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Find the area of the part of the plane 3x + 2y + z = 6 that lies in the first octant.
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Find the area of the part of the plane 3x + 2y + z = 6 that lies in the first octant.

User LeoShi
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I solved for z and made my parameters x = u y = v z=6-3u-2v. So I got the integral down do root(14) double integral 1.
User Eric Smith
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