Double Integral Problem- region of integration

**monster** · May 16th 2009, 12:54 AM

Have a double integral problem where it wants me to identify the region of integration and then evaluate using polar co-ords.
Given;

$ \int_{-2}^{2}\int_{0}^{\sqrt(4-y^2)} f(x,y) dx.dy $

I said region of integration is the half circle in the first and fourth quadrants, does this sound correct?

Having identified this region in order to integrate it with polar co-ords i need to specify a range for θ and i'm not sure how to specify θ's range for this region?

i was guessing -pi/2 <= θ <= pi/2 ;
but am really not sure any help here would be appreciated.

(i didn't give function f(x,y) as didnt think it is relevant and have no problems manipulating it for problem)

**Mush** · May 16th 2009, 01:29 AM

Originally Posted by monster

Have a double integral problem where it wants me to identify the region of integration and then evaluate using polar co-ords.
Given;

$ \int_{-2}^{2}\int_{0}^{\sqrt(4-y^2)} f(x,y) dx.dy $

I said region of integration is the half circle in the first and fourth quadrants, does this sound correct?

Having identified this region in order to integrate it with polar co-ords i need to specify a range for θ and i'm not sure how to specify θ's range for this region?

i was guessing -pi/2 <= θ <= pi/2 ;
but am really not sure any help here would be appreciated.

(i didn't give function f(x,y) as didnt think it is relevant and have no problems manipulating it for problem)

Yes, you are correct. The region is the semi-circle for positive x values (i.e., in the first and 4th quadrant). The radius of the circle is 2. And the angles to go between are $\frac{-\pi}{2} \to \frac{\pi}{2}$

So your new integration will be:

$ \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{2} f(r \cos(\theta), r \sin(\theta)) \times r \, dr \, d\theta$

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