→ hsnuyi :第二種 你z的積分範圍不覺得很奇怪嗎? 07/07 12:37
→ alasa15 :YES =.=... 07/07 12:50
Evaluate
∫∫∫ (x^2+y^2)^0.5 dV, where B is the volumn bounded by z=x^2+y^2
B and z=4
I do this problem in two ways, but i got two different answers.
Method I. ∫∫ (x^2+y2)0.5 * [4-x^2-y^2] dxdy
D
Method II. Let x=r cost, y=r sint, z=z => |J| = r
2π 2 2
∫ ∫ ∫ (r^2)^0.5 * r dy dr dt
0 0 0
which is correct?
(I think there are some mistakes in Method II..)
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※ 編輯: alasa15 來自: 163.14.32.105 (07/07 12:29)