The area (in sq units) bounded by the curves | vedclass

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(B) Given curves are $y^2=4x$ and $x^2=4y$.To find the intersection points, substitute $y = \frac{x^2}{4}$ into $y^2=4x$:$\left(\frac{x^2}{4}\right)^2

Vedclass · Accepted Answer

$\frac{16}{3}$

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(B) Given curves are $y^2=4x$ and $x^2=4y$.To find the intersection points, substitute $y = \frac{x^2}{4}$ into $y^2=4x$:$\left(\frac{x^2}{4}ight)^2 = 4x$$\frac{x^4}{16} = 4x$$x^4 = 64x$$x(x^3 - 64) = 0$Thus, $x = 0$ or $x = 4$.When $x = 0, y = 0$. When $x = 4, y = 4$.The intersection points are $O(0,0)$ and $A(4,4)$.The required area is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=4$:$	ext{Area} = \int_{0}^{4} \left( \sqrt{4x} - \frac{x^2}{4} ight) dx$$	ext{Area} = \int_{0}^{4} \left( 2\sqrt{x} - \frac{x^2}{4} ight) dx$$	ext{Area} = \left[ 2 \cdot \frac{x^{3/2}}{3/2} - \frac{x^3}{12} ight]_{0}^{4}$$	ext{Area} = \left[ \frac{4}{3}x^{3/2} - \frac{x^3}{12} ight]_{0}^{4}$$	ext{Area} = \left( \frac{4}{3}(4)^{3/2} - \frac{4^3}{12} ight) - (0)$$	ext{Area} = \left( \frac{4}{3} \cdot 8 - \frac{64}{12} ight)$$	ext{Area} = \frac{32}{3} - \frac{16}{3} = \frac{16}{3} 	ext{ sq units.}$

The area (in sq units) bounded by the curves $y^2=4x$ and $x^2=4y$ is

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