## Thomas' Calculus 13th Edition

$$\bar{z}=\frac{12}{5},, \ \bar{x}=\bar{y}=0,$$ \begin{align*} I_{x} &=\frac{7904}{105}\\ I_{y} &=\frac{4832}{63}\\ I_{z} &=\frac{256}{45} \end{align*}
Since \begin{align*} M&=4 \int_{0}^{1} \int_{0}^{1} \int_{4 y^{2}}^{4} d z d y d x\\ &=4 \int_{0}^{1} \int_{0}^{1}\left(4-4 y^{2}\right) d y d x\\ &=16 \int_{0}^{1} \frac{2}{3} d x\\ &=\frac{32}{3} \\ M_{x y}&=4 \int_{0}^{1} \int_{0}^{1} \int_{4 y^{2}}^{4} z d z d y d x\\ &=2 \int_{0}^{1} \int_{0}^{1}\left(16-16 y^{4}\right) d y d x\\ &=\frac{128}{5} \int_{0}^{1} d x\\ &=\frac{128}{5} \end{align*} Then by symmetry $$\bar{z}=\frac{12}{5},, \ \bar{x}=\bar{y}=0,$$ and \begin{align*} I_{x}&=4 \int_{0}^{1} \int_{0}^{1} \int_{4 y^{2}}^{4}\left(y^{2}+z^{2}\right) d z d y d x\\ &=4 \int_{0}^{1} \int_{0}^{1}\left[\left(4 y^{2}+\frac{64}{3}\right)-\left(4 y^{4}+\frac{64 y^{6}}{3}\right)\right] d y d x\\ &=4 \int_{0}^{1} \frac{1976}{105} d x\\ &=\frac{7904}{105}\\ I_{y}&=4 \int_{0}^{1} \int_{0}^{1} \int_{4 y^{2}}^{4}\left(x^{2}+z^{2}\right) d z d y d x\\ &=4 \int_{0}^{1} \int_{0}^{1}\left[\left(4 x^{2}+\frac{64}{3}\right)-\left(4 x^{2} y^{4}+\frac{64 y^{6}}{3}\right)\right] d y d x\\ &=4 \int_{0}^{1}\left(\frac{8}{3} x^{2}+\frac{128}{7}\right) d x\\ &=\frac{4832}{63}\\ I_{z}&=4 \int_{0}^{1} \int_{0}^{1} \int_{4 y^{2}}^{4}\left(x^{2}+y^{2}\right) d z d y d x\\ &=16 \int_{0}^{1} \int_{0}^{1}\left(x^{2}-x^{2} y^{2}+y^{2}-y^{4}\right) d y d x\\ &=16 \int_{0}^{1}\left(\frac{2 x^{2}}{3}+\frac{2}{15}\right) d x\\ &=\frac{256}{45} \end{align*}