三重积分求解

I = ∭ Ω z d x d y d z \displaystyle{I = \iiint_{\Omega}zdxdydz} I=Ωzdxdydz, 其中 Ω \Omega Ω 是由锥面 z = x 2 + y 2 z = \sqrt{x^2+y^2} z=x2+y2 与 平面 z = 1 z = 1 z=1 所围成的闭区域

⟹ ∫ 0 1 d z ∫ 0 2 π d θ ∫ 0 z z ∗ r d r \Longrightarrow \displaystyle{\int_{0}^{1}dz\int_{0}^{2\pi}d\theta\int_{0}^{z}z * rdr} 01dz02πdθ0zzrdr
⟹ ∫ 0 1 d z ∫ 0 2 π z ∗ r 2 2 ∣ 0 z d θ \Longrightarrow \displaystyle{\int_{0}^{1}dz\int_{0}^{2\pi} \frac{z * r^2}{2} \Big|_{0}^{z} d\theta} 01dz02π2zr20zdθ
⟹ 2 π ∗ z 4 8 ∣ 0 1 d z \Longrightarrow \displaystyle{2\pi * \frac{z^4}{8}\Big|_{0}^{1}dz} 2π8z401dz
⟹ 2 π ∗ 1 8 \Longrightarrow \displaystyle{2\pi * \frac{1}{8}} 2π81
= π 4 =\displaystyle{ \frac{\pi}{4}} =4π

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转载自blog.csdn.net/weixin_42290927/article/details/106930341