What is the Correct Domain of Integration for a Double Integral Problem?

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Homework Statement



question 2:
http://www.math.ubc.ca/~haber/courses/math253/Welcome_files/asgn5.pdf"

The Attempt at a Solution



So for part a) I tried to plot my domain of integration and ended up concluding it was an area bounded by y=0, y=4-x^2, and x=1. Is this okay?

In not too sure about part b) I'm just going off the graph here and so if that's wrong this will be wrong. Here is what i did I changed x domain from 1->(4-x^2) to 1->2 and I changed y domain from 0->3 to 0->4-x^2.

Is this correct?

Thanks!
 
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how could we integrate when the upper bound of the first integral is a polynomial x=4-x^2? Maybe I am misunderstanding something. Would it not have to be either a constant or a function of y?
 
The upper limit on the inner integral is surely a typo and probably is supposed to be x = 4 - y2.
 
I sure hope so. Thank you.
 
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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