If the area bounded by y = ax^2 and x = ay^2, | vedclass

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(B) The curves are $y = ax^2$ and $x = ay^2$ (or $y = \sqrt{x/a}$ for the upper branch).To find the intersection points,substitute $y = ax^2$ into $x

Vedclass · Accepted Answer

$\frac{1}{\sqrt{3}}$

Vedclass · Answer

(B) The curves are $y = ax^2$ and $x = ay^2$ (or $y = \sqrt{x/a}$ for the upper branch).To find the intersection points,substitute $y = ax^2$ into $x = ay^2$:$x = a(ax^2)^2 = a^3x^4$$x(a^3x^3 - 1) = 0$So,$x = 0$ or $x^3 = 1/a^3$,which gives $x = 1/a$.The intersection point $A$ is $(1/a, 1/a)$.The area $A$ is given by:$A = \int_0^{1/a} (\sqrt{x/a} - ax^2) dx = 1$$\int_0^{1/a} (\frac{1}{\sqrt{a}} x^{1/2} - ax^2) dx = 1$$[\frac{1}{\sqrt{a}} \cdot \frac{2}{3} x^{3/2} - \frac{a}{3} x^3]_0^{1/a} = 1$$\frac{1}{\sqrt{a}} \cdot \frac{2}{3} (\frac{1}{a})^{3/2} - \frac{a}{3} (\frac{1}{a})^3 = 1$$\frac{2}{3a^2} - \frac{1}{3a^2} = 1$$\frac{1}{3a^2} = 1$$a^2 = 1/3$Since $a > 0$,$a = \frac{1}{\sqrt{3}}$.

If the area bounded by $y = ax^2$ and $x = ay^2$,$a > 0$,is $1$,then $a = $

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