If the area bounded by the curve 2 y ^ 2 = 3x , lines x + y = 3, y...

Given:

$2{y}^{2} = 3x, x + y = 3\left(slopeis - 1\right) \text{ and } {\left(x - 3\right)}^{2} + {y}^{2} = 2$

Now plotting the diagram of the above function we have,

So, required area is

$A = {∫}_{0}^{\dfrac{3}{2}}\left(3 - y - \dfrac{2{y}^{2}}{3}\right)dy - \dfrac{\pi{\left(\sqrt{2}\right)}^{2} \times {45}^{°}}{{360}^{°}}$

$⇒ A = {\left[3y - \dfrac{{y}^{2}}{2}\right]}_{0}^{\dfrac{3}{2}} - \dfrac{2}{3}{\left[\dfrac{{y}^{3}}{3}\right]}_{0}^{\dfrac{3}{2}} - \dfrac{\pi}{4}$

$⇒ A = \dfrac{21}{8} - \dfrac{\pi}{4}$

So,

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

$⇒ \pi + 4A = \dfrac{21}{2}$

$⇒ 4\left(\pi + 4A\right) = 42$