# Which of the following double integrals would correctly solve this problem?

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?

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.
• 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. – abnry Dec 3 '13 at 21:12