Using the specified oriented surface, the surface integral is F=y³i+8j-xk is a downward-pointing normal is -53/60.
Given that,
We have to find using the specified oriented surface, calculate the surface integral: F=y³i+8j-xk is a downward-pointing normal that is a part of the plane x+y+z=1 in the octant x,y,z≥0.
We know that,
Finding the surface integral of [tex]\int {} \,\int\limits_s {F} \, ds[/tex]
Here,
The surface of the plane's part is x+y+z=1;
In the 1st octant, x,y,z≥0, and the oriented surface F=y³i+8j-xk.
The plane x + y + z = 1
z=1-x-y
dz/dx=-1
dz/dy=-1
So, the surface is oriented downward
dS= (dz/dzi+dz/dyj-k)dxdy
dS=(-i-j-k)dxdy
So, we get
[tex]\int {} \,\int\limits_s {F} \, ds[/tex]
[tex]\int {} \,\int\limits_s {F} \, (dz/dzi+dz/dyj-k)dxdy[/tex]
-53/60
Therefore, Using the specified oriented surface, the surface integral is F=y³i+8j-xk is a downward-pointing normal is -53/60.
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