1. Write both curves in usable form

The parabola is $y^2=2x \quad \Rightarrow \quad x=\frac{y^2}{2}.$

The line is $x+y=4 \quad \Rightarrow \quad x=4-y.$

Since both are conveniently written as $x$ in terms of $y$, we will integrate with respect to $y$.

2. Find the points of intersection

At intersection, $\frac{y^2}{2}=4-y.$

Multiply by $2$: $y^2=8-2y$ $y^2+2y-8=0.$

Factorizing, $(y+4)(y-2)=0.$

So, $y=-4,\;2.$

Now find corresponding $x$ values:

• For $y=-4$, $x=4-(-4)=8.$
• For $y=2$, $x=4-2=2.$

Thus, the curves intersect at $(8,-4) \text{ and } (2,2).$

3. Decide right curve and left curve

For a typical $y$ between $-4$ and $2$:

• Line: $x=4-y$
• Parabola: $x=\dfrac{y^2}{2}$

Take $y=0$:

• Line gives $x=4$
• Parabola gives $x=0$

So the line lies to the right and the parabola lies to the left.

Hence area is $A=\int_{-4}^{2} \left[(4-y)-\frac{y^2}{2}\right]dy.$

4. Evaluate the integral

$A=\int_{-4}^{2} \left(4-y-\frac{y^2}{2}\right)dy.$

Antiderivative: $\int \left(4-y-\frac{y^2}{2}\right)dy = 4y-\frac{y^2}{2}-\frac{y^3}{6}.$

Now apply limits: $A=\left[4y-\frac{y^2}{2}-\frac{y^3}{6}\right]_{-4}^{2}.$

At $y=2$: $4(2)-\frac{2^2}{2}-\frac{2^3}{6}=8-2-\frac{8}{6}=6-\frac{4}{3}=\frac{14}{3}.$

At $y=-4$: $4(-4)-\frac{(-4)^2}{2}-\frac{(-4)^3}{6}=-16-8+\frac{64}{6}=-24+\frac{32}{3}=\frac{-72+32}{3}=\frac{-40}{3}.$

Therefore, $A=\frac{14}{3}-\left(-\frac{40}{3}\right)=\frac{54}{3}=18.$

5. Final Answer

The required area is $\boxed{18}.$