## Calculus: Early Transcendentals 8th Edition

$\frac{166}{27}$
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: $$\bigg[\frac{92}{27}x-\frac{x^3}{3}\bigg]_{-1}^1\\ =\frac{166}{27}$$