If the area bounded by the curve 2y^2 = 3x,th | vedclass

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(A) The region is bounded by the parabola $x = \frac{2y^2}{3}$,the line $x = 3-y$,and the $x$-axis $(y=0)$.First,find the intersection of the parabola

Vedclass · Accepted Answer

$42$

Vedclass · Answer

(A) The region is bounded by the parabola $x = \frac{2y^2}{3}$,the line $x = 3-y$,and the $x$-axis $(y=0)$.First,find the intersection of the parabola and the line:$2y^2 = 3(3-y) \implies 2y^2 + 3y - 9 = 0$$(2y-3)(y+3) = 0$. Since $y \ge 0$,we have $y = \frac{3}{2}$.The area of the region bounded by the parabola,the line,and the $x$-axis is:$Area_{total} = \int_0^{3/2} ((3-y) - \frac{2y^2}{3}) dy = [3y - \frac{y^2}{2} - \frac{2y^3}{9}]_0^{3/2} = (3(\frac{3}{2}) - \frac{9}{8} - \frac{2}{9} \cdot \frac{27}{8}) = \frac{9}{2} - \frac{9}{8} - \frac{3}{4} = \frac{36-9-6}{8} = \frac{21}{8}$.The circle $(x-3)^2 + y^2 = 2$ has center $(3,0)$ and radius $\sqrt{2}$. The line $x+y=3$ passes through $(3,0)$ with slope $-1$,making an angle of $135^\circ$ with the positive $x$-axis. The portion of the circle inside the region is a sector of the circle.The area of the sector of the circle inside the region is $\frac{1}{8} \pi r^2 = \frac{1}{8} \pi (2) = \frac{\pi}{4}$.Thus,$A = \frac{21}{8} - \frac{\pi}{4}$.We need to find $4(\pi + 4A) = 4(\pi + 4(\frac{21}{8} - \frac{\pi}{4})) = 4(\pi + \frac{21}{2} - \pi) = 4(\frac{21}{2}) = 42$.

If the area bounded by the curve $2y^2 = 3x$,the lines $x+y=3$,$y=0$ and outside the circle $(x-3)^2 + y^2 = 2$ is $A$,then $4(\pi + 4A)$ is equal to $.........$.

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