MvCalc37

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The area of the plane A=\iint _{D}dxdy\, where D is the region bounded by the given parabola and the straight line.Now,we find out the limits of x and y.

The limits of y are from 0 to a and the limits of x are from 3a-y\, to {\frac  {y^{2}}{4a}}\,

Thus area A=\int _{0}^{a}[\int _{{3a-y}}^{{{\frac  {y^{2}}{4a}}}}dx]dy\,

=\int _{0}^{a}[x]_{{3a-y}}^{{{\frac  {y^{2}}{4a}}}}dy\,

=\int _{0}^{a}[{\frac  {y^{2}}{4a}}-3a+y]dy\,

=[{\frac  {y^{3}}{12a}}-3ay+{\frac  {y^{2}}{2}}]_{0}^{a}\,

={\frac  {a^{3}}{12a}}-3a^{2}+{\frac  {a^{2}}{2}}={\frac  {a^{2}}{12}}-3a^{2}+{\frac  {a^{2}}{2}}={\frac  {a^{2}-36a^{2}+6a^{2}}{12}}={\frac  {29a^{2}}{12}}\,

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