Using Stokes find the closed loop integral of $\vec{F}(x,y,z) = \left< yx-yz , 2x^2+z^2, y^3 \cos(xz)\right>$ on the square

Question

I am asked the following problem:

Using Stokes find the closed loop integral of $\vec{F}(x,y,z) = \left< yx-yz , 2x^2+z^2, y^3 \cos(xz)\right>$ on the square

$$\begin{cases} 0 \leq x \leq 2\\ 0 \leq y \leq 2\\ z=5 \end{cases} $$

My answer is not matching the textbook's answer.

My answer:

Evaluating the curl $\vec{\nabla} \wedge \vec{F}$ we get

$$\vec{\nabla} \wedge \vec{F} = \left< 3y^2\cos(xz)-2z , -y+y^3z\sin(xz) , 3x\right>$$

and since $f(x,y,z) = z-5$ the gradient is $\nabla f = \left< 0,0,1 \right>$

$$\nabla f \cdot \left( \nabla \wedge \vec{F} \right) = 3x$$

That said,

$$\int_{0}^{2} \int_{0}^{2} 3x \ dxdy = \cdots = 12$$

The problem is that the textbook's answer is completely different. Did I make a mistake somewhere?

Testbook's answer: $32$

Answer 1

In your z component of $\nabla \wedge F$ it should be $3x+z$ and the result follows.