They ask me to calculate this line integral and I want to make sure that I am doing this configuration correctly. So I tried to paramatise this surface by:

$ x = rcos ( theta), space y = rsin ( theta), space z = sqrt {1 – r} $

$ 0 leq r leq 1 $ Y $ 0 leq theta leq 2 pi $

Then I found:

$ Phi _ { theta} = -rsin ( theta) i + rcos ( theta) j + 0k, space Phi_ {r} = cos ( theta) i + sin ( theta) j + frac {1} {2 sqrt {1-r}} k $

what I would do:

$ Phi _ { theta} times Phi_ {r} = frac {1} {2} bigl ( frac {rcos ( theta)} { sqrt {1-r}} bigr) i + frac {1} {2} bigl ( frac {rsin ( theta)} { sqrt {1-r}} bigr) j -rk $

then my surface integral would be:

$ int_0 ^ {2 pi} int_0 ^ 1 ((rcos ( theta) + 3 (rsin ( theta)) ^ 5) i + (rsin ( theta) + 10 (rcos ( theta)) ( sqrt {1 – r})) j + ( sqrt {1 – r} – (rcos ( theta) rsin ( theta)) k) space cdot ( frac {1} {2} bigl ( frac {rcos ( theta)} { sqrt {1-r}} bigr) i + frac {1} {2} bigl ( frac {rsin ( theta)} { sqrt {1-r }} bigr) j -rk) $

Is this setting correct? I tried to reduce it after taking the product of points with the trigonometric identities, but I still have a rather complicated expression.