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I obtained two sets of boundary conditions.

Set 1:

$$x=-\sqrt{4-y^2}\quad (for\quad x<0)\quad to\quad x=\sqrt{4-y^2}\quad (for\quad x>0)\\y=-2\quad to\quad y=2$$

Set 2:

$$x=-2\quad to\quad x=2\\y=-\sqrt{4-x^2}\quad (for\quad y<0)\quad to\quad y=\sqrt{4-x^2}\quad (for\quad y>0)$$

This produces the following integrals:

$$\int_{-2}^{2}\int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}}(4-y)dxdy\\\int_{-2}^{2}\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}(4-y)dydx$$

So why aren't a, b, and c all correct? The correct answer is c. Why are a and b incorrect?

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For your first integral: The integrand is even with respect to $x$, the inner integral. This means you can write $\int_{-a}^a f(x) dx = 2 \int_0^a f(x)dx$. This will answer your question.

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  • $\begingroup$ And why can't you do the same for choices a and b? $\endgroup$ – user1251385 Dec 3 '13 at 20:56
  • $\begingroup$ The integrand is not even with respect to the inner integral, which in this case is with respect to $y$. That's why you cannot do the same thing. $\endgroup$ – abnry Dec 3 '13 at 21:12
  • $\begingroup$ I might what really matters is not the silly multiple choice answers, but the fact that you corrected set up the integral in the two possible ways (albeit without using the evenness of the integrand). $\endgroup$ – abnry Dec 3 '13 at 21:14
  • $\begingroup$ Aha, I get it now. Wow. That turned out a lot simpler than I thought it would be. $\endgroup$ – user1251385 Dec 3 '13 at 21:27

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