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chung

2007-12-11 1:22 am

I = lnlnln ( x^2/a^2 + y^2/b^2 + z^2/c^2 ) where x^2 + y^2 + z^2 <=1 , and a,b,c are constant.

(a)show I = lnlnln ( x^2/b^2+y^2/c^2+z^2/a^2 ) = lnlnln ( x^2/c^2+y^2/a^2+z^2/b^2 )

(b)hence find I by spherical coordinates........

只係識做(a) 都可以.

回答 (1)

用戶127502

2007-12-11 11:57 am

✔ 最佳答案

a.)
Actually because the function and domain is symmetric, you can change the roles of x,y,z by substiution: x=y, y=z, z=x and so on to obtain the result.

b.)
3I = add the 3 integrals = InInIn [1/a^2 + 1/b^2 + 1/c^2][x^2+y^2+z^2] dxdydz
then focus on InInIn (x^2+y^2+z^2) dxdydz

By spherical transformation,
The integral = 4pi/5
(Ref: http://en.wikipedia.org/wiki/Multiple_integral#Spherical_coordinates)

therefore,
3I = (1/a^2 + 1/b^2 + 1/c^2) 4pi/5
I = (1/a^2 + 1/b^2 + 1/c^2) 4pi/15 #

參考: me

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