MvCalc28

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The limits of x are from -y\, to {\sqrt  {y}}\, and y are from 0 to 2.

Therefore,I=\iint _{D}(1+x+y)dxdy=\int _{0}^{2}[\int _{{x=-y}}^{{{\sqrt  {y}}}}(1+x+y)dx]dy\,

Now \int _{{x=-y}}^{{{\sqrt  {y}}}}(1+x+y)dx=[x+{\frac  {x^{2}}{2}}+yx]_{{x=-y}}^{{{\sqrt  {y}}}}\,

=[{\sqrt  {y}}+{\frac  {y}{2}}+y{\sqrt  {y}}]-[-y+{\frac  {y^{2}}{2}}-y^{2}]\,

=[{\sqrt  {y}}+{\frac  {y}{2}}+y^{{{\frac  {3}{2}}}}+y+{\frac  {y^{2}}{2}}]\,

Hence I=\int _{0}^{2}[{\sqrt  {y}}+{\frac  {y}{2}}+y^{{{\frac  {3}{2}}}}+y+{\frac  {y^{2}}{2}}]dy\,

=[{\frac  {2}{3}}y^{{{\frac  {3}{2}}}}+{\frac  {y^{2}}{4}}+{\frac  {2}{5}}y^{{{\frac  {5}{2}}}}+{\frac  {y^{2}}{2}}+{\frac  {y^{3}}{6}}]_{0}^{2}\,

=[{\frac  {2}{3}}2^{{{\frac  {3}{2}}}}+1+{\frac  {2}{5}}2^{{{\frac  {5}{2}}}}+2+{\frac  {8}{6}}]\,

={\sqrt  {32}}({\frac  {1}{3}}+{\frac  {2}{5}})+3+{\frac  {4}{3}}\,

=4{\sqrt  {2}}({\frac  {11}{15}})+{\frac  {13}{3}}={\frac  {44}{15}}{\sqrt  {2}}+{\frac  {13}{3}}\,


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