Doubt in Stokes' theorem & line integral

Question

Use Stokes' Theorem to evaluate the line integral $$ F = -y^3 dx + x^3 dy - z^3 dz$$ where C is the intersection of the cylinder$$ x^2 + y^2 = 1$$ and the plane $$x+y +z=1$$

I solved it but I have 1 doubt in the answer given in the book.

SOLUTION:

$$ \operatorname{curl} F = 3\big(x^2+y^2\big) \,\hat{k} $$

Then taking projection on $x-y$ plane, eventually we get this answer (as per me)

$$ \int_0^{2\pi}\int_0^1\sqrt3r^2 r\,dr\,d\theta. $$

The solution in book says that because $x^2+y^2=1$ (because they lie on the circle), therefore, the answer would be $$ \int_0^{2\pi}\int_0^1\sqrt3 r\,dr\,d\theta. $$

But I think because we are talking about the area of circle, therefore $x,y$ don't lie on the circle.

Answer 1

Putting the question in vector form:

Find the line integral $\oint_C\vec{F}\cdot d\vec{r}$ where $\vec{F}=(-y^3,x^3,-z^3)$ and $C$ is the intersection of the cylinder $x^2+y^2=1$ and the plane $x+y+z=1$.

By Stokes' theorem, we have $$\oint_C\vec{F}\cdot d\vec{r}=\iint_S(\nabla\times\vec{F})\cdot\hat{n}dS$$ where $S$ is the surface enclosed by $C$ and $\hat{n}$ is the positive unit normal to $S$.

We have, $\nabla\times\vec{F}=(0,0,3x^2+3y^2)$ and $\hat{n}=(1/\sqrt{3},1/\sqrt{3},1/\sqrt{3})$. Thus

$$\iint_S(\nabla\times\vec{F})\cdot\hat{n}dS=\iint_R\frac{1}{\sqrt{3}}(3x^2+3y^2)\frac{dxdy}{1/\sqrt{3}}=3\iint_R(x^2+y^2)dxdy$$

where $R$ is the projection of $S$ on the $xy$-plane and note that $\iint_S\vec{F}\cdot\hat{n}dS=\iint_R\vec{F}\cdot\hat{n}\frac{dxdy}{|\hat{n}\cdot\hat{k}|}$. And changing to polar coordinates and integrating the integral, we get $\frac{3}{2}\pi$,, which turned out to be different from the book's as well as your solution.

Answer 2

By Stokes' theorem we have $$\oint_C\vec{F}\cdot d\vec{r}=\iint_E(\nabla\times\vec{F})\cdot\hat{n}\,{\rm d}\omega\ ,$$ whereby $\vec n$ has to be correctly oriented according to the orientation of $C=\partial E$. I'm assuming that $C$ goes counterclockwise when seen from above. In order to compute the integral over the elliptical disc $E$ we need a parametrization of $E$. I shall use polar coordinates $(r,\theta)$ in the $(x,y)$-plane and obtain the parametrization $$(r,\theta)\mapsto\vec r(r,\theta)=(r\cos\theta,r\sin\theta, 1-r\cos\theta-r\sin\theta)\ .$$ This gives $$\vec r_r=(\cos\theta,\sin\theta,-\cos\theta-\sin\theta),\qquad\vec r_\theta=r(-\sin\theta,\cos\theta,\sin\theta-\cos\theta)\ ,$$ so that the scalar surface element ${\rm d}\omega$ becomes $${\rm d}\omega=|\vec r_r\times\vec r_\theta|\>{\rm d}(r,\theta)=\sqrt{3}r\,{\rm d}(r,\theta)\ .$$ On the other hand $${\rm curl}\,\vec F=(0,0,3x^2+3y^2)=(0,0,3r^2),\qquad\vec n={1\over\sqrt{3}}(1,1,1)\ .$$ It follows that $$\iint_E(\nabla\times\vec{F})\cdot\hat{n}\,{\rm d}\omega=\int_0^1\int_0^{2\pi}3r^2\,{1\over\sqrt{3}}\sqrt{3}r\>d\theta\>dr=6\pi\int_0^1r^3\>dr={3\pi\over2}\ .$$ Computing the line integral directly gives the same value.