Double Integral - polar coordinates

[exidez]

Homework Statement

$$\displaystyle\int\int\sqrt{4-x^2-y^2} dA$$

$$R{(x,y)|x^2+y^2\leq4 .. 0\leq x}$$

The Attempt at a Solution

So far i have: $$\displaystyle\int^{\pi}_{0}\int^{r}_{0}\sqrt{4-r^2} rdrd\theta$$

Solving i get:
$$\displaystyle\int^{\pi}_{0}\frac{-1}{3}(4-r^2)^{\frac{3}{2}}+\frac{1}{3}(4)^{\frac{3}{2}}d\theta$$

Am i on the right track, and if so how do i integrate the third root term?edit//

ok i forgot the radius goes from 0 to 2... if i sub 2 into r, i can integrate it. But just to check, am i still doing this correctly?

 

[gabbagabbahey]

edit//

ok i forgot the radius goes from 0 to 2... if i sub 2 into r, i can integrate it. But just to check, am i still doing this correctly?

Yes, you should integrate from zero to 2.