Double integral of cos(x² + y²) in polar coordinates

[DavidAp]

∫∫cos(x^2 + y^2)dA, where R is the region that lies above the x-axis within the circle x^2 + y^2 = 9.

Answer: .5pi*sin(9)

My Work:
∫(0 ->pi) ∫(0 -> 9) cos(r^2) rdrdθ

u = r^2
du = 2rdr
dr = du/2r

.5∫(0 ->pi) ∫(0 -> 9) cos(u) dudθ
.5∫(0 ->pi) sin(u)(0 -> 9) dθ
.5∫(0 ->pi) sin(r^2)(0 -> 9) dθ
.5∫(0 ->pi) [sin(9^2) - sin(0)] dθ
.5∫(0 ->pi) sin(81) dθ
.5[sin(81) θ](0 -> pi)
.5pi*sin(81)

However, the answer is suppose to be .5pi*sin(9). Where did I go wrong? Am I not suppose to square r, and if not then doesn't that mean everything I did below the substitution doesn't work?

Thank you for your time and patience. I apologize for my integral notation, I'm not sure of a better way to type it.

 

[LCKurtz]

Your circle has radius 3, not 9. r should go from 0 to 3.

 

[DavidAp]

O.O! I love you forever, thank you!