The area included between the two curves y^2 | vedclass

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(D) To find the area between the curves $y^2 = 4ax$ and $x^2 = 4ay$,we first find their points of intersection.From $x^2 = 4ay$,we have $y = \frac{x^2

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$\frac{16}{3} a^2 	ext{ sq. unit}$

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(D) To find the area between the curves $y^2 = 4ax$ and $x^2 = 4ay$,we first find their points of intersection.From $x^2 = 4ay$,we have $y = \frac{x^2}{4a}$.Substituting this into $y^2 = 4ax$,we get $(\frac{x^2}{4a})^2 = 4ax$,which simplifies to $\frac{x^4}{16a^2} = 4ax$.Thus,$x^4 = 64a^3x$,which gives $x(x^3 - 64a^3) = 0$.So,$x = 0$ or $x = 4a$.The required area is the integral of the difference between the upper curve and the lower curve from $x = 0$ to $x = 4a$.The upper curve is $y = \sqrt{4ax} = 2\sqrt{a}\sqrt{x}$ and the lower curve is $y = \frac{x^2}{4a}$.Area $= \int_0^{4a} (2\sqrt{a}\sqrt{x} - \frac{x^2}{4a}) dx$$= [2\sqrt{a} \cdot \frac{x^{3/2}}{3/2} - \frac{x^3}{12a}]_0^{4a}$$= [\frac{4}{3}\sqrt{a} \cdot (4a)^{3/2} - \frac{(4a)^3}{12a}]$$= [\frac{4}{3}\sqrt{a} \cdot 8a\sqrt{a} - \frac{64a^3}{12a}]$$= \frac{32}{3}a^2 - \frac{16}{3}a^2 = \frac{16}{3}a^2 	ext{ sq. unit}$.

The area included between the two curves $y^2 = 4ax$ and $x^2 = 4ay$ is

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