Thread: Stokes

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.

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.

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.

Originally Posted by jedoob

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

2) z is now >=3.

In the plane x = 3, which is now the ground plane (cfr. the xy-plane z = 0 from last time), the equation is which happens to be the same circle as last time.

The equation can be rewritten explicitly in function of z, because we know z is positive. We then have which gives our r:

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.

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.

Originally Posted by jedoob

How do you avoid getting a zero surface integral?

Please help, thanks.

You don't, since the integral will equal 0...!