The area of the region common to the parabola | vedclass

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(B) Step $1$: Rewrite the equations as $y^2 = \frac{9}{4}x$ and $x^2 = \frac{16}{3}y$, which implies $y = \frac{3}{2}\sqrt{x}$ and $y = \frac{3}{16}x^

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$4$ sq. units

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(B) Step $1$: Rewrite the equations as $y^2 = \frac{9}{4}x$ and $x^2 = \frac{16}{3}y$, which implies $y = \frac{3}{2}\sqrt{x}$ and $y = \frac{3}{16}x^2$.Step $2$: Find the intersection points by substituting $y = \frac{3}{16}x^2$ into $4y^2 = 9x$: $4(\frac{3}{16}x^2)^2 = 9x \Rightarrow 4(\frac{9}{256}x^4) = 9x \Rightarrow \frac{9}{64}x^4 = 9x \Rightarrow x^4 = 64x$.Step $3$: Solving $x(x^3 - 64) = 0$ gives $x = 0$ and $x = 4$. The corresponding $y$ values are $0$ and $3$.Step $4$: The area $A$ is given by $\int_{0}^{4} (\frac{3}{2}\sqrt{x} - \frac{3}{16}x^2) dx$.Step $5$: $A = [\frac{3}{2} \cdot \frac{2}{3}x^{3/2} - \frac{3}{16} \cdot \frac{x^3}{3}]_{0}^{4} = [x^{3/2} - \frac{x^3}{16}]_{0}^{4} = (4^{3/2} - \frac{4^3}{16}) = (8 - 4) = 4$ sq. units.

The area of the region common to the parabolas $4y^2 = 9x$ and $3x^2 = 16y$ is...

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