The area of the region A = (x,y) : 0 y 6-x, y^2 4x-3, x 0 is:

Question

The area of the region $A = \{(x,y) : 0 \le y \le 6-x,\ y^2 \ge 4x-3,\ x \ge 0\}$ is:

Accepted Answer

$y \le 6-x$ (line), $x \le \dfrac{y^2+3}{4}$ (parabola). These meet where $\dfrac{y^2+3}{4} = 6-y \Rightarrow y^2+4y-21=0 \Rightarrow y=3$ (taking $y\ge0$).

For $0\le y\le3$: right boundary is $x=\dfrac{y^2+3}{4}$. For $3\le y\le6$: right boundary is $x=6-y$.

$$A = \int_0^3 \frac{y^2+3}{4}\,dy + \int_3^6(6-y)\,dy = \frac{1}{4}\left[\frac{y^3}{3}+3y\right]_0^3 + \left[6y-\frac{y^2}{2}\right]_3^6$$

$$= \frac{1}{4}(9+9) + \left(18-\frac{27}{2}\right) = \frac{9}{2}+\frac{9}{2} = 9$$

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