The area of the region {(x, y):|x-y| leq y le | vedclass

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(B) The given region is defined by $|x-y| \leq y \leq 4 \sqrt{x}$.This implies two inequalities: $y \geq |x-y|$ and $y \leq 4 \sqrt{x}$.From $y \geq |

Vedclass · Accepted Answer

$\frac{1024}{3}$

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(B) The given region is defined by $|x-y| \leq y \leq 4 \sqrt{x}$.This implies two inequalities: $y \geq |x-y|$ and $y \leq 4 \sqrt{x}$.From $y \geq |x-y|$,we have $-y \leq x-y \leq y$,which simplifies to $x \geq 0$ and $x \leq 2y$,or $y \geq \frac{x}{2}$.From $y \leq 4 \sqrt{x}$,we have $y^2 \leq 16x$ (for $y \geq 0$).To find the intersection points of $y = \frac{x}{2}$ and $y = 4 \sqrt{x}$,we set $\frac{x}{2} = 4 \sqrt{x} \Rightarrow x = 8 \sqrt{x} \Rightarrow x^2 = 64x \Rightarrow x(x-64) = 0$.Thus,the intersection points are at $x = 0$ and $x = 64$.For $x \in [0, 64]$,the curve $y = 4 \sqrt{x}$ lies above the line $y = \frac{x}{2}$.The area is given by $\int_0^{64} (4 \sqrt{x} - \frac{x}{2}) dx$.$= \left[ 4 \cdot \frac{x^{3/2}}{3/2} - \frac{x^2}{4} \right]_0^{64} = \left[ \frac{8}{3} x^{3/2} - \frac{x^2}{4} \right]_0^{64}$.$= \frac{8}{3} (64)^{3/2} - \frac{64^2}{4} = \frac{8}{3} (512) - \frac{4096}{4} = \frac{4096}{3} - 1024 = \frac{4096 - 3072}{3} = \frac{1024}{3}$.

The area of the region $\{(x, y):|x-y| \leq y \leq 4 \sqrt{x}\}$ is

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