## Calculus (3rd Edition)

The volume of the region is $\frac{4}{5}$.
Recall that the double integral defines the signed volume between the graph of $f\left( {x,y} \right)$ and the $xy$-plane. That is, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A$. Referring to the figure attached, we see that the solid region lies over the domain ${\cal D}$, which is a vertically simple region. The left and right boundaries of ${\cal D}$ are $x=-1$ and $x=1$, respectively. Whereas, the lower and upper boundaries are $y=0$ and $y = 1 - {x^2}$, respectively. Thus, the domain description of ${\cal D}$ is ${\cal D} = \left\{ {\left( {x,y} \right)| - 1 \le x \le 1,0 \le y \le 1 - {x^2}} \right\}$ From the figure attached, we see that the solid is bounded below by $z=1$ and bounded above by $z+y=2$. So, $z=2-y$. Thus, by Theorem 3, we have $f\left( {x,y} \right) = 2 - y - 1 = 1 - y$. We evaluate the volume of the solid as an iterated integral: $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{x = - 1}^1 \mathop \smallint \limits_{y = 0}^{1 - {x^2}} \left( {1 - y} \right){\rm{d}}y{\rm{d}}x$ $= \mathop \smallint \limits_{x = - 1}^1 \left( {\left( {y - \frac{1}{2}{y^2}} \right)|_{y = 0}^{1 - {x^2}}} \right){\rm{d}}x$ $= \mathop \smallint \limits_{x = - 1}^1 \left( {1 - {x^2} - \frac{1}{2}{{\left( {1 - {x^2}} \right)}^2}} \right){\rm{d}}x$ $= \mathop \smallint \limits_{x = - 1}^1 \left( {\frac{1}{2} - \frac{1}{2}{x^4}} \right){\rm{d}}x$ $= \left( {\left( {\frac{1}{2}x - \frac{1}{{10}}{x^5}} \right)|_{ - 1}^1} \right)$ $= \frac{1}{2} - \frac{1}{{10}} + \frac{1}{2} - \frac{1}{{10}}$ $= \frac{4}{5}$ So, the volume of the region is $\frac{4}{5}$.