Finding the area bounded by two curves when in terms of $x = y^2$?

Question

I can't seem to figure this problem out.

Find the area bounded by the curves $x=2y-y^2$ and $x=4-y^2$, in the first quadrant.

I am having difficulties with graphing the equations and coming up with the correct integral. Any help would be appreciated.

Answer 1

First, let's have a look to the graph of the parabolae:

Note that the $y$-axis corresponds to the abscissa whilst the $x$-axis correspond to the ordinates. You can easily plot these two functions by considering $x = x(y)$, as well as the intersection point $(y,x) = (2,0)$. Does this change the value of the enclosed area? Absolutely not.

Hence, the desired area is given by:

$$\color{blue}{A = \int^2_0 (x_1(y)-x_2(y)) \, \mathrm{d}y} \tag{1}$$

where $x_1(y) = 4-y^2$ and $x_2(y) = 2 y - y^2$.

A slightly more elegant way to do this is to parmetrize the enclosed region in the 1st quadrant as follows:

$$ \mathcal{R} = \{(x,y) \in \mathbb{R}^2/ 0 \leq y \leq 1, \ \ x_2(y) \leq x < x_1(y) \},$$ so:

$$ \color{blue}{A = \int_\mathcal{R} 1 \, \mathbf{d\mathcal{A}} = \int^1_0 \int^{x_1(y)}_{x_2(y)} \mathrm{d}x \, \mathrm{d}y} \tag{2}. $$

Of course, eq. (2) leads to eq. (1).

Cheers!