A line integral problem

Question

I tried to use stokes theorem in this problem but my answer doesn't match the correct answer.

Please help. This is my try pls help..

Answer 1

Stokes theorem says

$$\int_{C} F \cdot dr=\iint_{S} (\nabla \times F) \cdot \vec n dS$$

Where $C=\partial S$, here $\partial$ stands for the boundary.

Choose $S$ to be the plane $x+y=0$ which lies in $x^2+y^2+z^2 \leq 1$, clearly $\partial S=C$ where $C$ is the curve you want. In your example $\nabla \times F=\langle -1,-1,-1 \rangle$. How about unit the normal vector? We have $1x+1y+0z=0$. So the unit normal is $\frac{1}{\sqrt{2}}\langle 1,1,0 \rangle$. Actually with the correct orientation that goes with the problem it should be $\frac{1}{\sqrt{2}}\langle -1,-1,0 \rangle$.

So $(\nabla \times F) \cdot \vec n=\frac{2}{\sqrt{2}}=\sqrt{2}$.

Parametrizing the surface as $r(x,z)=\langle x,-x,z \rangle$ with $x^2+(-x)^2+z^2 \leq 1$. We obtain $dS=|r_x \times r_z| dA=\sqrt{2} dA$. From here we get,

$$\int_C F \cdot dr=\iint_{D} \sqrt{2} \sqrt{2} dA$$

Where $D$ is an ellipse (an the inside) $\frac{\sqrt{2}}{2}$ by $1$, centered at the origin. The ellipse is $(\frac{x}{\frac{\sqrt{2}}{2}})^2+(\frac{y}{1})^2 \leq 1$ from our parameterization.

Using the formula for the area of inside an ellipse $(\frac{x}{a})^2+(\frac{y}{b})^2=1$, $A=\pi ab$ we have,

$$=2 \iint_{D} dA=2(\pi)(\frac{\sqrt{2}}{2})(1)=\sqrt{2}\pi$$

Though this does not agree, I believe that the program you are using is at fault.