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 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 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 co÷rdinates, but it isn't necessary.
How do you go about doing a question using the method you applied previously but where curl F = 0.
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.