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Math Help - Vector Calculus

  1. #1
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    Vector Calculus

    Calculate the area of the part of the cone \ z^2 = x^2 + y^2 that is inside the cylinder  \ x^2 + y^2 \leq 2x, outside the cylinder \ x^2 + y^2 \leq 1 and above the xy plane.

    ___________________________________________

    Cylinder: (ucosv, usinv, u)

    Attempt:

    2x = 1 ---> x = 1/2 ---> y = \sqrt(3)/2

    ucos(v) = 1/2
    usen(v) = \sqrt(3)/2

    Solving I get: u = \sqrt(3)/2

    and

    \int_0^{2\pi} \int_0^{\sqrt{3}/3} u\sqrt{2}dudv

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  2. #2
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    Re: Vector Calculus

    That would be correct if the base in the xy-plane were the disk with center at (0, 0) with radius \sqrt{3}/2 (you have \sqrt{3}/3 but you mean \sqrt{3}/2). But in fact, it is the intersection of the two disks, x^2+ y^2\le 1, the circle with center at (0, 0) and radius 1, and x^2- 2x+ y^2= (x-1)^2+ y^2= 1. The points of intersection are (1/2, \sqrt{3}/2) and (1/2, -\sqrt{3}/2) which correspond to \theta= \pi/3, r= 1 and \theta=-\pi/3, r= 1 in polar coordinates.

    You will need to do this as three separate integrals. The main one will be with \theta going from -\pi/3 to \pi/3, r from 0 to 1. That leaves the two small sections between the straight lines from (0, 0) to (1/2, \sqrt{3}/2) and from (0, 0) to (1/2, -\sqrt{3}/2) and the circle (x-1)^2+ y^2= 1. Taking \theta from \pi/3 to \pi/2, r goes from 1 to the curve x^2+ y^2= 2x which, in polar coordinates is r^2= 2rcos(\theta) or r(r- 2cos(\theta)= 0 for r not 0, r= 2cos(\theta) so that r goes from 0 to 2cos(\theta). For the last piece, \theta goes from 3\pi/2 to 2\pi- \pi/3= 5\pi/3 and, for each \theta r goes from 0 to 2cos(\theta) also.
    Last edited by HallsofIvy; June 24th 2011 at 03:39 AM.
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  3. #3
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    Re: Vector Calculus

    Perfect. Thank you so much for the explanation.
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