Double Integrals and circles - Confirmation Wanted

Nima
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Hey, my Q is:

"Integrate f(x, y) = Sqrt(x^2 + y^2) over the region in the x-y plane bounded by the circles r = 1 and r = 4 in the upper half-plane".

Well, I firstly sketched out the region I get as my area in the x-y plane. I deduced that the ranges for x and y are:

0 <= x <= 4
Sqrt[1 - x^2] <= y <= Sqrt[16 - x^2]

1.) Is this right?
2.) How do I then calculate the integral of f(x, y) over this region? I know I'm doing a double integral but I don't see how I can separate my variables...

Thanks
 
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Have you covered polar coordinates?
 
TD said:
Have you covered polar coordinates?
Hi, no unfortunately I haven't covered polar co-ordinates yet.

mmm so yes I see that f(x, y) = r and now we have 2 circles with radii r = 4 and r = 1 respectively.

Could you explain to me how to do this Q if that's ok? Thanks.
 
Just think about it. from what pts are we integrating wrt the radius? Then, what angle to what angle are we integrating (wrt theta). drawing a picture is helpful.
 
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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