重積分∫∫[R] (4-x^2-y^2)^1/2 dA

2010-06-30 1:08 am
Let R be the region in the first quadrant that lies between the circles x^2 +y^2 =4and x^2 +y^2 =2x. Evaluate:
∫∫[R] (4-x^2-y^2)^1/2 dA

回答 (1)

2010-06-30 2:28 am
✔ 最佳答案
改為極坐標:
x^2+y^2=4 , r=2
x^2+y^2=2x, r^2=2rcosθ, then r=2cosθ
如下圖紅色區域

圖片參考:http://imgcld.yimg.com/8/n/AD04686329/o/161006290550713872554260.jpg

原積分=∫[0~π/2]∫[2cosθ~2] √(4-r^2) * rdr dθ
=∫[0~π/2] (-1/3)(4-r^2)^1.5 |[2cosθ~2] dθ
= (8/3)∫[0~π/2] (sinθ)^3 dθ
=(8/3)∫[0~π/2] [1-(cosθ)^2] sinθ dθ
=(8/3)[ -cosθ+(1/3)(cosθ)^3] for θ=0~π/2
=16/9


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