Evaluate Integral over conservative vector field

Question

Evaluate the line integral $$ \int\limits_C F \cdot dr $$ where $\DeclareMathOperator{grad}{grad}F= \grad f$, $f(x,y,z)=\sin(x)\cos(y)\,z$ and $C$ is the circle $x^2 +y^2=1$ and $z=3$.

I understand that the value of the integral is zero because am integrating a closed curve over a conservative vector field, however I am required to show the work. What I have done so far is finding the gradient which is the vector $$ F=(\cos(x)\cos(y)\,z, -\sin(x)\sin(y)\,z, \sin(x)\cos(y)) $$ and I've parametrized the circle as vector $$ r(t)= (\cos(t), \sin(t), 3) $$ but I am not sure how to express $F$ in terms of the parameterized curve.

Answer 1

You insert $$ r(t) = (x(t), y(t), z(t)) = (\cos(t), \sin(t), 3) $$ into your $F$: $$ F(t) = (3\cos(\cos(t))\cos(\sin(t)), -3\sin(\cos(t))\sin(\sin(t)), \sin(\cos(t))\cos(\sin(t))) $$

Answer 2

Given $f(x,y,z)=\sin(x)\cos(y)\,z$ and the path $C:x^2+y^2=1,\,z=3$ then \begin{eqnarray} dr&=&(dx,dy, 0)\\ \cos y&=&\cos\left(\sqrt{1-x^2}\right)\\ \sin x&=&\pm\sin\left(\sqrt{1-y^2}\right)\\ \end{eqnarray}

$$\nabla f=(\cos(x)\cos(y)\,z, -\sin(x)\sin(y)\,z, \sin(x)\cos(y))$$

\begin{eqnarray} \int_C \nabla f \cdot dr&=&\int_C z\cos x\cos y\,dx-z\sin x\sin y\,dy\\ &=&3\int_1^1\cos x\cos y\,dx-3\int_0^0\sin x\sin y\,dy\\ &=&\int_1^1\cos x\cos\left(\sqrt{1-x^2}\right)\,dx\pm\int_0^0\sin y\sin\left(\sqrt{1-y^2}\right)\,dy\\ &=&0 \end{eqnarray}