Exercicios Univesp Calculo 3

\[\eqalign{ & x = {y^2} \cr & 4 = {y^2} \cr & y = \pm \sqrt 4 \cr & y = \pm 2 \cr & \cr & M = \int_{ - 2}^2 {} \int_{{y^2}}^4 {D \cdot dx \cdot dy} \cr & M = \int_{ - 2}^2 {\int_{{y^2}}^4 {kx \cdot dx \cdot dy} } \cr & M = \int_{ - 2}^2 {\left[ {\dfrac{k}{2}{x^2}} \right]_{{y^2}}^4} \cdot dy \cr & M = \int_{ - 2}^2 {\left[ {8k - \dfrac{k}{2}{y^4}} \right]} \cdot dy \cr & M = \left[ {8ky - \dfrac{k}{{10}}{y^5}} \right]_{ - 2}^2 \cr & M = \left[ {16k + 16k - \dfrac{k}{{10}}{{\left( 2 \right)}^5} - \dfrac{k}{{10}}{{\left( 2 \right)}^5}} \right] \cr & M = \left[ {32k - \dfrac{{64k}}{{10}}} \right] \cr & M = \left[ {32k - \dfrac{{32k}}{5}} \right] \cr & M = \left[ {\dfrac{{160k}}{5} - \dfrac{{32k}}{5}} \right] \cr & M = \dfrac{{160k - 32k}}{5} \cr & M = \dfrac{{128k}}{5} }\]

Portanto, a resposta final é 128k/5.