Chapter 9 - Multiveriable Calculus - 9.6 Double Integrals - 9.6 Exercises - Page 514: 35

$$V = \frac{1}{4}\ln \frac{{17}}{8}$$

$$\eqalign{ & z = \frac{{xy}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}};\,\,\,\,\,\,\,1 \leqslant x \leqslant 2,\,\,\,\,\,1 \leqslant y \leqslant 4 \cr & {\text{by the equation just given}}{\text{, the volume is}} \cr & V = \iint\limits_R {f\left( {x,y} \right)}dxdy \cr & {\text{replacing the limits for the region }}R{\text{ and }}f\left( {x,y} \right) = z \cr & V = \int_1^4 {\int_1^2 {\frac{{xy}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}} dxdy} \cr & V = \int_1^4 {\left[ {\int_1^2 {\frac{{xy}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}} dx} \right]dy} \cr & {\text{solve the inner integral by treating }}y{\text{ as a constant}} \cr & = \int_1^2 {\frac{{xy}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}} dx \cr & {\text{write the denominator with negative exponent}} \cr & = \frac{1}{2}y\int_1^2 {{{\left( {{x^2} + {y^2}} \right)}^{ - 2}}\left( {2x} \right)} dx \cr & {\text{integrate by using the power rule }} \cr & = \frac{1}{2}y\left[ {\frac{{{{\left( {{x^2} + {y^2}} \right)}^{ - 1}}}}{{ - 1}}} \right]_1^2 \cr & = - \frac{1}{2}y\left[ {\frac{1}{{{x^2} + {y^2}}}} \right]_1^2 \cr & {\text{evaluating the limits in the variable }}x \cr & = - \frac{1}{2}y\left[ {\frac{1}{{{2^2} + {y^2}}} - \frac{1}{{{1^2} + {y^2}}}} \right] \cr & {\text{simplifying}} \cr & = \frac{1}{2}\left[ {\frac{y}{{{1^2} + {y^2}}} - \frac{y}{{4 + {y^2}}}} \right] \cr & = \frac{1}{{2\left( 2 \right)}}\left[ {\frac{{2y}}{{1 + {y^2}}} - \frac{{2y}}{{4 + {y^2}}}} \right] \cr & \cr & V = \int_1^4 {\left[ {\int_1^2 {\frac{{xy}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}} dx} \right]dy} \cr & V = \frac{1}{4}\int_1^4 {\left( {\frac{{2y}}{{1 + {y^2}}} - \frac{{2y}}{{4 + {y^2}}}} \right)dy} \cr & {\text{integrating}} \cr & V = \frac{1}{4}\left[ {\ln \left( {1 + {y^2}} \right) - \ln \left( {4 + {y^2}} \right)} \right]_1^4 \cr & {\text{using logarithmic properties}} \cr & V = \frac{1}{4}\left[ {\ln \left( {\frac{{1 + {y^2}}}{{4 + {y^2}}}} \right)} \right]_1^4 \cr & {\text{evaluating}} \cr & V = \frac{1}{4}\left[ {\ln \left( {\frac{{1 + {4^2}}}{{4 + {4^2}}}} \right) - \ln \left( {\frac{{1 + {1^2}}}{{4 + {1^2}}}} \right)} \right] \cr & V = \frac{1}{4}\left( {\ln \frac{{17}}{{20}} - \ln \frac{2}{5}} \right) \cr & V = \frac{1}{4}\ln \frac{{17/20}}{{2/5}} \cr & V = \frac{1}{4}\ln \frac{{17}}{8} \cr} $$