Question

求二重积分∫∫根号下（R^2 -X^2-Y^2）dxdy,其中积分区域D为圆周X^2+Y^2=RX. 简单 谢谢！！！！

Answer 1

按照下列从小到大的区间[-π/2→π/2]、[0→rcosθ]算出来的答案是对的，为什么不是区间从[π/2
->
-π/2][rcosθ→0]呢,我按这个区间算出来答案是(r³/3)[4/3-
π
]书上的区间都是从大到小的啊。∫∫
√(r²-x²-y²)
dxdy=∫∫
r√(r²-r²)
drdθ=∫[-π/2→π/2]
dθ∫[0→rcosθ]
r√(r²-r²)
dr=(1/2)∫[-π/2→π/2]
dθ∫[0→rcosθ]
√(r²-r²)
d(r²)=-(1/2)(2/3)∫[-π/2→π/2]
(r²-r²)^(3/2)
|[0→rcosθ]
dθ=(1/3)∫[-π/2→π/2]
(r³-r³|sinθ|³)
dθ=(2r³/3)∫[0→π/2]
(1-sin³θ)
dθ=(2r³/3)[∫[0→π/2]
1
dθ
-
∫[0→π/2]
sin³θ
dθ]=(2r³/3)[π/2
+
∫[0→π/2]
sin²θ
d(co...按照下列从小到大的区间[-π/2→π/2]、[0→rcosθ]算出来的答案是对的，为什么不是区间从[π/2
->
-π/2][rcosθ→0]呢,我按这个区间算出来答案是(r³/3)[4/3-
π
]书上的区间都是从大到小的啊。∫∫
√(r²-x²-y²)
dxdy=∫∫
r√(r²-r²)
drdθ=∫[-π/2→π/2]
dθ∫[0→rcosθ]
r√(r²-r²)
dr=(1/2)∫[-π/2→π/2]
dθ∫[0→rcosθ]
√(r²-r²)
d(r²)=-(1/2)(2/3)∫[-π/2→π/2]
(r²-r²)^(3/2)
|[0→rcosθ]
dθ=(1/3)∫[-π/2→π/2]
(r³-r³|sinθ|³)
dθ=(2r³/3)∫[0→π/2]
(1-sin³θ)
dθ=(2r³/3)[∫[0→π/2]
1
dθ
-
∫[0→π/2]
sin³θ
dθ]=(2r³/3)[π/2
+
∫[0→π/2]
sin²θ
d(cosθ)]=(2r³/3)[π/2
+
∫[0→π/2]
(1-cos²θ)
d(cosθ)]=(2r³/3)[π/2
+
cosθ
-
(1/3)cos³θ]
|[0→π/2]=(2r³/3)[π/2
-1
+
1/3]=(2r³/3)[π/2
-
2/3]=(r³/3)[π
-
4/3]

Answer 2

极坐标标
∫∫
√(R²-x²-y²)
dxdy
=∫∫
r√(R²-r²)
drdθ
=∫[-π/2→π/2]
dθ∫[0→Rcosθ]
r√(R²-r²)
dr
=(1/2)∫[-π/2→π/2]
dθ∫[0→Rcosθ]
√(R²-r²)
d(r²)
=-(1/2)(2/3)∫[-π/2→π/2]
(R²-r²)^(3/2)
|[0→Rcosθ]
dθ
=(1/3)∫[-π/2→π/2]
(R³-R³|sinθ|³)
dθ
=(2R³/3)∫[0→π/2]
(1-sin³θ)
dθ
=(2R³/3)[∫[0→π/2]
1
dθ
-
∫[0→π/2]
sin³θ
dθ]
=(2R³/3)[π/2
+
∫[0→π/2]
sin²θ
d(cosθ)]
=(2R³/3)[π/2
+
∫[0→π/2]
(1-cos²θ)
d(cosθ)]
=(2R³/3)[π/2
+
θ
-
(1/3)cos³θ]
|[0→π/2]
=(2R³/3)[π/2
+
π/2
-
0]
=2πR³/3
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