How to use Stokes theorem?

Question

Use Stokes theorem to evaluate $\int_{C} [ydx+y^{2}dy+(x+2z)dz]$ where $C$ is the curve of intersection of the sphere $x^{2}+y^{2}+z^{2}=a^{2}$ and the plane $y+z=a$ oriented counterclockwise as viewed from above.
I can't seem to understand the examples in the book. Can someone explain to me what this means and how to apply Stokes formula?

EDIT: I still dont understand what exactly hes talking about i have figured out wha ti am supposed to take the curl of at least (now that i know what thing the curl is suppsoed to be taken on).

Curl F = (0)dydz-(-1)dxdz+(-1)dxdy

from wha ti have been able to decifer i think n is supposed to be the vertor normal to the 2 sufaces $x^{2}+y^{2}+z^{2}=a^{2}$ and $y+z=a$ in which case

$n=\begin{bmatrix} e_{1}&e_{2}&e_{3} \\ 0&1&1 \\ 2x&2y&2z\\ \end{bmatrix}$

$n=(2z-2y)i+(2x)j-(2z)k$

thus $\int\int_{S}2xdxdz+2xdxdy$

$=(x^{2}z+x^{2}y)||_{S}$ which again makes no sense

when i calculate K below i get 1-1=0 which also makes no sense

Answer 1

We can rewrite out integral as

$$\oint_{C} P dx + Q dy + R dz$$

Where $P = y, Q = y^2, R = x + 2z$. We now compute

$$\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} = 0 - 0 = 0$$ $$\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} = 0 - 1 = -1$$ $$\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0 - 1 = -1$$

Stokes theorem tells us that

$$\oint_{C} P dx + Q dy + R dz = \\\int\int_{\mbox{interior of $C$}} \left(\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}\right) dydz - \left(\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}\right) dxdz+ \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dxdy \\ = \int\int_{\mbox{interior of $C$}} dxdz - dxdy \\ = \int\int_{\mbox{interior of $C$}}(0,1,-1) \cdot d\vec{A}$$

What now remains is to compute $d\vec{A} = \vec{n}dA$, where $\vec{n}$ is the normal vector to the interior of $C$, chosen with correct orientation (right hand rule). You can find this by computing the normal of the plane $y+z = a$. Then take the dot product with $(0,1,-1)$ to get some constant $K$, and you will have

$$ K\int\int_{\mbox{interior of $C$}} dA = K \times \mbox{Area of the interior of $C$, which is a disc}$$

Do you know how to proceed from here?

Answer 2

EDIT: Someone bumped this answer so i figured i should correct it...

Im leaving the picture up here my attempt at doing the question in both ways.

$\int_{C}[ydx+y^{2}dy+(x+2z)dz]$ where C is the curve of intersection of the sphere $x^{2}+y^{2}+z^{2}=a^{2}$ and the plane $y+z=a$

Stokes theorem makes very little sense to me but ill give my best shot at an answer both ways.

By a normal evaluation. $z=2*[a^{2}-(x^{2}+y^{2})]^{\frac {1}{2}}$

we have $z=a-y$ Subbing $\to x^{2}+y^{2}+(a-y)^{2}=a^{2}$

$x^{2}+y^{2}+a^{2} +y^{2}-2ay =a^{2}$

$x^{2} +2(y-\frac {a}{2})^2 -\frac {a^{2}}{2} =0$

$(\frac{2^{1/2}x}{a})^{2} + (\frac{2y}{a}-1)^2 =1$ $u=\frac{2^{1/2}x}{a}$ $v=(\frac{2y}{a}-1)$ $u^{2}+v^{2}=1$ = $\pi*1^{2} *2$ * $Det|u,v|$

$Det|u,v|= \begin{bmatrix} \frac {a}{2^{1/2}} & 0 \\ 0 & \frac {a}{2} \\ \end{bmatrix}$

Thus the answer should be $\frac{a^{2}\pi}{2^{1/2}}$

By Stokes.

First lets start with the theorem, $\int_{\partial S} F \cdot dx= \int \int_{S} (curl F) \cdot n dA$

$curl F = \begin{bmatrix} i & j& k \\ d1 & d2 & d3 \\ F1 & F2 & F3 \\ \end{bmatrix}$

$curl F =Det| \begin{bmatrix} i & j& k \\ d1 & d2 & d3 \\ y & y^{2} & (x+2z) \\ \end{bmatrix}|$ = $(0-0)i -(1-0)j+(0-1)k$

Clearly the norm of the surface pointed upward as viewed from above (n) is $Curl F = \begin{bmatrix} 0 \\ -1 \\ -1 \\ \end{bmatrix} n = \begin{bmatrix} 0 \\ 1 \\ 1 \\ \end{bmatrix}$

$Curl F \cdot n= -2$

Now we have $\int \int_{S} -2 dA$ using above dA over S = $\frac{a^{2}\pi}{2^{3/2}}$ so we have the $\frac{a^{2}\pi}{2^{3/2}} * -2 = -\frac{a^{2}\pi}{2^{1/2}}$ not really sure why this one is negative, its probably due to orientation but im not sure which way is right or even if either is correct.