Find a function f such that F=\triangledown fF(x,y)=x^2i+y^2jN

Anish Buchanan

Anish Buchanan

Answered question

2021-09-12

Find a function f such that F=
F(x,y)=x2i+y2j
C is the arc of the parabola y=2x2 from (-1,2) to (2,8)

Answer & Explanation

Liyana Mansell

Liyana Mansell

Skilled2021-09-13Added 97 answers

The x component of the given field is equal to x2. We first integrate with respect to x. It follows that:
f(x,y)=x33+g(y)
Take a partial derivative of f with respect to y to obtain that:
g(y)=y2=fy
Next, we integrate once again but this time with respect to y to get:
f(x,y)=y33+x33+C
Set C=0 since it doesnt matter for the fundamental theorem to be applied. We have:
f(x,y)=x3+y33
Result: f(x,y)=x3+y33

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