Consider a region R = (x,y) R^2 : x^2 y2x.

Question

Consider a region $R = \{(x,y)\in\mathbb R^2 : x^2\le y\le2x\}$. If a line $y=\alpha$ divides the area of region $R$ into two equal parts, then which of the following is true?

Accepted Answer

$x^2=2x \Rightarrow x=0,2$; total area $=\displaystyle\int_0^2(2x-x^2)\,dx = \dfrac43$.

Expressed with $x$ as function of $y$ (right boundary $x=\sqrt y$, left boundary $x=y/2$), the bisection condition is:

$$\int_0^\alpha\left(\sqrt y-\frac y2\right)dy = \int_\alpha^4\left(\sqrt y-\frac y2\right)dy$$

Evaluating both sides using $\displaystyle\int\sqrt y\,dy = \tfrac23y^{3/2}$ and $\displaystyle\int\tfrac y2\,dy=\tfrac{y^2}4$, and simplifying, gives:

$$8\alpha^{3/2}-3\alpha^2 = 8 \implies 3\alpha^2-8\alpha^{3/2}+8=0$$

Solutions · 1