Easy Integration: Finding Flux of a Sphere Surface | Double Integral Help

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JaysFan31

Homework Statement


I want to perform the following integration:
double integral of [(x^2)-(y^2)+2] dxdy where the function is subjected to the bound (x^2)+(y^2) greater than or equal to 2.

I'm trying to find the flux of a surface of a sphere (x^2)+(y^2)+(z^2)=9.

Homework Equations


Nothing, just rules of integration.

The Attempt at a Solution


Using Cartesian coordinates seems far too difficult. I could show you my work, but it's messy and complicated.
If I use parametrisation, then I get
double integral of [(r^2)cos^2(t)-(r^2)sin^2(t)+2) rdrdt
what are the bounds though?
t is between 0 and 2pi I'm pretty sure, but what about r?
It seems difficult since x^2+y^2 is greater than or equal to 2. This means that r^2 is greater than or equal to 2. Thus, it seems sqrt(2) is a lower bound, but what would the upper bound be?

Does this make any sense?
 
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For the trig part: Hint: http://www.sosmath.com/trig/Trig5/trig5/trig5.html
However, considering the bound you stated, it looks like your integral diverges.
If you integrate only where x>2 and y<1, you already get divergence.
So, to answer your question: No.
 
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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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