Find ${\iiint }_{E}ydV$, where $E$ is the solid bounded by the parabolic cylinder $z=x^2$ and the planes $y=0$ and $z=9-2y$
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Step 1 ：First, we need to find the limits of integration. The solid E is bounded by the parabolic cylinder $z=x^2$ and the planes $y=0$ and $z=9-2y$. So, we can express $x$ and $y$ in terms of $z$ as $x=\sqrt{z}$ and $y=\frac{9-z}{2}$.

Step 2 ：Now, we can set up the triple integral. The triple integral of $y$ over the solid E is given by ${\iiint }_{E}ydV={\int }_0^9{\int }_0^{\sqrt{z}}{\int }_0^{\frac{9-z}{2}}ydydxdz$.

Step 3 ：We first integrate with respect to $y$. The integral of $y$ with respect to $y$ is $\frac{1}{2}{y}^2$. Evaluating this from $0$ to $\frac{9-z}{2}$ gives $\frac{1}{2}{\left(\frac{9-z}{2}\right)}^2=\frac{1}{8}(9-z{)}^2$.

Step 4 ：Next, we integrate with respect to $x$. Since there is no $x$ in the integrand, the integral is simply $x\cdot \frac{1}{8}(9-z{)}^2$. Evaluating this from $0$ to $\sqrt{z}$ gives $\frac{1}{8}z(9-z{)}^2$.

Step 5 ：Finally, we integrate with respect to $z$. The integral of $\frac{1}{8}z(9-z{)}^2$ with respect to $z$ is $\frac{1}{8}{\int }_0^{9}z(9-z{)}^{2}dz$. This is a standard polynomial integral, which can be computed to be $\frac{1}{8}\cdot \frac{9^4}{4}=\frac{9^3}{4}$.

Step 6 ：Therefore, the triple integral of $y$ over the solid E is $\boxed{{\displaystyle \frac{9^3}{4}}}$.