So I have the following integral:
∫∞−∞∫∞−∞x2dxdy(1+√x2+y2)5
I know that converting the integral using polar coordinates gives:
∫∫r2cos2θ(1+r)5rdrdθ
I'm assuming r is going from 0 to infinity.
But what about θ?
Answer
Note that in order to cover R2, r extends from 0 to ∞ and θ spans an entire period of sin(θ) and cos(θ). (For example, the first quadrant alone is covered by θ∈[0,π/2], r∈[0,∞).)
Hence,
∫∞−∞∫∞−∞x2(1+√x2+y2)5dxdy=∫2π0∫∞0r2cos2(θ)(1+r)5rdrdθ=(∫2π0cos2(θ)dθ)⏟=π(∫∞0r3(1+r)5dr)⏟=1/4=π4
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