sqrt{3})[/itex]

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.

 

[Mark44]

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)= [itex]\pi(secx)^2\,-\,\pi(-2)^2[/itex]
=[itex]\pi(sec^2 x - 4)[/itex]

[tex]
\pi\int_{\frac{-\pi}{3}}^{\frac{\pi}{3}}\,(4-\sec^2x)\,dx
[/tex]

All in all I come out with [itex]\pi(4\pi/3 - \sqrt{3})[/itex]

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.

I think you probably made a sign error somewhere. Since your integrand is an even function, you can integrate between 0 and pi/3 and double that result.
This gives the right value.
[tex]2\pi\int_0^{\frac{\pi}{3}} (4-sec^2(x))~dx[/tex]

 

[mateomy]

AHHHH! I am usually super careful with keeping an eye out for the even/odd functions! THat makes it perfectly obvious, thanks a lot.