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 $y^2=4x$: $(\frac{x^2}{4})^2 = 4x \Rig

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$\frac{16}{3}$

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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 $y^2=4x$: $(\frac{x^2}{4})^2 = 4x \Rightarrow \frac{x^4}{16} = 4x \Rightarrow x^4 = 64x \Rightarrow x(x^3 - 64) = 0$. Thus,$x=0$ or $x=4$. When $x=0, y=0$ and when $x=4, y=4$. So,the intersection points are $O(0,0)$ and $A(4,4)$. The required area is the area between the two curves from $x=0$ to $x=4$: $	ext{Area} = \int_0^4 (y_{	ext{upper}} - y_{	ext{lower}}) dx = \int_0^4 (\sqrt{4x} - \frac{x^2}{4}) dx$. $= \int_0^4 (2\sqrt{x} - \frac{x^2}{4}) dx = [2 \cdot \frac{x^{3/2}}{3/2} - \frac{x^3}{12}]_0^4$. $= [\frac{4}{3}x^{3/2} - \frac{x^3}{12}]_0^4 = [\frac{4}{3}(4)^{3/2} - \frac{4^3}{12}] - [0]$. $= [\frac{4}{3}(8) - \frac{64}{12}] = \frac{32}{3} - \frac{16}{3} = \frac{16}{3} 	ext{ sq units}$.

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

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