解  由于 I = ∭ Ω x 2 a 2 d x d y d z + ∭ Ω y 2 b 2 d x d y d z + ∭ Ω z 2 c 2 d x d y d z , I=\displaystyle\iiint\limits_{\Omega}\cfrac{x^2}{a^2}\mathrm{d}x\mathrm{d}y\mathrm{d}z+\displaystyle\iiint\limits_{\Omega}\cfrac{y^2}{b^2}\mathrm{d}x\mathrm{d}y\mathrm{d}z+\displaystyle\iiint\limits_{\Omega}\cfrac{z^2}{c^2}\mathrm{d}x\mathrm{d}y\mathrm{d}z, I=Ω∭​a2x2​dxdydz+Ω∭​b2y2​dxdydz+Ω∭​c2z2​dxdydz,   其中 ∭ Ω x 2 a 2 d x d y d z = ∫ − a a x 2 a 2 d x ∬ D d y d z \displaystyle\iiint\limits_{\Omega}\cfrac{x^2}{a^2}\mathrm{d}x\mathrm{d}y\mathrm{d}z=\displaystyle\int^a_{-a}\cfrac{x^2}{a^2}\mathrm{d}x\displaystyle\iint\limits_{D}\mathrm{d}y\mathrm{d}z Ω∭​a2x2​dxdydz=∫−aa​a2x2​dxD∬​dydz，这里 D D D表示椭圆 y 2 b 2 + z 2 c 2 ⩽ 1 − x 2 a 2 \cfrac{y^2}{b^2}+\cfrac{z^2}{c^2}\leqslant1-\cfrac{x^2}{a^2} b2y2​+c2z2​⩽1−a2x2​，即 y 2 b 2 ( 1 − x 2 a 2 ) + z 2 c 2 ( 1 − x 2 a 2 ) ⩽ 1 , \cfrac{y^2}{b^2\left(1-\cfrac{x^2}{a^2}\right)}+\cfrac{z^2}{c^2\left(1-\cfrac{x^2}{a^2}\right)}\leqslant1, b2(1−a2x2​)y2​+c2(1−a2x2​)z2​⩽1,   其面积为 π ( b 1 − x 2 a 2 ) ( c 1 − x 2 a 2 ) = π b c ( 1 − x 2 a 2 ) . \pi\left(b\sqrt{1-\cfrac{x^2}{a^2}}\right)\left(c\sqrt{1-\cfrac{x^2}{a^2}}\right)=\pi bc\left(1-\cfrac{x^2}{a^2}\right). π⎝⎛​b1−a2x2​ ​⎠⎞​⎝⎛​c1−a2x2​ ​⎠⎞​=πbc(1−a2x2​).   故 ∭ Ω x 2 a 2 d x d y d z = ∫ − a a π b c a 2 x 2 ( 1 − x 2 a 2 ) d x = 4 15 π a b c . \displaystyle\iiint\limits_{\Omega}\cfrac{x^2}{a^2}\mathrm{d}x\mathrm{d}y\mathrm{d}z=\displaystyle\int^a_{-a}\cfrac{\pi bc}{a^2}x^2\left(1-\cfrac{x^2}{a^2}\right)\mathrm{d}x=\cfrac{4}{15}\pi abc. Ω∭​a2x2​dxdydz=∫−aa​a2πbc​x2(1−a2x2​)dx=154​πabc.   同理可得 ∭ Ω y 2 b 2 d x d y d z = 4 15 π a b c , ∭ Ω z 2 c 2 d x d y d z = 4 15 π a b c . \displaystyle\iiint\limits_{\Omega}\cfrac{y^2}{b^2}\mathrm{d}x\mathrm{d}y\mathrm{d}z=\cfrac{4}{15}\pi abc,\qquad\displaystyle\iiint\limits_{\Omega}\cfrac{z^2}{c^2}\mathrm{d}x\mathrm{d}y\mathrm{d}z=\cfrac{4}{15}\pi abc. Ω∭​b2y2​dxdydz=154​πabc,Ω∭​c2z2​dxdydz=154​πabc.   所以 I = 3 ⋅ 4 15 π a b c = 4 5 π a b c I=3\cdot\cfrac{4}{15}\pi abc=\cfrac{4}{5}\pi abc I=3⋅154​πabc=54​πabc。（这道题主要利用了对称性求解）

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