Area enclosed by curves.

Question

Find the area enclosed by the curves:

$x=4-y^2$ and $x=y-2$

$-3 \le y \le 2$

$y-2 \le x \le 4-y^2$

My multiple integral:

$\int_{-3}^{3}( \int_{y-2}^{4-y^2}dx)dy$

Is this right?

Answer 1

$x=4-y^2$ and $x=y-2$

$-3 \le y \le 2$

$y-2 \le x \le 4-y^2$

Yes, that is correct! A simple plot is always useful.

Careful though:

My multiple integral:

$\int_{-3}^{\color{red}{3}}( \int_{y-2}^{4-y^2}dx)dy$

The upper limit for $y$ should be $2$ instead of $3$, as you probably calculated correctly above.

Answer 2

$$\int_{-3}^{2}( \int_{y-2}^{4-y^2}dx)dy$$

$$\int_{-3}^{2}6-\int_{-3}^{2}y^2-\int_{-3}^{2} y =30-{35\over3}+{5\over2}={180-70+15\over6}={125\over6}$$