Answer
$0$
Work Step by Step
For a vector field to be Conservative, $\dfrac{\partial f_1}{\partial y}=\dfrac{\partial f_2}{\partial x}$ and $\dfrac{\partial f_2}{\partial z}=\dfrac{\partial f_3}{\partial y}$ and $\dfrac{\partial f_1}{\partial z}=\dfrac{\partial f_3}{\partial x}$
We are given that the force field as: $F(x,y, z)=(2xy+z^2, x^2, 2xz)$
So, we can see that $\dfrac{\partial f_1}{\partial y}=2x=\dfrac{\partial f_2}{\partial x}$ and $\dfrac{\partial f_2}{\partial z}=0=\dfrac{\partial f_3}{\partial y}$ and $\dfrac{\partial f_1}{\partial z}=2z=\dfrac{\partial f_3}{\partial x}$
Next, we will find the potential function for the given vector field as:
$\phi(x,y, z)=x^2 y+xz^2$
Also, the curve is $r(t)= (3 \cos t, 4 \cos t, 5 \sin t)$
Therefore, the integral can be expressed as:
$\int_C F \ dr=\phi[r(2\pi)]-\phi[r(0)] \\=\phi (3,4,0)-\phi (3,4,0)\\(3)^2(4)+0-[(3)^2(4)+0]\\=0$
