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.
In the plane x = 3, which is now the ground plane (cfr. the xy-plane z = 0 from last time), the equation is $\displaystyle x^2 + y^2 + 9 = 25 \Leftrightarrow x^2 + y^2 = 4^2 $ which happens to be the same circle as last time.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.
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.