Respuesta :
Answer:
[tex]\oint_C F \,dr = 0[/tex]
Step-by-step explanation:
In this problem, we have a vector field
[tex]F = \left \langle 2x-z^2,x^2+yz,-2xz +\frac{y^2}{2} \right \rangle[/tex].
We need to find the line integral
[tex]\oint_C F \,dr[/tex]
where [tex]C[/tex] is a circle
[tex]r(t) = \left \langle 2 \cos t, 4\sin t, 5 \cos t \right \rangle, \quad 0 \leq t \leq 2 \pi[/tex].
As we can see, the vector filed [tex]F[/tex] is defined [tex]\forall (x,y,z) \in \mathbb{R}^3[/tex] and its component functions have continuous partial derivatives.
First, we need to find the curl of the vector filed [tex]F[/tex].
[tex]\text{curl} F = \begin{vmatrix}i & j & k\\\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\2xy-z^2 & x^2 + yz & \frac{y^2}{2} - 2xz\\\end{vmatrix}[/tex]
Therefore,
[tex]\text{curl}F = i \left( \frac{\partial}{\partial y} \left( \frac{y^2}{2} - 2xz\right) - \frac{\partial}{\partial z} \left( x^2+yz\right)\right) - j \left( \frac{\partial}{\partial x} \left( \frac{y^2}{2} - 2xz\right) - \frac{\partial}{\partial z} \left( 2xy-z^2\right)\right)\\ \phantom{12356} +k \left( \frac{\partial}{\partial x} \left( x^2+yz\right) - \frac{\partial}{\partial y} \left( 2xy-z^2\right)\right)[/tex]
Now, we can easily calculate the needed partial derivatives and we obtain
[tex]\text{curl} F = i(y-y) - j(-2z + 2z) + k(2x-2x) = 0i + 0j+0k = \langle 0,0,0 \rangle[/tex]
So, the vector field [tex]F[/tex] is defined [tex]\forall (x,y,z) \in \mathbb{R}^3[/tex] , its component functions have continuous partial derivatives and [tex]\text{curl} F = 0[/tex] .Therefore, by a well-known theorem, [tex]F[/tex] is a conservative field.
Since [tex]C[/tex] is a closed path, we obtain that
[tex]\oint_C F \,dr = 0[/tex]
