Answer
$$2\pi $$
Work Step by Step
$$\eqalign{
& \int_{ - 3}^1 {\sqrt {3 - 2x - {x^2}} } dx;{\text{ }}u = x + 1 \cr
& {\text{Completing the square}} \cr
& 3 - 2x - {x^2} = 4 - {\left( {x + 1} \right)^2} \cr
& = \int_{ - 3}^1 {\sqrt {4 - {{\left( {x + 1} \right)}^2}} } dx \cr
& {\text{Let }}u = x + 1,{\text{ }}du = dx, \cr
& {\text{The new limits of integration are:}} \cr
& x = 1{\text{ }} \Rightarrow u = 1 + 1 = 2 \cr
& x = - 3{\text{ }} \Rightarrow u = - 3 + 1 = - 2 \cr
& {\text{Apply the substitution}} \cr
& \int_{ - 3}^1 {\sqrt {4 - {{\left( {x + 1} \right)}^2}} } dx = \int_{ - 2}^2 {\sqrt {4 - {u^2}} } du \cr
& {\text{The integral of the form }}\int_{ - a}^a {\sqrt {{a^2} - {u^2}} du} ,{\text{ represents a semicircle}} \cr
& {\text{centered at the origin with radius }}a,{\text{ then the geometry formula}} \cr
& {\text{for the area is }}\int_{ - a}^a {\sqrt {{a^2} - {u^2}} du} = \frac{1}{2}\pi {a^2},{\text{ therefore}} \cr
& {\text{ }} = \frac{1}{2}\pi {\left( 2 \right)^2} \cr
& {\text{ }} = 2\pi \cr} $$
