Answer
$\frac{166}{27}$
Work Step by Step
We can begin by getting the equation for the elliptic paraboloid in terms of $z$. Doing this, we get:
$$z=1-\frac{x^2}{4}-\frac{y^2}{9}$$
The volume under the curve in region $R$ can be expressed by:
$$\iint_R\bigg(1-\frac{x^2}{4}-\frac{y^2}{9}\bigg)dA$$
Rewriting as an iterated integral, we get:
$$\int_{-1}^{1}\int_{-2}^{2}\bigg(1-\frac{x^2}{4}-\frac{y^2}{9}\bigg)\,dy\,dx$$
Solving the inner integral, we get:
$$\int_{-1}^{1}\bigg[y-\frac{x^2y}{4}-\frac{y^3}{27}\bigg]_{-2}^{2}dx\\
=\int_{-1}^1\bigg(\frac{92}{27}-x^2\bigg)dx$$
Integrating again, we get:
$$\bigg[\frac{92}{27}x-\frac{x^3}{3}\bigg]_{-1}^1\\
=\frac{166}{27}$$
