Respuesta :
Answer:
a) [tex]f (x,y,z)= xy^2\sin(z)[/tex]
b) [tex]\int_C F \cdot dr =0[/tex]
Step-by-step explanation:
Recall that given a function f(x,y,z) then [tex]\nabla f = (\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z})[/tex]. To find f, we will assume it exists and then we will find its form by integration.
First assume that F = [tex]\nabla f[/tex]. This implies that
[tex]\frac{\partial f}{\partial x} = y^2\sin(z)[/tex] if we integrate with respect to x we get that
[tex] f(x,y,z) = xy^2\sin(z) + g(y,z)[/tex] for some function g(y,z). If we take the derivative of this equation with respect to y, we get
[tex] \frac{\partial f}{\partial y} = 2xy\sin(z) + \frac{\partial g}{\partial y}[/tex]
This must be equal to the second component of F. Then
[tex]2xy\sin(z) + \frac{\partial g}{\partial y}=2xy\sin(z)[/tex]
This implies that [tex]\frac{\partial g}{\partial y}=0[/tex], which means that g depends on z only. So [tex] f(x,y,z) = xy^2\sin(z) + g(z)[/tex]
Taking the derivative with respect to z and making it equal to the third component of F, we get
[tex] xy^2\cos(z)+\frac{dg}{dz} = xy^2\cos(z)[/tex]
which implies that [tex]\frac{dg}{dz}=0[/tex] which means that g(z) = K, where K is a constant. So
[tex]f (x,y,z)= xy^2\sin(z)[/tex]
b) To evaluate [tex]\int_C F \cdot dr [/tex] we can evaluate it by using f. We can calculate the value of f at the initial and final point of C and the subtract them as follows.
[tex]\int_C F \cdot dr = f(r(\pi))-f(r(0))[/tex]
Recall that [tex]r(\pi) = (\pi^2, 0, \pi)[/tex] so [tex]f(r(\pi)) = \pi^2\cdot 0 \cdot \sin(\pi) = 0[/tex]
Also [tex]r(0) = (0, 0, 0)[/tex] so [tex]f(r(0)) = 0^2\cdot 0 \cdot \sin(0) = 0[/tex]
So [tex]\int_C F \cdot dr =0[/tex]
