Surface Integrals

**adam_leeds** · May 8th 2011, 03:26 AM

$\iint_s (x+y) d\sigma$ , 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?

**Danny** · May 8th 2011, 06:06 AM

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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