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transform to polar coordinates and evaluate

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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

Oct 16th, 2017

+∞–∞+∞–∞ e –√(x2 + y2) (x2 + y2)3 dxdy = transform to polar coordinates

0 dθ ∫+∞0 e–rr rdr = 2π ∫+∞0 e–rr7dr = 2π Γ(8) = 2π 7! = 10080π

Here Γ is the gamma-function.

Apr 7th, 2015

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

Apr 7th, 2015

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)

Apr 7th, 2015

Sorry, forgot the coefficient in the exponent:

+∞–∞+∞–∞ e –2√(x2 + y2) (x2 + y2)3 dxdy = transform to polar coordinates

0 dθ ∫+∞0 e–2rr rdr = 2π ∫+∞0 e–2rr7dr = 2π Γ(8)/27 = 2π 7! / 128 = 10080π / 128 = 315π /4.

Apr 7th, 2015

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

Apr 7th, 2015

Here's my other one:

both integrals go from infinity to 0

SS e^-(x^2+y^2)  (x^2) dxdy

Apr 7th, 2015

+∞0+∞0 e –(x2 + y2) x2 dxdy = transform to polar coordinates

π/20 cos2θdθ ∫+∞0 e–r2r rdr = ∫ π/20 (1+cos2θ)/2 dθ ∫+∞0 e–r2r3dr = substitute u = r2

(π/8) ∫+∞0 e–uudu = π/8.

Apr 7th, 2015

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Oct 16th, 2017
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Oct 16th, 2017
Oct 17th, 2017
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