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$$\int_{-a}^{a}\int_{0}^{b\cdot \sqrt{1-\frac{x^2}{a^2}}}2x^2ydydx=\int_{-a}^{a}2b^2(x^2-x^4/a^2) dx =2b^2(2/3a^2-2a^5/5a^2)=\frac{8}{15}a^3b^2$$ But the actually answer in $\frac{4}{15}a^3b^2$. Which step did I go wrong?

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  • $\begingroup$ $y$ integrates to $y^2/2$, eliminating the original factor $2$ $\endgroup$ – David Peterson Mar 19 '16 at 10:27
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The first equality should have been $$\int_{-a}^{a}\int_{0}^{b\cdot \sqrt{1-\frac{x^2}{a^2}}}2x^2ydydx=\int_{-a}^{a}b^2(x^2-x^4/a^2)dx$$

since the antiderivative of $2y$ is $y^2$. That's why you get $2$ times the desired result.

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