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

Given:

$2y^2=3x, x+y=3\left(\mathrm{slope} \mathrm{is} -1\right) \& {\left(x-3\right)}^2+y^2=2$

Now plotting the diagram of the above function we have,

So, required area is

$A={\int }_0^{\frac{3}{2}}\left(3-y-\frac{2y^2}{3}\right)dy-\frac{\mathrm{\pi }{\left(\sqrt{2}\right)}^2\times {45}^{°}}{{360}^{°}}$

$\Rightarrow A={\left[3y-\frac{y^2}{2}\right]}_0^{\frac{3}{2}}-\frac{2}{3}{\left[\frac{y^3}{3}\right]}_0^{\frac{3}{2}}-\frac{\mathrm{\pi }}{4}$

$\Rightarrow A=\frac{21}{8}-\frac{\mathrm{\pi }}{4}$

So,

$A=\frac{1}{4}\left(\frac{21}{2}-\mathrm{\pi }\right)$

$\Rightarrow \mathrm{\pi }+4A=\frac{21}{2}$

$\Rightarrow 4\left(\mathrm{\pi }+4A\right)=42$