Question

求解一道极坐标下的二重积分

求z=3x^2+3y^2和z=4-x^2-y^2之间的体积

Answer 1

先求交界
3x^2+3y^2=4-x^2-y^2
x^2+y^2=1
画出图，4-x^2-y^2在3x^2+3y^2之上
所以所求区域为{(x,y,z)|x^2+y^2<=1,3x^2+3y^2<=z<=4-x^2-y^2}
体积=∫∫{x^2+y^2<=1}(∫{3x^2+3y^2<=z<=4-x^2-y^2}dz)dxdy
=∫∫{x^2+y^2<=1}(z|[3x^2+3y^2,4-x^2-y^2])dxdy
=4∫∫{x^2+y^2<=1}(1-x^2-y^2)dxdy
令x=rcost,y=rsint
得到
=4∫[0,2π]dt∫[0,1](1-r^2)rdr
(r是极坐标Jacobian矩阵行列式)
=4*2π*(r-r^3/3)|[0,1]
=8π*(1-1/3)
=16π/3