Find the area bounded by  x = -y^2  and  y = x+2 using a double integral?

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Apr 2, 2017

Bounded Area = $\frac{9}{2}$

Explanation:

Based on the sketch. we are looking for a double integral solution to calculate the area bounded by the curves:

$x = - {y}^{2}$
$y = x + 2 = > x = y - 2$ The points of intersection are the solution of the equation:

$x = - {\left(x + 2\right)}^{2}$
$\therefore x = - \left({x}^{2} + 4 x + 4\right)$
 :. x^2+5x+4 = 0  :. (x+1)(x+4) = 0  :. x=-1, -4#

The corresponding $y$-coordinates are:

$x = - 1 \implies y = 1$
$x = - 4 \implies y = - 2$

Giving the coordinates $\left(- 1 , 1\right)$ and $\left(- 4 , - 2\right)$

If in the above diagram we look at an infinitesimally thin horizontal strip (in black) then the limits for $x$ and $y$ are:

$x$ varies from $y - 2$ to $- {y}^{2}$
$y$ varies from $- 2$ to $1$

And so we can represent the bounded are by the following double integral:

$A = \int {\int}_{R} \mathrm{dA}$
$\setminus \setminus \setminus = {\int}_{-} {2}^{1} \setminus {\int}_{y - 2}^{- {y}^{2}} \setminus \mathrm{dx} \setminus \mathrm{dy}$

We can calculate the inner integral:

${\int}_{y - 2}^{- {y}^{2}} \setminus \mathrm{dx} = {\left[x\right]}_{y - 2}^{- {y}^{2}}$

$\text{ } = \left(- {y}^{2}\right) - \left(y - 2\right)$
$\text{ } = - {y}^{2} - y + 2$
$\text{ } = - \left({y}^{2} + y - 2\right)$

And so:

$A = {\int}_{-} {2}^{1} \setminus {\int}_{y - 2}^{- {y}^{2}} \setminus \mathrm{dx} \setminus \mathrm{dy}$
$\setminus \setminus \setminus = {\int}_{-} {2}^{1} - \left({y}^{2} + y - 2\right) \setminus \mathrm{dy}$
$\setminus \setminus \setminus = - {\int}_{-} {2}^{1} {y}^{2} + y - 2 \setminus \mathrm{dy}$
$\setminus \setminus \setminus = - {\left[{y}^{3} / 3 + {y}^{2} / 2 - 2 y\right]}_{-} {2}^{1}$
$\setminus \setminus \setminus = - \left\{\left(\frac{1}{3} + \frac{1}{2} - 2\right) - \left(- \frac{8}{3} + 2 + 4\right)\right\}$
$\setminus \setminus \setminus = - \left\{\left(- \frac{7}{6}\right) - \left(\frac{10}{3}\right)\right\}$
$\setminus \setminus \setminus = \frac{9}{2}$