MvCalc36

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Given domains are 3x=4-y^{2},x=y^{2}\,. we have to find the limits for evaluating the integration A=\iint _{D}dxdy\,

Solving the given equations for y, we get 3y^{2}=4-y^{2},4y^{2}=4,y^{2}=1,y=1,-1\,

Hence the limits of y are from -1 to 1.

The limits of x are from y^{2}\, to {\frac  {4-y^{2}}{3}}\,

Therefore the area of the plane A=\int _{{-1}}^{{1}}[\int _{{x=y^{2}}}^{{{\frac  {4-y^{2}}{3}}}}dx]dy\,

=\int _{{-1}}^{{1}}[x]_{{x=y^{2}}}^{{{\frac  {4-y^{2}}{3}}}}dy\,

=\int _{{-1}}^{{1}}[{\frac  {4-y^{2}}{3}}-y^{2}]dy\,

=\int _{{-1}}^{{1}}[{\frac  {4}{3}}-{\frac  {4y^{2}}{3}}]dy\,

=[{\frac  {4}{3}}y-{\frac  {4y^{3}}{9}}]_{{-1}}^{{1}}=[{\frac  {4}{3}}-{\frac  {4}{9}}]-[-{\frac  {4}{3}}+{\frac  {4}{9}}]\,

={\frac  {8}{3}}-{\frac  {8}{9}}={\frac  {16}{9}}\,

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