Calculating line integrals using Stokes theorem

Question

Use Stokes' Theorem to find the exact value of the line integral $$\int_{C}(y\:dx+z^2\:dy+x\:dz)$$ Where $C$ is the curve of intersection of the plane $2x + z = 0$ and the ellipsoid $x^2 + 5y^2 + z^2 = 1$, oriented counterclockwise as seen from above.

What i tried

Using Stokes theorem, i first find $curl F$ which gives $<-2z,-1,-1>$, however, what im unsure of is that curve $C$ is not directly given and i assume that the ellipsoid represent curve $C$ and in order to find the normal i have to parametrise the ellipsoid $x^2 + 5y^2 + z^2 = 1$ to find the normal, which gives $<x/z,5y/z,1>$ and then i do a dot product of $curl F$ and $n$ to set up the double integral to evulate the line integral. Also Curve C as mentioned is the intersection between the plane and the ellipsoid, hence giving a circle with radius $r=1/5$ hence i have to integrate over this circle.Im unsure whether my assumption is correct though. Could anyone explain. Thanks

Answer 1

As answered by Martin-Blaz We can find the projection of $C$ onto the $xy$-plane, call this Region $D$. $D= \displaystyle\big\{(x,y)| \ x^2+y^2\leq\frac{1}{5} \big\}$

Since this region is simply in the $xy$-plane, we can easily use $\textbf{k}$ as our unit normal vector.

Now we can evaluate the integral \begin{align}\oint\limits_C\textbf{F}\centerdot d\textbf{r} &=\iint \limits_S(\nabla \times\textbf{F}) \ d\textbf{S} \\ &= \iint\limits_D (\nabla \times\textbf{F})\centerdot\textbf{k} \ d\textbf{A} \end{align} Now first note that \begin{align}(\nabla \times\textbf{F})&=\begin{vmatrix}i&j&k\\\frac{\partial}{\partial x} & \frac{\partial}{\partial y}& \frac{\partial}{\partial z} \\ y&z^2&x\end{vmatrix} \\&= \left\langle \frac{\partial}{\partial y}(x)-\frac{\partial}{\partial z}(z^2), \frac{\partial}{\partial z}(y)-\frac{\partial}{\partial x}(x), \frac{\partial}{\partial x}(z^2)-\frac{\partial}{\partial y}(y)\right\rangle \\&= \left\langle -2z,-1,-1\right\rangle\end{align} So \begin{align}(\nabla\times\textbf{F})\centerdot\textbf{k}&=\left\langle-2z,-1,-1 \right\rangle \centerdot \left\langle 0,0,1\right\rangle \\ &=-1\end{align}

Thus we can solve our integral as follows \begin{align}\iint\limits_D (\nabla \times\textbf{F})\centerdot\textbf{k} \ d\textbf{A} &=\iint\limits_{x^2+y^2\leq\frac{1}{5}}(-1)d\textbf{A} \\ &=\underbrace{- \int\limits_{0}^{2\pi}\int \limits_{0}^{\sqrt{\frac{1}{5}}}r \ dr \ d\theta}_\text{In Polar Coordinates}\end{align}

Answer 2

Hint: eliminating $z$ between $x^2 + 5y^2 + z^2 = 1$ and $2x+z=0$ the result is the projection of $C$ on the plane $XY$.