There is a double integral, both goes from infinity to negative infinity SS (e^-2(x^2+y^2)^(1/2) (x^2+y^2)^3)dx dy

∫^{+∞}_{–∞} ∫^{+∞}_{–∞} e ^{–√(x2 + y2)} (x^{2} + y^{2})^{3} dxdy = transform to polar coordinates

∫ ^{2π}_{0} dθ ∫^{+∞}_{0 }e^{–r}r^{6·} rdr = 2π ∫^{+∞}_{0 }e^{–r}r^{7}dr = 2π Γ(8) = 2π 7! = 10080π

Here Γ is the gamma-function.

He also mentioned using reduction formulas so we didn't have to integrate by parts seven times. I know the answer is suppose to be 315pi/8

I'm sorry but I have to have step by step. I can do this much:

SS e^-2(r^2cos^2theta+r^2sin^2theta)^(1/2) (r^2cos^2theta+r^2sin^2theta)^3 (r)(dr)(dtheta)

Sorry, forgot the coefficient in the exponent:

∫^{+∞}_{–∞} ∫^{+∞}_{–∞} e ^{–2√(x2 + y2)} (x^{2} + y^{2})^{3} dxdy = transform to polar coordinates

∫ ^{2π}_{0} dθ ∫^{+∞}_{0 }e^{–2r}r^{6·} rdr = 2π ∫^{+∞}_{0 }e^{–2r}r^{7}dr = 2π Γ(8)/2^{7} = 2π 7! / 128 = 10080π / 128 = 315π /4.

Use the Pythagorean identity: x^2 + y^2 = r^2cos^2 theta + r^2sin^2 theta = r^2

Here's my other one:

both integrals go from infinity to 0

SS e^-(x^2+y^2) (x^2) dxdy

∫^{+∞}_{0} ∫^{+∞}_{0} e ^{–(x2 + y2)} x^{2} dxdy = transform to polar coordinates

∫ ^{π/2}_{0} cos^{2}θdθ ∫^{+∞}_{0 }e^{–r2}r^{2·} rdr = ∫ ^{π/2}_{0} (1+cos2θ)/2 dθ ∫^{+∞}_{0 }e^{–r2}r^{3}dr = substitute u = r^{2}

(π/8) ∫^{+∞}_{0 }e^{–u}udu = π/8.

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