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Math Help - Calculus 3--parametric surfaces/surface area

  1. #1
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    Calculus 3--parametric surfaces/surface area

    1) Eliminate the parameters to obtain an equation in rectangular coordinates, and describe the surface.
    x=ucos(v), y=(u^2), z=usin(v) for 0<=u<=2 and 0<=v<(2pi)
    2) Find the area of the given surface.
    The portion of the surface z=2x + (y^2) that is above the triangular region with vertices (0,0), (0,1), (1,1).
    For this one I know to use a double integral I just need help setting up the region R to use on the integrals.

    Thanks.
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  2. #2
    Super Member Showcase_22's Avatar
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    1).

    x=ucos(v)
    y=u^2
    z=usin(v)

    x^2+z^2=u^2cos^2(v)+u^2sin^2(v)

    x^2+z^2=u^2

    x^2+z^2=y^2

    x^2-y^2=z^2

    Fixing z gives the equation of a hyperbola (ie. hyperbolas as contours).

    The graph has to be two cones "point to point" if that makes sense (ie. two cones, touching each other at the point).

    2).

    I'm not too good at these, but I think you should have this:

    \int_0^1 \int_x^1 2x+y^2 \ dydx
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