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(Solved): Find the flux of the given vector field F across the given surface S in the given direction: 4 ...


Find the flux of the given vector field ?F across the given surface S in the given direction:

4. \( \overrightarrow{\mathrm{F}}=\langle x y, 0,-z\rangle ; \mathrm{S} \) is the surface of the cone \( z=\sqrt{x^{2}+y^{2}}

4. \( \overrightarrow{\mathrm{F}}=\langle x y, 0,-z\rangle ; \mathrm{S} \) is the surface of the cone \( z=\sqrt{x^{2}+y^{2}}, 0 \leq z \leq 1 \); outward direction.


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