If someone could guide me through the process of solving this integration problem that would be very helpful.
It's an applications problem dealing with the volume generated by rotating a curve by an axis. I'll go strait to the problem: I have to rotate the region bounded by the graphs y=secx, y=1, x=-1, and x=1 with reference to the x-axis.
The integral I came with is:\[\int\limits_{-1}^{1}((1+\sec(x))^2-1)dx\] Before proceeding is the integrand correct?
always graph it first...
Thanks for the advice but I think I already graphed it and I'm struggling with the integral the graph suggests to be the answer for this problem. I don't think we have covered the indefinite integral of sec(x) in our class.
ok, I just have to do it myself to make sure you got the right setup I can't just "see" the answer lol, still human...|dw:1333607809980:dw|
the way I see it the outer radius of each disk is \(\sec x\) and the inner is \(y=1\) going about x this gives the integral\[\pi\int \sec^2x-1dx\]so I think you had it wrong...
this integral really should not be a problem; you should know the integral of sec^2xdx
bounds are -1 to 1 of course...
Did you take into account the distance between secx and y=0?
yes, that is the outer radius\[r_o=y=\sec x\]
the inner radius is\[r_i=y=1\]
I think I see how you approach this, I saw the outer disk as secx+1 and the inner disk as 1. I mean their radii.
you set up like you were revolving around the line y=-1
ok, not quite, but almost like that you sort of mixed the two ideas...
Yeah I see what you mean, I was thrown off by the graph of secx and y=1 so close together.
I really appreciate your help and time. Thanks for everything
which is why I always say to draw the graph do it by hand, then check with wolfram or a graphing calculator to be sure when you can visualization is the key to these problems
One last question though, wouldn't the inner disk be a cylinder?
yes, here it is with a few rings drawn in|dw:1333608686746:dw|
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