Verify Stoke's theorem for this surface

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Homework Statement


The problem and its solution are attached in TheProblemAndSolution.jpg.

Homework Equations


Stoke's theorem: ∮_C F ⋅ dr = ∮_C (FT^) dS = ∫∫_S (curl F) ⋅ n^ dS

The Attempt at a Solution


In the solution attached in the TheProblemAndSolution.jpg file, I don't understand what's going on with the integral that has |n^k^| on a denominator.

Could someone please add the steps that are skipped by the solution?
 

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  • TheProblemAndSolution.jpg
    TheProblemAndSolution.jpg
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An easier way would be to say:

$$\oint_C \vec F \cdot d \vec r = \iint_S \text{curl}(\vec F) \cdot d \vec S = \iint_D \left[(z^2 + x) \hat i - (z+3) \hat k \right] \cdot (\vec r_x \times \vec r_y) \space dA = \iint_D \left[(4 + x) \hat i - 5 \hat k \right] \cdot (\vec r_x \times \vec r_y) \space dA$$

Where ##\vec r(x,y) = x \hat i + y \hat j + 2 \hat k##.

A simple switch to polar co-ordinates afterwards would clean that up nicely.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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