MvCalc38

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The area of the plane is A=\iint _{D}dxdy\,

Now finding the limits from the given domains,we get

4a(x+a)=4b(b-x),4ax+4a^{2}=4b^{2}-4bx,x=b-a\, that is the limits of x are from 0 to b-a\,

The limits of y are {\sqrt  {4a(x+a)}}\, to {\sqrt  {4b(b-x)}}\,

Now area A=\int _{{0}}^{{b-a}}[\int _{{{\sqrt  {4a(x+a)}}}}^{{{\sqrt  {4b(b-x)}}}}dy]dx\,

=\int _{{0}}^{{b-a}}[y]_{{{\sqrt  {4a(x+a)}}}}^{{{\sqrt  {4b(b-x)}}}}dx\,

=\int _{{0}}^{{b-a}}[{\sqrt  {4b(b-x)}}-{\sqrt  {4a(x+a)}}]dx\,

=[-{\frac  {4}{3b}}(b^{2}-bx)^{{{\frac  {3}{2}}}}-{\frac  {4}{3a}}(ax+a^{2})^{{{\frac  {3}{2}}}}]_{{0}}^{{b-a}}\,

=[-{\frac  {4}{3b}}(b^{2}-b^{2}+ab)^{{{\frac  {3}{2}}}}-{\frac  {4}{3a}}(ab)^{{{\frac  {3}{2}}}}]-[-{\frac  {4}{3b}}(b^{3})-{\frac  {4}{3a}}(a^{3})]\,

=[-{\frac  {4(ab)^{{{\frac  {3}{2}}}}}{3b}}-{\frac  {4(ab)^{{{\frac  {3}{2}}}}}{3a}}]+{\frac  {4b^{2}}{3}}+{\frac  {4a^{2}}{3}}\,

={\frac  {8(a+b){\sqrt  {ab}}}{3}}\, on simplifying the above.

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