Stokes' Theorem to evaluate integral

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Use Stokes' Theorem to evaluate $\int\limits_C (-y^3dx + x^2dy-z^3dz)$ where C is the interaction of cylinder $x^2+y^2=1$ and plane $x+y+z=1$.

The surface on which the integral is to be done should be an ellipse according to me. What should be the integral limits for such an ellipse ?

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Let $\vec F(x,y,z)=-\hat x y^3+\hat y x^2-\hat z z^3$ and $C$ be the contour written parametrically as

$$\begin{align} x&=\cos(\phi)\\\\ y&=\sin(\phi)\\\\ z&=1-\sin(\phi)-\cos(\phi) \end{align}$$

for $0\le \phi <2\pi$.

We could proceed to calculate the line integral of $\vec F(x,y,z)$ directly as

$$\begin{align}\oint_C \vec F(x,y,z)\cdot \,d\vec \ell&=\underbrace{\int_0^{2\pi}\sin^4(\phi)\,d\phi}_{=3\pi/4}\\\\ & +\underbrace{\oint_C\cos^3(\phi)\,d\phi}_{=0}\\\\ &-\underbrace{\oint_C (1-\sin(\phi)-\cos(\phi))^3(\sin(\phi)-\cos(\phi))\,d\phi}_{=0}\\\\ &=3\pi/4 \tag 1\end{align}$$

The mechanics of directly evaluating the integrals on the right-hand side of $(4)$ are straightforward and include exploiting symmetry and using trigonometric identities. We next proceed to evaluate the line integral in $(1)$ by invoking Stokes' Theorem.


From Stokes' Theorem, we have

$$\oint_C \vec F(x,y,z)\cdot \,d\vec \ell=\int_{S}\nabla \times \vec F(x,y,z)\cdot \hat n\,dS$$

where $C$ is a contour that bounds the open surface $S$.

Here, the Curl of $\vec F(x,y,z)$ is

$$\nabla \times \vec F(x,y,z)=\hat z (2+3x)x$$

One surface that is bounded by $C$ is the surface located by $\vec R(u,v)=\hat x u+\hat y v +\hat z(1-u-v)$, where $u\in [-1,1]$ and $v\in [-\sqrt{1-u^2},\sqrt{1-u^2}]$. The vector surface differential $\hat n\,dS$ can be found as

$$\begin{align} \hat n\,dS&= \left(\frac{\partial \vec R(u,v)}{\partial u}\times \frac{\partial \vec R(u,v)}{\partial v}\right)\,du\,dv\\\\ &=(\hat x +\hat y+\hat z)\,du\,dv \end{align}$$

Therefore, we find that

$$\begin{align} \int_S \nabla \times \vec F(x,y,z)\,dS&=\int_{-1}^1\int_{-\sqrt{1-u^2}}^{\sqrt{1-u^2}} (2u+3u^2)\,du\,dv\\\\ &=12\int_0^1 u^2\sqrt{1-u^2}\,du\\\\ &=3\pi/4 \end{align}$$

which agrees with the result obtained directly!