解析学

質問

解法を教えてください。
１８は解法はわかっていて答えを確認したいので、
答えをお願いします。

回答No.1

muturajcp

18)
y=(1-x^{2/3})^{3/2}(sint)^3
↓
V=∬∫_{x^{2/3}+y^{2/3}+z^{2/3}≦1}dxdydz
=2∬_{x^{2/3}+y^{2/3}≦1}[(1-x^{2/3}-y^{2/3})^{3/2}]dxdy
=4∫_{-1～1}{∫_{0～(1-x^{2/3})^{3/2}}[(1-x^{2/3}-y^{2/3})^{3/2}]dy}dx
=24∫_{0～1}{(1-x^{2/3})^3}{∫_{0～π/2}{(cost)^4}(sint)^2dt}dx
=3∫_{0～1}{(1-x^{2/3})^3}{∫_{0～π/2}{(2+cos2t-2cos4t-cos6t)/4}dt}dx
=(3π/4)∫_{0～1}{(1-x^{2/3})^3}dx
=(3π/4)∫_{0～1}(1-3x^{2/3}+3x^{4/3}-x^2)dx
=16π/35

19)
x^2+y^2=z^2≦2y
0≦x^2≦2y-y^2
0≦y≦2
z^2≦2y≦4
0≦z≦2
だから
x=zsint
y=zcost
と置ける
z≦2cost
r=(zsint,zcost,z)
r_z=(sint,cost,1)
r_t=(zcost,-zsint,0)
|r_z×r_t|=|(zsint,zcost,-z)|=z√2
D={(z,t)|0≦z≦2cost,-π/2≦t≦π/2}
↓
S=∫_{x^2+y^2=z^2≦2y,0≦z}dS
=∬_{D}|r_z×r_t|dtdz
=√2∫_{-π/2～π/2}(∫_{0～2cost}zdz)dt
=√2∫_{-π/2～π/2}([(z^2)/2]_{0～2cost})dt
=√2∫_{-π/2～π/2}[2(cost)^2]dt
=2√2∫_{0～π/2}[1+cos(2t)]dt
=2√2[t+(sin2t)/2]_{0～π/2}
=π√2