KindofSlow
Junior Member
- Joined
- Mar 5, 2010
- Messages
- 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.
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
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.
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