Consider the integral I = doubleintegral_R^2 e^-(x^2 = y^2) dA= integral_-infini
ID: 2878508 • Letter: C
Question
Consider the integral I = doubleintegral_R^2 e^-(x^2 = y^2) dA= integral_-infinity^infinity integral_-infinity^infinity e^-(x^2 = y^2) dy dx = lim_n rightarrow infinity doubleintegral_D_a e^-9x^2 + y^2)dA where D_a is the disk with radius a and center the origin, Prove that integral_-infinity^infinity integral_-infinity^infinity e^-(x^2 + y^2) dA = pi An equivalent definition of I is doubleintegral_R^2 e^-(x^2 + y^2) dA = lim_a rightarrow infinity doubleintegral_Sa e^-(x^2 + y^2)dA where S_a is the square with vertices (plusminus am plusminus a). Use this to show that integral_-infinity^infinity e^-x^2 dx integral_-infinity^infinity e^-y^2 dy = piExplanation / Answer
in polar coordinates
x=rcos, y=rsin
x2+y2=r2
x2+y2=a2
0<=<=2,0<=r<=a
dA =r dr d
[- to] [- to ]e-(x^2+y^2) dy dx
=lima->[0 to 2] [0 to a] e-(r^2)r dr d
=lima->[0 to 2][0 to a] (-1/2)e-(r^2) d
=lima->[0 to 2] (-1/2)(e-(a^2)-e-(0^2)) d
=lima->[0 to 2] (-1/2)(e-(a^2)-1) d
=lima->[0 to 2] (1/2)(1-e-(a^2)) d
=lima->[0 to 2] (1/2)(1-e-(a^2))
=lima-> (1/2)(1-e-(a^2))(2-0)
=lima-> (1-e-(a^2))
=(1-0)
=
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b)[- to] [- to ]e-(x^2+y^2) dy dx=
=lima->[-a to a] [-a to a]e-x^2e-y^2 dy dx=
=lima->[-a to a] e-x^2dx[-a to a]e-y^2 dy =
=[- to ] e-x^2dx[-to ]e-y^2 dy =