Answer
$V = \frac{166}{27}$
Work Step by Step
$V=\int^{2}_{-2}\int^{1}_{-1} (1-\frac{1}{4}x^{2} - \frac{1}{9}y^{2}) dxdy$
$V=\int^{2}_{-2}[x-\frac{1}{12}x^{3} - \frac{1}{9}xy^{2}]^{x=1}_{x=-1} dy$
$V=\int^{2}_{-2}[1-\frac{1}{12}-\frac{1}{9}y^{2}] - [-1+\frac{1}{12}+\frac{1}{9}y^{2}]dy$
$V=\int^{2}_{-2} \frac{11}{6} - \frac{2}{9}y^{2}dy$
$V=[\frac{11}{6}y - \frac{2}{27}y^{3}]^{2}_{-2}$
$V=[\frac{11}{6}(2) - \frac{2}{27}(2)^{3}]-[\frac{11}{6}(-2) - \frac{2}{27}(-2)^{3}]=\frac{166}{27}$
