The area (in sq. units) of the region described by $A=\{(x,y): \frac{y^2}{2} \leq x \leq y+4\}$ is $18$.
The area (in sq. units) of the region described by$A=\{(x,y): y^2 \leq 2x \; \textrm{and} \; y \geq 4x-1\}$ is $\frac{9}{32}$.
Similarly I solved a few questions of such type. I observed that bounded area between $y^2=4ax$ and $y=mx+c$ is coming out to be $72\displaystyle\bigg|\frac{a^2}{m^3}\bigg|$. While this formula is working in many cases, it is also failing in some cases. Could someone help me in finding conditions for which this formula can be applied if at all it can be applied.
Also I found that area bounded between $y^2=4ax$ and $y=mx$ is $\displaystyle \frac{8}{3} \bigg|\frac{a^2}{m^3}\bigg|$ which is working fine.
Note: I found out answers to first two questions by finding their points of intersection and then using integrals and I had derived the formula for $y^2=4ax$ and $y=mx$ which is pretty easy. So by observation I deduced formula $72\displaystyle\bigg|\frac{a^2}{m^3}\bigg|$ for $y^2=4ax$ and $y=mx+c$.
I also thought the logic may be hidden in quadratic equation formed by solving $y^2=4ax$ and $y=mx+c$ and tried to use the fact that discriminant of that quadratic must be perfect square just out of intuition which gave me the condition that $a^2-amc$ must be perfect square but even that isn't working in all cases.
The formula given in my comment above, for the area between parabola $y^2=4ax$ and line $y=mx+c$, can be found, after some tedious calculations, from this integral along $y$: $$ area=\int_{y_1}^{y_2}\left({y-c\over m}-{y^2\over4a}\right)dy, $$ where $y_1$ and $y_2$ (with $y_1\le y_2$) are the ordinates of the intersections between line and parabola, i.e. the solutions of the resolvent equation $$ my^2-4ay+4ac=0. $$ But the same result can also be found in a more geometric way, by using a famous result due to Archimedes of Syracuse:
In the example in figure below: $$ area={4\over3}area_{ABC}={4\over3} \cdot{1\over2}CM\cdot(AH+BK). $$ But: $$ y_M={y_A+y_B\over2}={2a\over m},\quad x_M={y_M-c\over m}={2a-mc\over m^2},\quad x_C={y_M^2\over 4a}={a\over m^2}, $$ hence: $$ CM=|x_M-x_C|={|a-mc|\over m^2} $$ and, from the resolvent equation: $$ AH+BK=|y_B-y_A|={4\over |m|}\sqrt{a(a-mc)}. $$ Inserting those values into the formula given above one finally gets: $$ area={8\over 3|m|^3}\sqrt{a(a-mc)^3}. $$