The area (in sq.

Question

The area (in sq. units) of the region $\{(x,y) : x\ge0,\ x+y\le3,\ x^2\le4y,\ y\le1+\sqrt x\}$ is:

Accepted Answer

The relevant boundary curves meet as follows: $y=1+\sqrt x$ and $y=3-x$ meet at $x=1$ ($y=2$); $y=3-x$ meets $x^2=4y$ at $x=2$ ($y=1$).

$$\text{Area} = \int_0^1(1+\sqrt x)\,dx + \int_1^2(3-x)\,dx - \int_0^2\frac{x^2}{4}\,dx$$

$$= \left[x+\frac{2}{3}x^{3/2}\right]_0^1 + \left[3x-\frac{x^2}{2}\right]_1^2 - \left[\frac{x^3}{12}\right]_0^2 = \frac52$$

Solutions · 1