Consider the space region bounded below by the right angled cone z=sqrt(x^2+y^2), and above by the sphere
x^2+y^2+z^2=2.
These two surfaces intersect in a horizontal circle, let T be the horizontal disk having this circle as boundary, S the spherical cap forming the upper surface, and U the cone forming the lower surface.

Orient S,T,U "upwards", so the normal vector has a positive k-component.

QUESTION 1)For each of the three surfaces, determine geometrically (without calculation) whether the flux of the vector field F= -x*i-y*j is positive or negative.

QUESTION 2) Calculate the flux of F ACROSS EACH SURFACE (with the upwards orientation) .

Solution Preview

1.)
Because, S, T and U are upward, therefore, the area vector for cone (U)
will be diverging from the axis of cone (z-axis) and normal to the surface, for sphere (S) will be converging towards the origin and for circle (T) it will be in only in +ve z direction.
F = -x*i-y*j will always be in a plane parallel to x-y plane and directed towards the z axis, therefore for cone (U) the flux (F.A) will be -ve, for sphere (S) it will be +ve and for circle (T) it will be zero. --Answer

2.)
For ...

Solution Summary

This shows how to work with the flux of given vector fields.

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