I have been struggling with this for an hour

[Muteki]

If you have the Thomas Calculus Alternate Edition, it's problem 11 in section 5.3

I need to find the volume by revolving the region about the given line.

The region in the first quadrant bounded above by y = sqr(x), below by curve y = sec x tan x, and on the left by the y-axis, about the line y = sqr(2).


Thanks for any help you can offer.

EDIT: Here is the work I have done so far.

I graphed the equation, and have set up the integral to look like

V(x) = pi * the integral from 0 to pi/4 of (sqr(2) - sec x tan x ) ^2 dx

I'm not sure what to do with the sec x tan x. I have tried reducing it down to either just sin or cos, but everything ends up at a dead end for me.

 

[morson]

\(\displaystyle \L\ \frac{d}{dx}\ (secx) = secx tanx\)

 

[Opalg]

sec x tan x is the derivative of sec x, and sec^2 x tan^2 x is the derivative of (1/3)tan^3 x.

 

[Muteki]

\(\displaystyle \L\ \frac{d}{dx}\ (secx) = secx tanx\)

Sorry if I didn't make it clear, but the whole thing is squared which was throwing me off.

sec x tan x is the derivative of sec x, and sec^2 x tan^2 x is the derivative of (1/3)tan^3 x.

Thank you very much. That is what I needed

EDIT: Sorry I'm new to these forums and all, but is there anyway to give you guys karma anything like that?

 

[morson]

\(\displaystyle \L\ \int sec^2(x) tan^2(x)\ dx = \int sec^2(x) u^2\ dx = \int u^2\ du\), where \(\displaystyle u = tan(x)\)

... if you're interested in knowing its derivation.