If the area bounded by the curve 2y^2=3x, the lines x+y=3, y=0, and outside the circle (x-…

Question

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 \_\_\_\_\_\_\_\_.

Answer: 42

Accepted Answer

The parabola $y^2=\tfrac32x$ meets $x+y=3$ where $2y^2+3y-9=0\Rightarrow(2y-3)(y+3)=0\Rightarrow y=\tfrac32$ (rejecting $y=-3$).

The circle is centred $(3,0)$ with radius $\sqrt2$, and the line $x+y=3$ passes through its centre. Computing the area between the line ($x=3-y$) and the parabola ($x=\tfrac23y^2$) for $y\in[0,\tfrac32]$, and removing the portion lying inside the circle (a quarter-disc of area $\pi/4$):

$$A=\int_0^{3/2}\left[(3-y)-\frac{2y^2}3\right]dy-\frac\pi4=\frac{21}8-\frac\pi4.$$

$$4A+\pi=\frac{21}2\ \Longrightarrow\ 4(\pi+4A)=4\cdot\frac{21}2=42.$$

Solutions · 1