Let $\vec{F}= 2y \hat{i}+ x^2 \hat{j} + xy \hat{k}$ and let C be the curve of intersection of the plane $x + y + z = 1$ and the cylinder $x^2 + y^2 = 1$. Then find the value of $$\left|\oint \vec{F}.d\vec{r}\right|$$
My question is can I use Stokes’ Theorem in here? (Why or why not?) If yes, then which surface should I consider for evaluating the Normal vector? (That of the plane or that of the cylinder?). Also, kindly provide me with the solution of this problem. Thanks :)
$\iint \nabla \times F\ dA = \oint f\cdot dr$
$\nabla \times F = (\frac{\partial}{\partial y} xy - \frac{\partial}{\partial z} x^2,\frac{\partial}{\partial z} 2y- \frac{\partial}{\partial x} xy,\frac{\partial}{\partial x} x^2 - \frac{\partial}{\partial y} 2y) = (x, -y, 2x-2)$
..... you can skip much of what follows, I am complicating things more than I need to ....
the normal to the surface is $(\frac {1}{\sqrt 3},\frac 1{\sqrt 3},\frac 1{\sqrt 3})$
$\iint \frac {3x - y - 2}{\sqrt 3} \ dS$
Parameterize that surface?
$u = \frac {x-y}{\sqrt 2}\\ v = \frac {x+y - 2z}{\sqrt 6} \\ w = \frac {x+y+z}{\sqrt 3}$
This coordinate transformation is orthonormal.
jacobian $\ dx\ dy\ dz = du\ dv\ dw$
In these coordinates $w = 0$
The semi-major axis of this elipse is at the point $(0,\sqrt {3},0)$ and the semi-minor axis is at $(1,0,0)$
or
$x = \frac{1}{\sqrt 2} u + \frac {1}{\sqrt 6} v + \frac{1}{\sqrt 3} w\\ y = -\frac{1}{\sqrt 2} u + \frac {1}{\sqrt 6} v + \frac{1}{\sqrt 3} w\\ z = -\frac {2}{\sqrt 6} v + \frac{1}{\sqrt 3} w$
$x^2 + y^2 = 1$ with $x+y+z = 0$ becomes $u^2 + \frac {v^2}3 = 1$ with $w= 0$
Putting it together we have
$\iint \frac {a_1 u + a_2 v - 2}{\sqrt 3} \ du \ dv$
We don't need to work out what the linear coefficients of u,v are because whatever they are, over this ellipse:
$\iint a_1 u + a_2 v\ du\ dv = 0$
$-\frac 2{\sqrt 3} A$
$A$ is the area of the ellipse. $A = \pi\sqrt 2$
The integral evaluates to $-2\pi$
.... Resume here, I don't have the heart do delete it .......
instead of the coordinate transformation, we can say:
$x = x\\y = z\\ z= -x-y\\dS = (-\frac {\partial z}{\partial x},-\frac {\partial z}{\partial y}, 1)\ dx\ dy$
$\iint (x, -y, 2x-2)\cdot (1,1,1)\ dx\ dy\\ \iint x -y+ 2x-2\ dx\ dy$
$\iint x -y+ 2x \ dx\ dy$ are odd functions evaluated over a region that is symmetric.
$\iint -2 \ dx\ dy = -2\pi$
How did this compare to calculating it directly?
$(x,y,z) = (\cos t, \sin t, 1 - \cos t - \sin t) dr = (-\sin t, \cos t, 1 + \sin t - \cos t)$
$F\cdot dr = (2\sin t, \cos^2 t, \cos t\sin t)\cdot(-\sin t, \cos t, 1 + \sin t - \cos t)\\ -2\sin^2t + \cos^3 t + \cos t\sin t + \cos t\sin^2 t - \cos^2 t\sin t$
That is going to be a snap to integrate. Most of those terms evaluate to 0
$\int_0^{2\pi} -2\sin^2t + \cos^3 t + \cos t\sin t + \cos t\sin^2 t - \cos^2 t\sin t \ dt\\ \int_0^{2\pi} -2\sin^2t\ dt\\ -2\pi$