Area of a hemisphere(surface integral)

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Homework Statement


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Homework Equations


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The Attempt at a Solution


I was finally able to find the parameters and do a cross product, i can double integrate but I am unsure what to do in this circumstance(with the absolute value thing as well).
 

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clurt said:

Homework Statement


Please view attached


Homework Equations


Please view attached


The Attempt at a Solution


I was finally able to find the parameters and do a cross product, i can double integrate but I am unsure what to do in this circumstance(with the absolute value thing as well).

||N|| is the same thing as the length of the vector N. Use a little trig.
 
Clurt, when you post in the homework forum, you are expected to show your attempt to solve the problem up to the point where you get stuck. You're asking about what do about the "absolute value thing" in
$$\iint_S \|N(u,v)\|\mathrm du\mathrm dv,$$ where N is a function you have already found. The obvious first step is to look up the definition of ##\|\ \|## in your book and try to use it. We won't have a lot to say until you have done that.
 
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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