Denote by D the region volume is to be calculated. Volume D is VD =∭ D d x d y d z \underset { D }{ \iiint } dxdydz D ∭ d x d y d z
x = r cos 4 φ sin 4 θ y = r sin 4 φ sin 4 θ z = r cos 4 θ \begin{matrix} x=r\cos ^{ 4 }{ \varphi \sin ^{ 4 }{ \theta } } \\ y=r\sin ^{ 4 }{ \varphi \sin ^{ 4 }{ \theta } } \\ z=r\cos ^{ 4 }{ \theta } \end{matrix} x = r cos 4 φ sin 4 θ y = r sin 4 φ sin 4 θ z = r cos 4 θ
x + y + z = r = 1 \sqrt { x } +\sqrt { y } +\sqrt { z } =\sqrt { r } =1 x + y + z = r = 1
The image of the region D is the region Δ = { ( r , φ , θ ) : 0 ≤ r ≤ 1 , 0 ≤ φ ≤ π 2 , 0 ≤ θ ≤ π 2 } =\left\{ \left( r,\varphi ,\theta \right) :\quad 0\le r\le 1,\quad 0\le \varphi \le \frac { \pi }{ 2 } ,0\le \theta \le \frac { \pi }{ 2 } \right\} = { ( r , φ , θ ) : 0 ≤ r ≤ 1 , 0 ≤ φ ≤ 2 π , 0 ≤ θ ≤ 2 π }
∭ D d x d y d z = ∭ Δ ∣ ∂ ( x , y , z ) ∂ ( r , φ , θ ) ∣ d r d φ d θ \underset { D }{ \iiint } dxdydz\quad =\underset { \Delta }{ \iiint } \quad \left| \frac { \partial (x,y,z) }{ \partial (r,\varphi ,\theta ) } \right| drd\varphi d\theta D ∭ d x d y d z = Δ ∭ ∣ ∣ ∂ ( r , φ , θ ) ∂ ( x , y , z ) ∣ ∣ d r d φ d θ ∂ ( x , y , z ) ∂ ( r , φ , θ ) = d e t ( cos 4 φ sin 4 θ − 4 r cos 3 φ sin φ sin 4 θ 4 r cos 4 φ sin 3 θ cos θ sin 4 φ sin 4 θ 4 r sin 3 φ cos φ sin 4 θ 4 r sin 4 φ sin 3 θ cos θ cos 4 θ 0 − 4 r cos 3 θ sin θ ) \frac { \partial (x,y,z) }{ \partial (r,\varphi ,\theta ) } =det\left( \begin{matrix} \cos ^{ 4 }{ \varphi \sin ^{ 4 }{ \theta } } & -4r\cos ^{ 3 }{ \varphi \sin { \varphi } \sin ^{ 4 }{ \theta } } & 4r\cos ^{ 4 }{ \varphi \sin ^{ 3 }{ \theta } \cos { \theta } \quad } \\ \sin ^{ 4 }{ \varphi \sin ^{ 4 }{ \theta } } & 4r\sin ^{ 3 }{ \varphi \cos { \varphi } \sin ^{ 4 }{ \theta } } & 4r\sin ^{ 4 }{ \varphi \sin ^{ 3 }{ \theta \cos { \theta } } } \\ \cos ^{ 4 }{ \theta } & 0 & -4r\cos ^{ 3 }{ \theta } \sin { \theta } \end{matrix} \right) \quad ∂ ( r , φ , θ ) ∂ ( x , y , z ) = d e t ⎝ ⎛ cos 4 φ sin 4 θ sin 4 φ sin 4 θ cos 4 θ − 4 r cos 3 φ sin φ sin 4 θ 4 r sin 3 φ cos φ sin 4 θ 0 4 r cos 4 φ sin 3 θ cos θ 4 r sin 4 φ sin 3 θ cos θ − 4 r cos 3 θ sin θ ⎠ ⎞
