Can anyone help:

Verify stokes theorem for the following vector field:

A = (1+x(y^2))i + (x+2xyz+2)j + (sin z)k

and the surface defined by (x^2)+(y^2)+(z^2)=25 where z>=3.

Thanks.

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- Mar 28th 2006, 08:19 AMjedoobStokes
Can anyone help:

Verify stokes theorem for the following vector field:

A = (1+x(y^2))i + (x+2xyz+2)j + (sin z)k

and the surface defined by (x^2)+(y^2)+(z^2)=25 where z>=3.

Thanks. - Mar 28th 2006, 08:48 AMTD!
It can be done in exactly the same fashion as this problem you posted a while ago :)

It's completely the same question, method, manner of integration: only another area and vector field. - Mar 28th 2006, 11:47 AMjedoobstill having problems
Hello TD,

I am still confused as to how you would do this question for two reasons:

1) how do you recreate the vector r that you had when you answered my question last time?

2) z is now >=3.

Please help. Thanks. - Mar 28th 2006, 11:56 AMTD!Quote:

Originally Posted by**jedoob**

The equation can be rewritten explicitly in function of z, because we know z is positive. We then have $\displaystyle z = \sqrt {25 - x^2 - y^2 }$ which gives our r: $\displaystyle \left( {x,y,\sqrt {25 - x^2 - y^2 } } \right)$

The fact that the minimum z is now 3 just means that you have to replace z by 3 where you replaced it by 0 last time.

Note that it may be interesting to convert to other coördinates, but it isn't necessary. - Apr 3rd 2006, 02:20 AMjedoobcurl F = 0????
Hey,

How do you go about doing a question using the method you applied previously but where curl F = 0.

For example:

Verify stoke for:

F(x,y,z) = (x^2)i + (y^2)j + (z^2)k

z=squareRoot[(x^2)+(y^2)] below the plane z=1.

How do you avoid getting a zero surface integral?

Please help, thanks. - Apr 3rd 2006, 02:28 AMTD!Quote:

Originally Posted by**jedoob**