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Math Help - Surface Integrals

  1. #1
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    Surface Integrals

    \iint_s (x+y) d\sigma, where S is the portion of the plane x + 2y - 3z = 4 within 0 \leqslant x \leqslant 1, 1 \leqslant y \leqslant 2

    You can't convert to polar co-ordinates as its not a circle, right?
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  2. #2
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    You could but why would you want to. For your integral I think it best to use x and y so from your surface

    z =\frac{x}{3} + \frac{2y}{3} - \frac{4}{3}

    so dS = \sqrt{1 + z_x^2 + z_y^2} = \sqrt{\frac{14}{9}} so the integral becomes

    \sqrt{\frac{14}{9}}\iint_R (x+y) dA

    where R is the region bound by the square - not really amenable to polar coordinates but definitely to cartesian coords.
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