Question

WIll award medal
stokes theorem

Answer 1

@iambatman

Answer 2

Im just curious on the parameterized function r(r,theta) for the z component. i think it is r^2 but idk

Answer 3

for the surface you know z = x^2 + y^2
and you know
x^2 + y^2 = 1
you can replace x^2 + y^2 with z
Z = 1

Answer 4

1 is your z component not r^2
because r ^2 is saying that your radius is changing and that would be false in this problem.

Answer 5

r(t) = (cost,sint,1)

Answer 6

z=1 and dz =0. makes it much simpler than it might look

Answer 7

and then for would r(r,theta) be <rcos(theta),r,rsin(theta)>????

Answer 8

@Austin6i6

Answer 9

the line integral is
\[\int\limits <x^2z^2,\ y^2z^2,\ xyz>•<dx,dy,dz>\]but z = 1, dz = 0 so
\[\int\limits\limits <x^2, y^2, xy>•<dx,dy,0> = \int\limits\limits x^2 dx + y^2 dy\]
and the line defines x vs y as thus:
\[x^2 + y^2 = 1\]
we can now paramaterise -- there is absolutely no need to do so -- but we can place this in trig space. radius = 1, so x = cos(t) and y = sin(t).

dx = -sin(t) dt ; and dy = cos(t) dt.

Answer 10

\[\int\limits_{t = 0}^{2 \pi } \cos^2 t \ (-sin t) + \sin^2 t \ \cos t \ dt = \frac{1}{3} \left| \sin^3 t = \cos^3 t \right|^{2 \pi}_{0} = 0\]