Solving a Double Integral with k Converging to pi

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First attatchment is an integral I have been given.

Using different values of k I have to find out what value the integral converges too.

What I want to know is does this mean integrating the volume of circle with radius 1.

Shown in formula in the second attachment, I have also arranged it using polar coordinates.

And if that is right, I have found but only using mathcad that if k = 1/infinity the value converges to pi.

Is that right, because how I suppose to integrate the function shown using fractions to k, surely this is too complex to do by hand.

Cheers Ash
 

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And how do import pictures from mathcad into the post without having to make them attachments.

Cheers
 
A circle of radius 1 doesn't have a "volume"! What it means is that you are to integrate the given function over the interior of the circle.

And k can't "equal 1 over infinity" because infinity is not a number. Do you mean k= 0?

Yes, polar coordinates is the way to go here. You are aware, are you not, that sin^2(\theta)+ cos^2(\theta)= 1, so that x2+ y2= r2? That simplifies your integral a great deal!
 
Yes sorry I meant area, not volume,

and yes i know it simplifies, i was just asking whether I had interpreted the problem correctly.

and yes by 1/infinity I did mean zero.

I was meaning as k gets smaller it converges to pi.

So is the function that i posted the right interpretation of the problem?

Cheers Ash
 
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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