Answer
$$\frac{{625}}{4}\pi $$
Work Step by Step
$$\eqalign{
& \iint\limits_R {\left( {{x^2} + {y^2}} \right)}dA \cr
& {\text{Let the region be }} \cr
& R = \left\{ {\left( {x,y} \right):0 \leqslant y \leqslant \sqrt {25 - {x^2}} ,{\text{ }} - 5 \leqslant x \leqslant 5} \right\} \cr
& \iint\limits_R xdA = \int_{ - 5}^5 {\int_0^{\sqrt {25 - {x^2}} } {\left( {{x^2} + {y^2}} \right)} } dydx \cr
& {\text{By symmetry}} \cr
& = 2\int_0^5 {\int_0^{\sqrt {25 - {x^2}} } {\left( {{x^2} + {y^2}} \right)} } dydx \cr
& {\text{Integrate with respect to }}y \cr
& = 2\int_0^5 {\left[ {{x^2}y + \frac{{{y^3}}}{3}} \right]_0^{\sqrt {25 - {x^2}} }} dx \cr
& = 2\int_0^5 {\left[ {{x^2}\sqrt {25 - {x^2}} + \frac{{{{\left( {\sqrt {25 - {x^2}} } \right)}^3}}}{3}} \right]} dx \cr
& = 2\int_0^5 {\left[ {{x^2}\sqrt {25 - {x^2}} + \frac{{\left( {25 - {x^2}} \right)\sqrt {25 - {x^2}} }}{3}} \right]} dx \cr
& {\text{Integrate by using a CAS}} \cr
& = \frac{{625}}{4}\pi \cr} $$
