What would you like to ask?Find integral from 0 to infinity of integral from 0 to infinity of 1 / ((1 + x^2 +

Find integral from 0 to infinity of integral from 0 to infinity of 1 / ((1 + x^2 + y^2)^2) dydx?

we're learning double integrals and such and I'm pretty sure you're supposed to substitute r since you've got x^2 and y^2 there begging to be cos^2 and sin^2
shows steps please
wolframalpha says the answer is pi/4, but i'm not sure how to get there

IYou've on the right track with the idea to switch to polar coordinates.
x = r∙cos(φ)
y = r∙sin(φ)
By this coordinate transformation the differential area changes to
dxdy = r drdφ
The denominator of the integrand changes to
1 + x² + y² = 1 + r²∙cos²(φ) + r²∙sin²(φ) = 1 + r²
φ is the counterclockwise measured angle with positive x-axis. Since we integrate only over the area where both x and y are positive, i,e, the first quadrant, φ may vary between 0 and π/2.
The upper limit for both x and y is infinity, to you need to integrate from r=0 to infinity.

Hence,
∫ [0;∞] ∫ [0;∞] 1/(1 + x² + y²)² dxdy
= ∫ [0;π/2] ∫ [0;∞] r/(1 + r²)² drdφ

let u = 1 + r²
=> du = 2∙r dr
=>
∫ r/(1 + r²)² dr = (1/2) ∙ ∫ 2∙r/(1 + r²)² dr = (1/2) ∙ ∫ 1/u² du = -(1/2)∙(1/u) = -(1/2)∙(1/(1 + r³))

Hence,
∫ [0;π/2] ∫ [0;∞] r/(1 + r²)² drdφ
= ∫ [0;π/2] -(1/2)∙( lim|r→∞|{ 1/(1+ r²) - 1/(1 + 0²) dφ
= ∫ [0;π/2] -(1/2)∙( 0 - 1) dφ
= ∫ [0;π/2] (1/2) dφ
= (1/2)∙( π/2 - 0)
= π/4
q.e.d.