∫ 0 1 ∫ 0 1 − x 2 ∫ x 2 + y 2 2 − x 2 − y 2 x y d z d y d x d z d y d x = ρ 2 sin θ d φ d θ d ρ x = ρ cos φ sin θ , y = ρ sin φ sin θ , z = ρ cos θ x 2 + y 2 = z x 2 + y 2 + z 2 = 2 z 2 ρ 2 = 2 ρ 2 cos θ cos θ = 1 2 , θ = π 4 2 − x 2 − y 2 = z 2 − x 2 − y 2 = z 2 2 = x 2 + y 2 + z 2 ρ 2 = 2 , ρ = 2 1 − x 2 = y 1 = x 2 + y 2 x 2 + y 2 = ρ 2 sin 2 θ ( cos 2 φ + sin 2 φ ) = ρ 2 sin 2 θ 1 = ρ 2 sin 2 θ sin 2 θ = 1 2 , θ = π 4 x = 0 , ρ cos φ cos θ = 0 , φ = π 2 ρ sin φ cos θ = 1 2 sin φ sin ( π 4 ) = 1 φ = π 2 ∫ 0 1 ∫ 0 1 − x 2 ∫ x 2 + y 2 2 − x 2 − y 2 x y d z d y d x = ∫ 0 π 2 ∫ 0 π 4 ∫ 0 2 ( ρ 2 cos φ sin φ sin 2 θ ⋅ ρ 2 sin θ ) d ρ d θ d φ = ∫ 0 2 ρ 4 d ρ ∫ 0 π 2 sin ( 2 φ ) 2 d φ ∫ 0 π 4 sin 3 θ d θ = − ρ 5 5 ∣ 0 2 ⋅ cos ( 2 φ ) 4 ∣ 0 π 2 ⋅ − 3 cos θ + cos ( 3 θ ) 3 4 ∣ 0 π 4 = − ρ 5 5 ∣ 0 2 ⋅ cos ( 2 φ ) 4 ∣ 0 π 2 ⋅ − 9 cos θ + cos ( 3 θ ) 12 ∣ 0 π 4 = 4 2 5 ⋅ 1 4 ⋅ ( − 3 2 8 − 2 24 + 9 12 − 1 12 ) = 2 5 ⋅ ( 2 3 − 3 2 8 − 2 24 ) = 1 5 ⋅ ( 2 2 3 − 5 6 ) = 2 2 15 − 1 6 \displaystyle\int_0^1 \int_0^{\sqrt{1 - x^2}}\int_{\sqrt{x^2 + y^2}}^{\sqrt{2 - x^2 - y^2}} \,xy \mathrm{d}z\mathrm{d}y\mathrm{d}x\\
\, \mathrm{d}z\, \mathrm{d}y\, \mathrm{d}x = \rho^2\sin{\theta}\mathrm{d}\varphi\, \mathrm{d}\theta\, \mathrm{d}\rho\\
x = \rho\cos{\varphi}\sin{\theta}, y = \rho\sin{\varphi}\sin{\theta}, z = \rho\cos{\theta}\\
\sqrt{x^2 + y^2} = z\\
x^2 + y^2 + z^2 = 2z^2\\
\rho^2 = 2\rho^2\cos{\theta}\\
\cos{\theta} = \frac{1}{\sqrt{2}}, \theta = \frac{\pi}{4}\\
\sqrt{2 - x^2 - y^2} = z\\
2 - x^2 - y^2 = z^2\\
2 = x^2 + y^2 + z^2\\
\rho^2 = 2, \rho = \sqrt{2}\\
\sqrt{1 - x^2}= y\\
1 = x^2 + y^2\\
x^2 + y^2 = \rho^2\sin^2{\theta}(\cos^2{\varphi} + \sin^2{\varphi}) = \rho^2\sin^2{\theta}\\
1 = \rho^2\sin^2\theta\\
\sin^2\theta = \frac{1}{2}, \theta = \frac{\pi}{4}\\
x = 0, \rho\cos\varphi\cos\theta = 0, \varphi = \frac{\pi}{2}\\
\rho\sin\varphi\cos\theta = 1\\
\sqrt{2}\sin\varphi\sin\left(\frac{\pi}{4}\right) = 1\\
\varphi = \frac{\pi}{2}\\
\begin{aligned}
