MvCalc29

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The ordinates are x=2a and x-axis. X-axis means y=0.Given parabola x^{2}=4ay\, implies x^{2}=0\,,hence the limits of x are from 0 to a.

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

Therefore I=\iint _{D}xydxdy=\int _{0}^{a}[\int _{{y=0}}^{{{\frac  {x^{2}}{4a}}}}xydy]dx\,

Now \int _{{y=0}}^{{{\frac  {x^{2}}{4a}}}}xydy=[{\frac  {xy^{2}}{2}}]_{{0}}^{{{\frac  {x^{4}}{4a}}}}={\frac  {(x)({\frac  {x^{4}}{16a^{2}}})}{2}}={\frac  {x^{5}}{32a^{2}}}\,

HenceI=\int _{0}^{a}[{\frac  {x^{6}}{192a^{2}}}]dx=[{\frac  {x^{4}}{32a}}]_{0}^{a}={\frac  {a^{4}}{192}}\,

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