Answer
$$\frac{{25}}{6}\pi $$
Work Step by Step
$$\eqalign{
& \int_{ - 5/3}^{5/3} {\sqrt {25 - 9{x^2}} dx} ;{\text{ }} \cr
& {\text{Let }}u = 3x,{\text{ }}du = 3dx,{\text{ }}dx = \frac{1}{3}du \cr
& {\text{The new limits of integration are:}} \cr
& x = 5/3{\text{ }} \Rightarrow u = 3\left( {\frac{5}{3}} \right) = 5 \cr
& x = - 5/3{\text{ }} \Rightarrow u = 3\left( { - \frac{5}{3}} \right) = - 5 \cr
& {\text{Apply the substitution}} \cr
& \int_{ - 5/3}^{5/3} {\sqrt {25 - 9{x^2}} dx} = \int_{ - 5}^5 {\sqrt {25 - {u^2}} \left( {\frac{1}{3}} \right)} du \cr
& {\text{ }} = \frac{1}{3}\int_{ - 5}^5 {\sqrt {{{\left( 5 \right)}^2} - {u^2}} } du \cr
& {\text{The integral of the form }}\int_{ - a}^a {\sqrt {{a^2} - {u^2}} du} ,{\text{ represents a semicircle}} \cr
& {\text{centered at the origin with radius }}a,{\text{ then the geometry formula}} \cr
& {\text{for the area is }}\int_{ - a}^a {\sqrt {{a^2} - {u^2}} du} = \frac{1}{2}\pi {a^2},{\text{ therefore}} \cr
& {\text{ }} = \frac{1}{3}\left( {\frac{1}{2}\pi {{\left( 5 \right)}^2}} \right) \cr
& {\text{ }} = \frac{{25}}{6}\pi \cr} $$
