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Math Help - Surface Area of revolving object

  1. #1
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    Surface Area of revolving object

    Problem:
    The arc of the curve y=arcsec(x) from x=2/sqrt(3) to x=2 is revolved about the line y=pi/2.

    Could someone help me set up the integrals to the find surface area by integrating with respect to both x and y? (I could solve myself after I get it set up)
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  2. #2
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    Draw yourself the graph and the curve. Now convince yourself the rotating \sec^{-1}(x) around the line y=\pi/2 is the same as moving the curve down \pi/2 units and rotating about y=0. When we move the curve down the new curve is \sec^{-1} x - \pi / 2. Thus, find the surface area of revolution of this new curve using the formula that you learned from x=2/\sqrt{3} to x=2.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    Draw yourself the graph and the curve. Now convince yourself the rotating \sec^{-1}(x) around the line y=\pi/2 is the same as moving the curve down \pi/2 units and rotating about y=0. When we move the curve down the new curve is \sec^{-1} x - \pi / 2. Thus, find the surface area of revolution of this new curve using the formula that you learned from x=2/\sqrt{3} to x=2.
    Well the formula I have is S = \int_{c}^{d} 2 * pi * g(x) sqrt{1+g'(x)^2} but where does it mean by integrating with respect to x and with respect to y?
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  4. #4
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    2\pi \int_a^b f(x) \sqrt{1+[f'(x)]^2} \ dx.

    As I explain a=2/\sqrt{3} and b=2. But instead of f(x) = \sec^{-1} x we have rather f(x) = \sec^{-1} x - \frac{\pi}{2}.

    Now continue.
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