Question

Find the flux of f = x3 i y3 j z3 k out of the closed surface z = x2 y2 , 0 ≤ z ≤ 5 oriented outward. g

Answer 1

The flux of the given function f = x³i + y³j + z³k is given by f',

f' = 3(x²i + y²j + z²k)

Given, f = x³i + y³j + z³k

we have to find the flux of the given function f = x³i + y³j + z³k

On partial differentiating the function, we get

f' = 3x²i + 3y²j + 3z²k

f' = 3(x²i + y²j + z²k)

So, the flux of the given function f = x³i + y³j + z³k is given by f',

f' = 3(x²i + y²j + z²k)

Hence, the flux of the given function f = x³i + y³j + z³k is given by f',

f' = 3(x²i + y²j + z²k)

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