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For (i), is that I have to show $curl F = 0$ ?

For (ii) and (iii), what should I do in order to find the potential function and work done? Also, is the answer $4$ for (iii)?


(i) That's correct.

(ii) You want $\nabla f = (2x \cos y -2z^3, 3 + 2ye^z-x^2\sin y,y^2e^z - 6xz^2) = F$. Then you need to integrate

$$\frac{\partial f}{\partial x} = 2x \cos y -2z^3, \frac{\partial f}{\partial y}=3 + 2ye^z-x^2\sin y,\frac{\partial f}{\partial z}=y^2e^z - 6xz^2 $$ to find $f$.

(iii) Evaluate $\int F\dot \ dr = f(B) - f(A)$.

Edit: $$\begin{align} \frac{\partial f}{\partial x} &= 2x \cos y -2z^3 \Rightarrow \int \frac{\partial f}{\partial x} dx = \int (2x \cos y -2z^3) dx \\&= x^2\cos y - 2x z^3 + g(y,z) \\ &\Rightarrow f(x,y,z) = x^2 \cos y - 2x z^3 + g(y,z) \end{align}$$

Now you need to find the function $g(y,z)$, to do so derivate with respect to $y$ and compare to $\frac{\partial f}{\partial y}$. You need to do the same to the other partials.

  • $\begingroup$ @user110858 If you have any question, feel free to ask. $\endgroup$ – Aaron Maroja Nov 12 '14 at 12:40
  • $\begingroup$ Why I have to integrate the above equations? I am still confused. $\endgroup$ – user110858 Nov 14 '14 at 9:12
  • $\begingroup$ In order to find $f(x,y,z)$, we have by definition that $\nabla f = \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y},\frac{\partial f}{\partial z} \rangle$ and we want $\nabla f = F$. $\endgroup$ – Aaron Maroja Nov 14 '14 at 9:15
  • $\begingroup$ So, is the answer $2xsinx +2cosx -2xz^3 +3y +y^2 e^z +x^2 cosy + Constant$ ? $\endgroup$ – user110858 Nov 14 '14 at 9:42
  • $\begingroup$ You need to be careful, I'll do the first integration. $\endgroup$ – Aaron Maroja Nov 14 '14 at 9:48

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