Expert: Alon Mandes - 10/8/2009

Question
Alon how are you?

I am getting my intergration incorrect with double intergrals
the questions wants to evulate the intergral

(double intergral sign A)(1+x+cos((x^2+y^2))dxdy

where the area of intergration A is 1<=x^2+y^2<=4
I know that this is a dount shape from 1 to 2 but each time i work it work it out i think my intergration is wrong.

Thanks for your time
Kayne

Answer
Hello Kayne, I'm fine thank you. The issue here is to transform to polar coordinates. Because
the integral of cos(x^2) is not immediate at all. SO, the transformation will be :
x=rcosØ & y=rsinØ . We must keep in mind that dxdy=rdrdØ ("The Jacobian"). The dount area will
become, 1≤x^2+y^2≤4 --> 1≤r≤4 & 0≤Ø≤2π . Hence,

∫∫1+x+cos((x^2+y^2))dxdy = ∫∫ [1+rcosØ+cos(r^2)]rdrdØ =
A                          A

2π  4
∫  ∫ r+r^2cosØ+rcos(r^2)drdØ = ∫∫rdrdØ + ∫∫r^2cosØdrdØ + ∫∫rcos(r^2)drdØ =
0 1

2π  4        2π     4         2π  4
∫dØ∫rdr  +  ∫cosØdØ∫r^2dr  +  ∫dØ∫rcos(r^2)dr =
0  1        0      1          0  1

2π(16-1)/2  +  [sin(2π)-sin(0)]*(64-1)/3  +  2π[sin(16)-sin(1)]/2 =
15π+0+π[sin(16)-sin(1)] .

Alon.