Evaluating 2D Integrals: f(x,y)=min(x,y)

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Homework Statement


Evaluate the following definite two-dimensional integrals over the specified domains of integration.

f(x,y)=min(x,y), over the region {(x,y) : 0 \leq x \leq 2, 0 \leq y \leq 1}


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The Attempt at a Solution


I'm not even sure where to start because I'm not sure what the problem even means by f(x,y)=min(x,y). HELP!
 
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min(x,y) means minimum of x and y. For example, in the region x < y, f(x,y) = x.
 
So if I integrate first from 0 \leq y \leq 1 then the x bounds, my f(x,y)=y? I'm a little confused over this. I don't remember going over this in calculus and this homework is supposed to be a calculus review.
 
Divide the region {(x,y) : 0 < x < 2, 0 < y < 1} into two parts, one where f(x,y) = x, and one where f(x,y) = y. Then do the usual double integration for the two regions seperately.
 
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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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