triple integration question . limit problem(substitution) I have to evaluate the following integral-
$$\iiint x^2y \,dx \,dy \,dz$$ over the region bounded by $ \frac{x^2}1+\frac{y^2}4+\frac{z^2}9=1.$
I take the limits as -
for x: $-1$ to $1$
for y: $-2(1-x^2)^{\frac 12})$ to $2((1-x^2)^{\frac 12})$
for z: $-3((1-x^2-(\frac {y^2}4))^{\frac 12})$ to $3((1-x^2-(\frac {y^2}4))^{\frac 12})$
I easily evaluated the $dz$ portion
But in the $dy$ portion , i substituted $y^2=t$ and both the upper and lower limits became the same giving the answer 0 . What should be the substitution in the y step?

$$K=\iiint_E x^2y \,\mathrm dx \,\mathrm dy \,\mathrm dz$$ where $E$ is the ellipsoid $E=\left\{ (x,y,z)\in\Bbb R^3: \frac{x^2}1+\frac{y^2}4+\frac{z^2}9=1\right\}$
So we have
$$
|z|\le 3,\quad |x|\le \sqrt{1-\frac{z^2}{9}},\quad |y|\le 2 \sqrt{1-\frac{x^2}{1}-\frac{z^2}{9}}
$$
and 
$$
\begin{align}
K&={\Large\int}_{-3}^3{\Large\int}_{-\sqrt{1-\frac{z^2}{9}}}^\sqrt{1-\frac{z^2}{9}}{\Large\int}_{-2 \sqrt{1-\frac{x^2}{1}-\frac{z^2}{9}}}^{2 \sqrt{1-\frac{x^2}{1}-\frac{z^2}{9}}} \;x^2y\,\mathrm dy \,\mathrm dx \,\mathrm dz=0
\end{align}
$$
because $y$ is a odd function integrated over a symmetric interval.