\int_0^1 \int_0^{\sqrt{1 - x^2}}\int_{\sqrt{x^2 + y^2}}^{\sqrt{2 - x^2 - y^2}}xy \,\mathrm{d}z\mathrm{d}y\mathrm{d}x &= \int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{4}}\int_0^{\sqrt{2}} (\rho^2\cos\varphi\sin\varphi\sin^2\theta\cdot\rho^2\sin\theta)\, \mathrm{d}\rho\,\mathrm{d}\theta\,\mathrm{d}\varphi
\\&=\int_0^{\sqrt{2}} \rho^4\, \mathrm{d}\rho \int_0^{\frac{\pi}{2}} \frac{\sin(2\varphi)}{2}\,\mathrm{d}\varphi \int_0^{\frac{\pi}{4}}\sin^3{\theta}\, \mathrm{d}\theta
\\&=-\frac{\rho^5}{5}\biggr\vert_0^{\sqrt{2}} \cdot\frac{\cos(2\varphi)}{4}\biggr\vert_0^{\frac{\pi}{2}} \cdot\frac{-3\cos{\theta} + \frac{\cos(3\theta)}{3}}{4}\biggr\vert_0^{\frac{\pi}{4}}
\\&=-\frac{\rho^5}{5}\biggr\vert_0^{\sqrt{2}} \cdot\frac{\cos(2\varphi)}{4}\biggr\vert_0^{\frac{\pi}{2}} \cdot\frac{-9\cos{\theta} + \cos(3\theta)}{12}\biggr\vert_0^{\frac{\pi}{4}}
\\&=\frac{4\sqrt{2}}{5} \cdot\frac{1}{4} \cdot\left(-\frac{3\sqrt{2}}{8} - \frac{\sqrt{2}}{24} + \frac{9}{12} - \frac{1}{12}\right)
\\&=\frac{\sqrt{2}}{5}\cdot\left(\frac{2}{3}-\frac{3\sqrt{2}}{8} - \frac{\sqrt{2}}{24}\right)
\\&=\frac{1}{5}\cdot\left(\frac{2\sqrt{2}}{3} - \frac{5}{6}\right)
\\&=\frac{2\sqrt{2}}{15} - \frac{1}{6}
\end{aligned} ∫ 0 1 ∫ 0 1 − x 2 ∫ x 2 + y 2 2 − x 2 − y 2 x y d z d y d x d z d y d x = ρ 2 sin θ d φ d θ d ρ x = ρ cos φ sin θ , y = ρ sin φ sin θ , z = ρ cos θ x 2 + y 2 = z x 2 + y 2 + z 2 = 2 z 2 ρ 2 = 2 ρ 2 cos θ cos θ = 2 1 , θ = 4 π 2 − x 2 − y 2 = z 2 − x 2 − y 2 = z 2 2 = x 2 + y 2 + z 2 ρ 2 = 2 , ρ = 2 1 − x 2 = y 1 = x 2 + y 2 x 2 + y 2 = ρ 2 sin 2 θ ( cos 2 φ + sin 2 φ ) = ρ 2 sin 2 θ 1 = ρ 2 sin 2 θ sin 2 θ = 2 1 , θ = 4 π x = 0 , ρ cos φ cos θ = 0 , φ = 2 π ρ sin φ cos θ = 1 2 sin φ sin ( 4 π ) = 1 φ = 2 π ∫ 0 1 ∫ 0 1 − x 2 ∫ x 2 + y 2 2 − x 2 − y 2 x y d z d y d x = ∫ 0 2 π ∫ 0 4 π ∫ 0 2 ( ρ 2 cos φ sin φ sin 2 θ ⋅ ρ 2 sin θ ) d ρ d θ d φ = ∫ 0 2 ρ 4 d ρ ∫ 0 2 π 2 sin ( 2 φ ) d φ ∫ 0 4 π sin 3 θ d θ = − 5 ρ 5 ∣ ∣ 0 2 ⋅ 4 cos ( 2 φ ) ∣ ∣ 0 2 π ⋅ 4 − 3 cos θ + 3 c o s ( 3 θ ) ∣ ∣ 0 4 π = − 5 ρ 5 ∣ ∣ 0 2 ⋅ 4 cos ( 2 φ ) ∣ ∣ 0 2 π ⋅ 12 − 9 cos θ + cos ( 3 θ ) ∣ ∣ 0 4 π = 5 4 2 ⋅ 4 1 ⋅ ( − 8 3 2 − 24 2 + 12 9 − 12 1 ) = 5 2 ⋅ ( 3 2 − 8 3 2 − 24 2 ) = 5 1 ⋅ ( 3 2 2 − 6 5 ) = 15 2 2 − 6 1
#332078 on Oct 2022
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