$y^2 + 2y = x$

$y^2 + 2y + 1 = x + 1$

$(y + 1)^2 = x + 1$ → rightward parabola, vertex at (-1, -1), LR = 1

when

$x = 3$

$(y + 1)^2 = 3 + 1$

$y + 1 = \pm 2$

$y = 1 \, \text{ and } \, -3$

Using Horizontal Strip

$\displaystyle A = {\int_{y_1}}^{y_2} (x_R - x_L) \, dy$

$\displaystyle A = {\int_{-3}}^{\,1} [ \, 3 - (y^2 + 2y) \, ] \, dy$

$A = \left[ 3y - \frac{1}{3}y^3 - y^2 \right]_{-3}^1$

$A = [ \, 3(1) - \frac{1}{3}(1^3) - 1^2 \, ] - [ \, 3(-3) - \frac{1}{3}(-3)^3 - (-3)^2 \, ]$

$A = \frac{5}{3} - (-9)$

$A = \frac{32}{3} \, \text{unit}^2$ *answer*