changing and evaluating double integral

[mathstresser]

Evaluate the given integral by changing to polar coordinates.

SS (4-x^2-y^2)^(1/2) dA
where R= {(x,y) | x^2+y^2<=4, x>=0}

Is what I have for the following anywhere close? What do I need to add? or do differently?

I change the integral to (4-r^2)^(1/2) and the region is {r^2 <= 4, rcosx>=0
}

S(?,?) S(0,4) [(4-r^2)^ (1/2) rdrdx]


12

 

[galactus]

\(\displaystyle \L\\\int_{0}^{2{\pi}}\int_{0}^{2}\sqrt{4-r^{2}}rdrd{\theta}\)

 

[mathstresser]

Thanks!