Part of Plane 3x+2y+z=6 in First Octant

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SUMMARY

The discussion focuses on the calculation of the area of the part of the plane defined by the equation 3x + 2y + z = 6 that lies in the first octant. The initial integration limits proposed were incorrect, as the domain of integration is a triangle in the x-y plane rather than a rectangle. The correct approach requires the limits for y to depend on x, ensuring accurate computation of the area. The final answer should reflect this triangular domain, leading to the correct area calculation.

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

[tex]A(s)=\int_0^2\int_0^3\sqrt{14}dydy[/tex]

Are my limits not correct? B/c my answer is just off by a little.

me: 6\sqr(t14), answer: 3\sqrt(14)
 
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No, not correct. The domain of integration is a triangle in the x-y plane, not a rectangle. So the y limits should depend on x (or vice versa).
 
Dick said:
No, not correct. The domain of integration is a triangle in the x-y plane, not a rectangle. So the y limits should depend on x (or vice versa).
dope! Gotcha, thanks.
 

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