Rewrite the given curves

The parabola is $2y^2=3x \implies x=\frac{2}{3}y^2.$

The line is $x+y=3 \implies x=3-y.$

Also, $y=0$ is the $x$-axis.

The circle is $(x-3)^2+y^2=2,$ centered at $(3,0)$ with radius $\sqrt{2}$.

Find the region bounded by parabola, line, and $y=0$

The parabola and line intersect where $\frac{2}{3}y^2=3-y.$ Multiply by $3$: $2y^2=9-3y$ $2y^2+3y-9=0.$ Solving, $y=\frac{-3\pm 9}{4}.$ So, $y=\frac{3}{2},\,-3.$ Since the region is with $y=0$, we use $0\le y\le \frac{3}{2}$.

For this strip,

• left boundary: $x=\frac{2}{3}y^2$
• right boundary: $x=3-y$

Hence total bounded area is $\int_0^{3/2}\left[(3-y)-\frac{2}{3}y^2\right]dy.$

Understand the part outside the circle

We need the area inside the above bounded region but outside the circle.

First check where the line and parabola meet the circle.

Intersection of line and circle

Put $x=3-y$ into circle: $((3-y)-3)^2+y^2=2$ $y^2+y^2=2$ $2y^2=2 \implies y=\pm 1.$ In our region, $y=1$. So point is $(x,y)=(2,1).$

Intersection of parabola and circle

Put $x=\frac{2}{3}y^2$ into circle: $\left(\frac{2}{3}y^2-3\right)^2+y^2=2.$ Checking $y=1$, $x=\frac{2}{3},$ and $\left(\frac{2}{3}-3\right)^2+1=\left(-\frac{7}{3}\right)^2+1=\frac{49}{9}+1>2,$ so no relevant upper intersection there.

More importantly, in the bounded region, the circle cuts off the right-hand portion near $(3,0)$, and this happens for $0\le y\le 1$.

For the upper semicircle, $y=\sqrt{2-(x-3)^2} \iff x=3\pm\sqrt{2-y^2}.$ Since the bounded region lies to the left of $x=3$, relevant circle boundary is $x=3-\sqrt{2-y^2}.$

For a fixed $y\in[0,1]$:

• region inside the original bounded figure runs from $x=\frac{2}{3}y^2$ to $x=3-y$
• inside the circle corresponds to $x\ge 3-\sqrt{2-y^2}$ up to $x=3-y$

Therefore, the part outside the circle is:

• for $0\le y\le 1$: from $x=\frac{2}{3}y^2$ to $x=3-\sqrt{2-y^2}$
• for $1\le y\le \frac{3}{2}$: whole strip from $x=\frac{2}{3}y^2$ to $x=3-y$

So, $A=\int_0^1\left[3-\sqrt{2-y^2}-\frac{2}{3}y^2\right]dy+\int_1^{3/2}\left[3-y-\frac{2}{3}y^2\right]dy.$

Evaluate the second integral

$I_2=\int_1^{3/2}\left(3-y-\frac{2}{3}y^2\right)dy.$

Antiderivative: $3y-\frac{y^2}{2}-\frac{2}{9}y^3.$

Hence $I_2=\left[3y-\frac{y^2}{2}-\frac{2}{9}y^3\right]_1^{3/2}.$

At $y=\frac{3}{2}$:

Evaluate the first integral

$I_1=\int_0^1\left(3-\frac{2}{3}y^2\right)dy-\int_0^1\sqrt{2-y^2}\,dy.$

First part:

At $y=1$: $\frac{1}{2}\cdot 1+\sin^{-1}\left(\frac{1}{\sqrt2}\right)=\frac12+\frac\pi4.$

At $y=0$: $0$.

Hence $\int_0^1\sqrt{2-y^2}\,dy=\frac12+\frac\pi4.$

Therefore, $I_1=\frac{25}{9}-\left(\frac12+\frac\pi4\right)=\frac{41}{18}-\frac\pi4.$

Compute $A$

$A=I_1+I_2=\left(\frac{41}{18}-\frac\pi4\right)+\frac{25}{72}.$

$\frac{41}{18}=\frac{164}{72},$ so

Find $4(\pi+4A)$

First, $4A=4\left(\frac{21}{8}-\frac\pi4\right)=\frac{21}{2}-\pi.$

Then, $\pi+4A=\pi+\frac{21}{2}-\pi=\frac{21}{2}.$

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

Compare with stored answer

Derived answer is $42$, which matches the stored correct answer.