Here, $y = \frac{1}{2}x^{2}$, $a = 0$ and $b = 2$. So, we have
\[M_{x} = \rho \int_{0}^{2} \frac{1}{2}x^{2} \, dx = \rho \int_{0}^{2} \frac{1}{8}x^{4} \, dx = \rho \left[\frac{1}{40}x^{5}\right]_{0}^{2} = \rho \left(\frac{2^{5}}{40}\right) = \frac{4}{5}\rho.\]
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