Answer
$$V = 1$$
Work Step by Step
$$\eqalign{
& z = \frac{1}{{{{\left( {x + 1} \right)}^2}{{\left( {y + 1} \right)}^2}}} \cr
& {\text{The volume of the solid is given by}} \cr
& V = \iint\limits_R zdA,{\text{ }}z = \frac{1}{{{{\left( {x + 1} \right)}^2}{{\left( {y + 1} \right)}^2}}} \cr
& {\text{Let the region be }} \cr
& R = \left\{ {\left( {x,y} \right):0 \leqslant y \leqslant \infty ,{\text{ }}0 \leqslant x \leqslant \infty } \right\} \cr
& V = \iint\limits_R zdA = \int_0^\infty {\int_0^\infty {\frac{1}{{{{\left( {x + 1} \right)}^2}{{\left( {y + 1} \right)}^2}}}} } dydx \cr
& V = \int_0^\infty {\frac{1}{{{{\left( {x + 1} \right)}^2}}}\left[ {\int_0^\infty {\frac{1}{{{{\left( {y + 1} \right)}^2}}}} dy} \right]} dx{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Integrate }}\int_0^\infty {\frac{1}{{{{\left( {y + 1} \right)}^2}}}} dy{\text{ with respect to }}y \cr
& \int_0^\infty {\frac{1}{{{{\left( {y + 1} \right)}^2}}}} dy = - \mathop {\lim }\limits_{b \to \infty } \int_0^b {\frac{1}{{y + 1}}} dy \cr
& = - \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{1}{{b + 1}} - \frac{1}{{0 + 1}}} \right] \cr
& = - \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{1}{{b + 1}} - 1} \right] \cr
& = - \left[ {\frac{1}{{\infty + 1}} - 1} \right] \cr
& = 1 \cr
& {\text{Substitute the previous result in }}\left( {\bf{1}} \right) \cr
& V = \int_0^\infty {\frac{1}{{{{\left( {x + 1} \right)}^2}}}\left[ 1 \right]} dx \cr
& V = \int_0^\infty {\frac{1}{{{{\left( {x + 1} \right)}^2}}}} dx \cr
& {\text{By symmetry }}\int_0^\infty {\frac{1}{{{{\left( {x + 1} \right)}^2}}}} dx = 1,{\text{ then}} \cr
& V = 1\left( 1 \right) \cr
& V = 1 \cr} $$
