# Help with the bounds on this double integral

• Aug 17th 2008, 07:32 PM
Help with the bounds on this double integral
double integral (D) of y^3(x^2+y^2)^(-3/2) DXDY where (D) is the region determined by the ocnditions .5 less than or equal to y less than or equal to 1 and x^2+y^2 less than or equal to 1.
• Aug 17th 2008, 10:50 PM
mr fantastic
Quote:

double integral (D) of y^3(x^2+y^2)^(-3/2) DXDY where (D) is the region determined by the ocnditions .5 less than or equal to y less than or equal to 1 and x^2+y^2 less than or equal to 1.

Have you drawn the region of integration? It's the interior of the unit circle that lies above the line y = 1/2.

One possible set up is $\int_{x = -\sqrt{3}/2}^{x = \sqrt{3}/2} \int_{y=1/2}^{y=+\sqrt{1 - x^2}} y^3 (x^2 + y^2)^{-3/2} \, dy \, dx$.

Another is $\int_{x = - \sqrt{1 - y^2}}^{x=+\sqrt{1 - y^2}} \int_{y = 1/2}^{y=1} y^3 (x^2 + y^2)^{-3/2} \, dx \, dy$.

But you should consider switching to polar coordinates. In which case you'll need to express the line y = 1/2 in polar coordinates. Note that $\frac{\pi}{6} \leq \theta \leq \frac{5 \pi}{6}$.
• Aug 18th 2008, 03:55 AM
could I get a tip on how to integrate this as well, just not seeing it
as much help on this as possible would be greatly appreciated.
• Aug 18th 2008, 04:04 AM
mr fantastic
Quote:

as much help on this as possible would be greatly appreciated.

Have you drawn a sketch of the region?

Switching to polar coordinates: $I = \int_{\theta = \pi/6}^{5 \pi/6} \int_{r = 1/(2 \sin \theta)}^{r=1} r \sin^3 \theta \, dr \, d\theta$.

If you're still stuck now, you'll need to say what part(s) of this expression you're struggling to get.
• Aug 18th 2008, 06:03 AM
customary to switch to polar here?
I punched it in my calculator and got xy/sqrt(x^2+y^2). I just wanted to make sure that I knew how to do it both ways. Thanks very much for all your help.
• Aug 18th 2008, 06:09 AM
mr fantastic
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