The area of the region enclosed by the curves y = x^2-4x+4 and y^2 = 16-8x is:

Question

The area of the region enclosed by the curves $y = x^2-4x+4$ and $y^2 = 16-8x$ is:

Accepted Answer

$y=(x-2)^2$ and $y^2=16-8x$. Substituting $u=x-2$, $y=u^2$: $u^4 = 16-8(u+2) = -8u \Rightarrow u(u^3+8)=0 \Rightarrow u=0$ or $u=-2$.

So the curves meet at $(2,0)$ and $(0,4)$. Expressing $x$ in terms of $y$: from the parabola $x = 2-\dfrac{y^2}{8}$, and from the other curve (left branch, since $x<2$ near $y=4$) $x = 2-\sqrt y$.

$$\Delta x = \left(2-\frac{y^2}{8}\right)-(2-\sqrt y) = \sqrt y - \frac{y^2}{8}$$

$$A = \int_0^4\left(\sqrt y - \frac{y^2}{8}\right)dy = \left[\frac23 y^{3/2}\right]_0^4 - \frac18\left[\frac{y^3}{3}\right]_0^4 = \frac{16}{3}-\frac{8}{3} = \frac{8}{3}$$

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