# Thread: Help with the bounds on this double integral

1. ## 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.

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 $\frac{\pi}{6} \leq \theta \leq \frac{5 \pi}{6}$.

3. ## 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.

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.

5. ## 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.