Answer
$$25$$
Work Step by Step
$$\eqalign{
& \iint\limits_R xdA \cr
& {\text{Let the region be }} \cr
& R = \left\{ {\left( {x,y} \right):0 \leqslant y \leqslant \sqrt {25 - {x^2}} ,{\text{ }} - 5 \leqslant x \leqslant 0} \right\} \cr
& \iint\limits_R xdA = \int_0^3 {\int_{4y/3}^{\sqrt {25 - {y^2}} } x } dxdy \cr
& {\text{Integrate with respect to }}x \cr
& = \int_0^3 {\left[ {\frac{{{x^2}}}{2}} \right]_{4y/3}^{\sqrt {25 - {y^2}} }} dy \cr
& = \int_0^3 {\left[ {\frac{{{{\left( {\sqrt {25 - {y^2}} } \right)}^2}}}{2} - \frac{{{{\left( {4y/3} \right)}^2}}}{2}} \right]} dy \cr
& = \int_0^3 {\left( {\frac{{25 - {y^2}}}{2} - \frac{{8{y^2}}}{9}} \right)} dy \cr
& = \int_0^3 {\left( {\frac{{25}}{2} - \frac{{25{y^2}}}{{18}}} \right)} dy \cr
& {\text{Integrate}} \cr
& = \left[ {\frac{{25}}{2}y - \frac{{25{y^3}}}{{54}}} \right]_0^3 \cr
& = \left[ {\frac{{25}}{2}\left( 3 \right) - \frac{{25{{\left( 3 \right)}^3}}}{{54}}} \right] - 0 \cr
& = 25 \cr} $$