= − 16 r 2 cos 3 θ cos 3 φ sin 3 φ sin 7 θ =-16{ r }^{ 2 }\cos ^{ 3 }{ \theta } \cos ^{ 3 }{ \varphi } \sin ^{ 3 }{ \varphi } \sin ^{ 7 }{ \theta } = − 16 r 2 cos 3 θ cos 3 φ sin 3 φ sin 7 θ
VD = ∫ 0 1 ∫ 0 π 2 ∫ 0 π 2 16 r 2 cos 3 θ cos 3 φ sin 3 φ sin 7 θ d r d φ d θ = 16 3 ⋅ 1 12 ⋅ 1 40 = 1 90 \int _{ 0 }^{ 1 }{ \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \int _{ 0 }^{ \frac { \pi }{ 2 } }{ 16{ r }^{ 2 } } \cos ^{ 3 }{ \theta } \cos ^{ 3 }{ \varphi } \sin ^{ 3 }{ \varphi } \sin ^{ 7 }{ \theta } drd\varphi d\theta \quad =\frac { 16 }{ 3 } \cdot \frac { 1 }{ 12 } \cdot } \frac { 1 }{ 40 } } =\frac { 1 }{ 90 } ∫ 0 1 ∫ 0 2 π ∫ 0 2 π 16 r 2 cos 3 θ cos 3 φ sin 3 φ sin 7 θ d r d φ d θ = 3 16 ⋅ 12 1 ⋅ 40 1 = 90 1
∫ 0 π 2 cos 3 φ sin 3 φ d φ = ∫ 0 π 2 cos 2 φ sin 3 φ d ( sin φ ) = ∫ 0 1 ( 1 − t 2 ) t 3 d t = 1 4 − 1 6 = 1 12 ∫ 0 π 2 cos 3 θ sin 7 θ d θ = ∫ 0 π 2 cos 2 θ sin 7 θ d ( sin θ ) = ∫ 0 1 ( 1 − t 2 ) t 7 d t = 1 8 − 1 10 = 1 40 \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \cos ^{ 3 }{ \varphi } \sin ^{ 3 }{ \varphi } d\varphi } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \cos ^{ 2 }{ \varphi } \sin ^{ 3 }{ \varphi } d\left( \sin { \varphi } \right) } =\int _{ 0 }^{ 1 }{ \left( 1-{ t }^{ 2 } \right) { t }^{ 3 } } dt=\frac { 1 }{ 4 } -\frac { 1 }{ 6 } =\frac { 1 }{ 12 } \quad \\ \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \cos ^{ 3 }{ \theta \sin ^{ 7 }{ \theta d\theta } } } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \cos ^{ 2 }{ \theta \sin ^{ 7 }{ \theta d\left( \sin { \theta } \right) } } } =\int _{ 0 }^{ 1 }{ \left( 1-{ t }^{ 2 } \right) { t }^{ 7 }dt } =\frac { 1 }{ 8 } -\frac { 1 }{ 10 } =\frac { 1 }{ 40 } ∫ 0 2 π cos 3 φ sin 3 φ d φ = ∫ 0 2 π cos 2 φ sin 3 φ d ( sin φ ) = ∫ 0 1 ( 1 − t 2 ) t 3 d t = 4 1 − 6 1 = 12 1 ∫ 0 2 π cos 3 θ sin 7 θ d θ = ∫ 0 2 π cos 2 θ sin 7 θ d ( sin θ ) = ∫ 0 1 ( 1 − t 2 ) t 7 d t = 8 1 − 10 1 = 40 1
∫ 0 1 16 r 2 d r = 16 3 \int _{ 0 }^{ 1 }{ 16{ r }^{ 2 } } dr=\frac { 16 }{ 3 } ∫ 0 1 16 r 2 d r = 3 16
ANSWER The volume is 1 90 \frac{1}{90} 90 1
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