Verify Stokes theorem for F =(y^2 + x^2 - x^2)i + (z^2 + x^2 - y^2)j + (x^2 + y^2 - z^2)k over the portion of the surface x^2 + y^2 -2ax + az = 0

Question

While evaluating the integral we get hard to evaluate integrals. What can we do to simplify this?

Answer 1

F =(y^2 + x^2 - x^2)i = TYPO?

Answer 2

if as you wrote
curl
i component = dFz/dy - dFy/dz = 2 y - 2z
j component = dFx/dz - dFz/dx = 0 - 2x = -2x
k component = dFy/dx - dFx/dy= 2x + 2y

Answer 3

Yes there's a typo :

Verify Stokes theorem for F =(y^2 + z^2 - x^2)i + (z^2 + x^2 - y^2)j + (x^2 + y^2 - z^2)k over the portion of the surface x^2 + y^2 -2ax + az = 0

While evaluating the integral we get hard to evaluate integrals. What can we do to simplify this?

Answer 4

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

Answer 5

surface:
x^2 - 2 a x + y^2 + az= 0
x^2 -2ax + a^2 + y^2 + az-a^2 = 0
(x-a)^2 + y^2 = a (a-z)
circles at every height z, no go for z>a
line integral around circle at height z and surface integral through the circle.

Answer 6

Thank you! This makes it much clearer!

Answer 7

You are welcome, good luck :)