The Vector field $F$ is given by $\mathbf{F}=\left\langle e^{y-z}, 0,0\right\rangle$. Consider the square $S$ with vertices $(9,0,4),(9,9,4),(0,9,4) \text {, and }(0,0,4) \text {. }$
I need to verify Stokes' theorem. I have done the line integral part which i got it as:
$$\oint_{C} \vec{F} \cdot \overrightarrow{d l}=81\left(e^{-5}+e^{4}\right)$$
Now we have $$\vec{\nabla}\times \vec{F}=-e^{y-z}(\hat{j}+\hat{k})$$
Also the unit outward normal to the square $S$ which is in $z=4$ plane is $\hat{n}=\hat{k}$. So we have:
$$\iint_{S}(\vec{\nabla} \times \vec{F}) \cdot \hat{n} d s=\iint_{S}-e^{y-z} d s=-e^{-4}\iint_{S}e^ydydx=-e^{-4} \int_{x=0}^{9} \int_{y=0}^{9} e^{y} d y d x=9\left(e^{5}-e^{-4}\right)$$
Where i went wrong?
Best Answer
Your line integral is wrong. It should be:
$\oint_C \vec{F} \cdot d\vec{l}=\int_0^9 e^{9-4} dx + \int_9^0 e^{0-4} dx=9(e^5-e^{-4})$
(You only have to consider the line integrals from (0,9,4) to (9,9,4) and (9,0,4) to (0,0,4) since the field is perpendicular to the path of integration on the other two sides of the square.)