MvCalc25

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Solving the equations for x from the given y=x,y^{2}=4x\, we get x=0,4\,

Hence the limits of x are from 0 to 4.

Similarly,the limits of y are from y=x\, to y=2{\sqrt  {x}}\,

Therefore I=\int _{0}^{4}\int _{{x}}^{{2{\sqrt  {x}}}}(x^{2}+y^{2})dxdy\,

=I=\int _{0}^{4}[\int _{{y=x}}^{{y=2{\sqrt  {x}}}}(x^{2}+y^{2})dy]dx\,

Now \int _{{y=x}}^{{2{\sqrt  {x}}}}(x^{2}+y^{2})dy=[x^{2}y+{\frac  {y^{3}}{3}}]_{{X}}^{{2{\sqrt  {x}}}}\,

=[2x^{{{\frac  {5}{2}}}}+{\frac  {8}{3}}x^{{{\frac  {3}{2}}}}-{\frac  {4}{3}}x^{3}]\,

Therefore,I=\int _{{x=0}}^{{4}}[2x^{{{\frac  {5}{2}}}}+{\frac  {8}{3}}x^{{{\frac  {3}{2}}}}-{\frac  {4}{3}}x^{3}]dx\,

=[{\frac  {4}{7}}x^{{{\frac  {7}{2}}}}+{\frac  {16}{15}}x^{{{\frac  {5}{2}}}}-{\frac  {1}{3}}x^{4}]_{0}^{4}\,

=[{\frac  {4}{7}}4^{{{\frac  {7}{2}}}}+{\frac  {16}{15}}4^{{{\frac  {5}{2}}}}-{\frac  {1}{3}}4^{4}]\,

=[{\frac  {4}{7}}(128)+{\frac  {16}{15}}(32)-{\frac  {256}{3}}]\,

={\frac  {7680+3584-8960}{105}}={\frac  {2304}{105}}={\frac  {768}{35}}\,


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