Verifying Stokes' Theorem: F=(x^2,xy,-z^2)

[imana41]

this Q want to check Stokes' theorem ? for [URL]http://latex.codecogs.com/gif.latex?F=(x^2,xy,-z^2)[/URL] and surface [URL]http://latex.codecogs.com/gif.latex?x^2+y^2+z^2=1[/URL] and [URL]http://latex.codecogs.com/gif.latex?z\geqslant%200[/URL]
i should equal [PLAIN]http://latex.codecogs.com/gif.latex?\oint%20Mdx+ndy+pdz
with [PLAIN]http://latex.codecogs.com/gif.latex?\int%20\int%20curlF.n.d\sigma

i know : curlF=(0,0,y) and n=(x,y,z) and [URL]http://latex.codecogs.com/gif.latex?d\sigma%20=\frac{dxdy}{z}[/URL] and [PLAIN]http://latex.codecogs.com/gif.latex?\int%20\int%20curlF.n.d\sigma = [URL]http://latex.codecogs.com/gif.latex?\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}ydydx=\frac{4}{3}[/URL]

but for [PLAIN]http://latex.codecogs.com/gif.latex?\oint%20Mdx+ndy+pdz if i get x=cost ,y=sint and z=0 the answer is [URL]http://latex.codecogs.com/gif.latex?\int_{0}^{\pi/2}x^2dx+xydy-z^2dz=0[/URL] but i can't find my problem??

 

[HallsofIvy]

Why have you not changed $x^2dx+ xydy- z^2dz$ into terms of t?

if x= cos(t) then dx= -sin(t)dt, if y= sin(t) then dy= cos(t)dt, and if z= 0, then dz= 0.
Now, what us $x^2dx+ xydy- z^2dz$ in terms of t?

 

[grey_earl]

$$\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} y dy = 0$$

 

[imana41]

$$\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} y dy = 0$$

i can't say [URL]http://latex.codecogs.com/gif.latex?\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}ydydx[/URL]

= [URL]http://latex.codecogs.com/gif.latex?4\int_{0}^{1}\int_{0}^{\sqrt{1-x^2}}ydydx=\frac{4}{3}[/URL]

 

[imana41]

Why have you not changed $x^2dx+ xydy- z^2dz$ into terms of t?

if x= cos(t) then dx= -sin(t)dt, if y= sin(t) then dy= cos(t)dt, and if z= 0, then dz= 0.
Now, what us $x^2dx+ xydy- z^2dz$ in terms of t?

 

[grey_earl]

i can't say

=

No, you only have
$$\int_{-a}^a f(x) \mathrm{d} x = 2 \int_0^a f(x) \mathrm{d} x$$ when f(-x) = f(x), since then (set t = -x in the first integral that follows)
$$\int_{-a}^a f(x) \mathrm{d} x = \int_{-a}^0 f(x) \mathrm{d} x + \int_0^a f(x) \mathrm{d} x = - \int_a^0 f(-t) \mathrm{d} t + \int_0^a f(x) \mathrm{d} x = \int_0^a [ f(x) + f(-x)] \mathrm{d} x = 2 \int_0^a f(x) \mathrm{d} x$$.
If you have f(x) = -f(-x), then f(x)+f(-x) = 0, so your integral yields zero. This happens for your inner integral, where f(y) = y.

 

[imana41]

thanks for your help