Double Integrals

[never_lose]

Evaluate the double integral.
$\displaystyle \int \int_{\Omega} (4 - y^2) dx dy$

$\displaystyle \Omega$ being the bounded region between $\displaystyle y^2 = 2x$ and $\displaystyle y^2 = 8 - 2x$


I'm not sure how to set this up. The book sets it up like this:

$\displaystyle \int_{-2}^{2} \int_{\frac{1}{2}y^2}^{4-\frac{1}{2}y^2} (4 - y^2) dx dy$

I can kinda see how they got the -2 to 2 for the y integral, because solving the system for x, you get x = 2.

Then solving the first equation for y you get:

$\displaystyle y = \pm \sqrt {2x}$

$\displaystyle -\sqrt {2x} \le y \le \sqrt {2x}$

$\displaystyle -2 \le y \le 2$

I have no idea how they got $\displaystyle \int_{\frac{1}{2}y^2}^{4-\frac{1}{2}y^2}$, though.

 

[tkhunny]

Draw your two curves on the same coordinate axes. Start on the y-axis (somewhere between y = -2 and y = 2) and move in the positive x-direction. Which curve do you first encounter? This is the lower x-limit. Which curve do you last encounter? This is the upper x-limit.