Surface Integral

KindofSlow

Junior Member
Joined
Mar 5, 2010
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90
The problem is:
Evaluate surface integral ∫∫ 6xy dS where S is the portion of the plane x+y+z=1 that lies in front of the yz plane.
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My only question is regarding the area of "D" on the yz plane and the resultant limits of integration.
The answer in the book shows 0≤ y ≤1 and 0≤ z ≤(1-y), which is all in the first octant where x,y,z are all positive.
Since there are many points (everywhere that x is positive and either y or z are negative) that fulfill x+y+z=1 and also lie in front of the yz plane.
I think either:
1. The problem should state "...that lies in front of the yz plane in the first octant where x,y,z are are all positive."
or 2. I am wrong about something.
I hope my explanation makes sense.
Any illumination will be greatly appreciated.
Thank you
 
I think that "in front of the yz plane" means the first octant.

\(\displaystyle z=1-x-y\)

R is the region enclosed by \(\displaystyle x+y=1, \;\ x=0, \;\ y=0\)

\(\displaystyle \displaystyle 6\int\int xyzdS=6\int\int_{R}xy(1-x-y)\sqrt{3}\)

\(\displaystyle \displaystyle=6\sqrt{3}\int_{0}^{1}\int_{0}^{1-x}(xy-x^{2}y-xy^{2})dydx\)
 
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