To evaluate the double integral and find the volume of the solid bounded by $z=1-2y^2-3x^2$ and the $xy$-plane, we need to set up and solve the integral.
The volume can be calculated by integrating the height function $f(x,y)$ over the region $D$ in the $xy$-plane. In this case, the height function is given by $f(x,y)=1-2y^2-3x^2$.
Thus, the volume can be expressed as:
$V={\iint }_{D}f(x,y)dA$
where $dA$ represents the area element.
To determine the limits of integration for $x$ and $y$, we need to find the region $D$ in the $xy$-plane that corresponds to the solid bounded by the surface and the $xy$-plane.
From the equation $z=1-2y^2-3x^2$, we can see that the surface intersects the $xy$-plane when $z=0$:
$0=1-2y^2-3x^2$
Simplifying this equation, we obtain:
$2y^2+3x^2=1$
This equation represents an ellipse in the $xy$-plane.
To determine the limits of integration, we need to find the range of $x$ and $y$ values that correspond to this ellipse.
Let's solve for $y$ in terms of $x$:
$2y^2=1-3x^2$
$y^2=\frac{1-3x^2}{2}$
$y=\pm \sqrt{\frac{1-3x^2}{2}}$
Since the ellipse is symmetric about the $x$-axis, we only need to consider the positive square root.
The limits of integration for $x$ will be determined by the range of $x$ values that correspond to the ellipse. To find these limits, we need to solve the following equation:
$2y^2+3x^2=1$
Substituting $y=\sqrt{\frac{1-3x^2}{2}}$, we have:
$2\left(\frac{1-3x^2}{2}\right)+3x^2=1$
Simplifying further, we obtain:
$1-3x^2+3x^2=1$
$-3x^2+3x^2=0$
This equation holds true for any value of $x$. Therefore, the limits of integration for $x$ are $-\infty$ to $+\infty$.
Next, we need to determine the limits of integration for $y$. Since the ellipse is symmetric about the $x$-axis, the range of $y$ values will be from $0$ to the positive $y$ value of the ellipse.
Therefore, the limits of integration for $y$ are $0$ to $\sqrt{\frac{1-3x^2}{2}}$.
Now, we can set up the double integral to calculate the volume:
$V={\iint }_D(1-2y^2-3x^2)dA$
$V={\int }_{-\infty }^{+\infty }{\int }_0^{\sqrt{\frac{1-3x^2}{2}}}(1-2y^2-3x^2)dydx$
To evaluate this integral, we first integrate with respect to $y$:
$V={\int }_{-\infty }^{+\infty }{[y-\frac{2}{3}{y}^3-3x^{2}y]}_0^{\sqrt{\frac{1-3x^2}{2}}}dx$
Simplifying the expression further, we have:
$V={\int }_{-\infty }^{+\infty }[\sqrt{\frac{1-3x^2}{2}}-\frac{2}{3}{\left(\sqrt{\frac{1-3x^2}{2}}\right)}^3-3x^2\sqrt{\frac{1-3x^2}{2}}]dx$
Finally, we can integrate with respect to $x$:
$V={[\frac{1}{3}{\left(\frac{1-3x^2}{2}\right)}^{3/2}-\frac{2}{15}{\left(\frac{1-3x^2}{2}\right)}^{5/2}-\frac{3}{5}{\left(\frac{1-3x^2}{2}\right)}^{3/2}{x}^2]}_{-\infty }^{+\infty }$
Since the limits of integration for $x$ are $-\infty$ to $+\infty$, the expression evaluates to:
$V={[\frac{1}{3}{\left(\frac{1-3x^2}{2}\right)}^{3/2}-\frac{2}{15}{\left(\frac{1-3x^2}{2}\right)}^{5/2}-\frac{3}{5}{\left(\frac{1-3x^2}{2}\right)}^{3/2}{x}^2]}_{-\infty }^{+\infty }$
Substituting the limits, we get:
$V=(\frac{1}{3}{\left(\frac{1-3·(-\infty {)}^2}{2}\right)}^{3/2}-\frac{2}{15}{\left(\frac{1-3·(-\infty {)}^2}{2}\right)}^{5/2}-\frac{3}{5}{\left(\frac{1-3·(-\infty {)}^2}{2}\right)}^{3/2}·(-\infty {)}^2)-(\frac{1}{3}{\left(\frac{1-3·(+\infty {)}^2}{2}\right)}^{3/2}-\frac{2}{15}{\left(\frac{1-3·(+\infty {)}^2}{2}\right)}^{5/2}-\frac{3}{5}{\left(\frac{1-3·(+\infty {)}^2}{2}\right)}^{3/2}·(+\infty {)}^2)$
Since $(-\infty {)}^2$ and $(+\infty {)}^2$ both evaluate to $\infty$, we have:
$V=(\frac{1}{3}{\left(\frac{1}{2}\right)}^{3/2}-\frac{2}{15}{\left(\frac{1}{2}\right)}^{5/2}-\frac{3}{5}{\left(\frac{1}{2}\right)}^{3/2}·\infty )-(\frac{1}{3}{\left(\frac{1}{2}\right)}^{3/2}-\frac{2}{15}{\left(\frac{1}{2}\right)}^{5/2}-\frac{3}{5}{\left(\frac{1}{2}\right)}^{3/2}·\infty )$
Since any finite number multiplied by $\infty$ is still $\infty$, we can simplify further:
$V=(\frac{1}{3}{\left(\frac{1}{2}\right)}^{3/2}-\frac{2}{15}{\left(\frac{1}{2}\right)}^{5/2}-\infty )-(\frac{1}{3}{\left(\frac{1}{2}\right)}^{3/2}-\frac{2}{15}{\left(\frac{1}{2}\right)}^{5/2}-\infty )$
Simplifying the expression inside the parentheses, we have:
$V=(\frac{1}{3}{\left(\frac{1}{2}\right)}^{3/2}-\frac{2}{15}{\left(\frac{1}{2}\right)}^{5/2})-(\frac{1}{3}{\left(\frac{1}{2}\right)}^{3/2}-\frac{2}{15}{\left(\frac{1}{2}\right)}^{5/2})$
Both terms inside the parentheses cancel each other out, leaving us with:
$V=0$
Therefore, the volume of the solid bounded by $z=1-2y^2-3x^2$ and the $xy$-plane is $0$.