The area of the region {(x, y): xy leq 8, 1 l | vedclass

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(B) The region is bounded by $xy = 8$ (or $x = 8/y$),$y = 1$,and $y = x^2$ (or $x = \sqrt{y}$ for $x > 0$).To find the intersection point of $x = 8/y$

Vedclass · Accepted Answer

$16 \log_e 2 - \frac{14}{3}$

Vedclass · Answer

(B) The region is bounded by $xy = 8$ (or $x = 8/y$),$y = 1$,and $y = x^2$ (or $x = \sqrt{y}$ for $x > 0$).To find the intersection point of $x = 8/y$ and $x = \sqrt{y}$,we set $\frac{8}{y} = \sqrt{y}$,which implies $y^{3/2} = 8$,so $y = 4$.The region is bounded by $y$ from $1$ to $4$,where the right boundary is $x = 8/y$ and the left boundary is $x = \sqrt{y}$.Thus,the required area is $\int_1^4 \left(\frac{8}{y} - \sqrt{y}\right) dy$.$= [8 \ln|y| - \frac{2}{3} y^{3/2}]_1^4$$= (8 \ln 4 - \frac{2}{3} \cdot 4^{3/2}) - (8 \ln 1 - \frac{2}{3} \cdot 1^{3/2})$$= (8 \cdot 2 \ln 2 - \frac{2}{3} \cdot 8) - (0 - \frac{2}{3})$$= 16 \ln 2 - \frac{16}{3} + \frac{2}{3}$$= 16 \ln 2 - \frac{14}{3}$.

The area of the region $\{(x, y): xy \leq 8, 1 \leq y \leq x^2\}$ is

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