用对称性计算三重积分 (3x^2+5y^2+7z^2)dxdydz，V：x^2+y^2+z^2≤R^2, z≥0 要有过程。在线等，谢谢！

Question

如题所述

Answer 1

令
x=psinucosv
y=psinusinv
z=pcosu
积分区域变为0<=p<=R,0<u<π/2,0<v<2π
∫∫∫(3x^2+5y^2+7z^2)dxdydz
=∫∫∫(3p^2sin^2ucos^2v+5p^2sin^2usin^2v+7p^2cos^2u)p^2cosusinudpdudv
=∫∫(3sin^3ucosucos^2v+5sin^3ucosusin^2v+7sinucos^3u)dudv∫p^4dp
=(R^5/5)∫[3cos^2v∫sin^3udsinu+5sin^2v∫sin^3udsinu-7∫cos^3udcosu]dv
=(R^5/5)∫[(3/4)cos^2v+(5/4)sin^2v+7/4]dv
=(R^5/5)∫[(3/8)(1+cos2v)+(5/8)(1-cos2v)+7/4]dv
=11πR^5/10追问

不对啊，答案是6πR^5

追答

不好意思，做变量代换是错了点。
∫∫∫(3x^2+5y^2+7z^2)dxdydz
=∫∫∫(3p^2sin^2ucos^2v+5p^2sin^2usin^2v+7p^2cos^2u)p^2sinudpdudv
=∫∫(3sin^3cos^2v+5sin^3usin^2v+7sinucos^2u)dudv∫p^4dp
=(R^5/5)∫[3cos^2v∫(cos^2u-1)dcosu+2sin^2v∫(cos^2u-1)dcosu-7∫cos^2udcosu]dv
=(R^5/5)∫[2cos^2v+(10/3)sin^2v+7/3]dv
=(R^5/5)∫[((1+cos2v)+(5/3)(1-cos2v)+7/3]dv
=(R^5/5)(1+5/3+7/3)*2π
=2πR^5这次不会错了，如果有问题肯定是答案或者题目抄错了。

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