Find $\iiint_{E} y d V$, where $E$ is the solid bounded by the parabolic cylinder $z=x^{2}$ and the planes $y=0$ and $z=9-2 y$
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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 $\iiint_{E} y d V = \int_{0}^{9} \int_{0}^{\sqrt{z}} \int_{0}^{\frac{9-z}{2}} y dy dx dz$.

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 $\boxed{\frac{9^{3}}{4}}$.