Answer
$$2\pi $$
Work Step by Step
$$\eqalign{
& \int_0^4 {\sqrt {x\left( {4 - x} \right)} } dx \cr
& {\text{completing the square for }}4x - {x^2} \cr
& 4x - {x^2} = - \left( {{x^2} - 4x + 4} \right) + 4 \cr
& 4x - {x^2} = 4 - {\left( {x - 2} \right)^2} \cr
& \int_0^4 {\sqrt {x\left( {4 - x} \right)} } dx = \int_0^4 {\sqrt {4 - {{\left( {x - 2} \right)}^2}} } dx \cr
& \cr
& {\text{Write in terms of }}\theta \cr
& {\text{substitute }}x - 2 = 2\sin \theta ,{\text{ }}\,\,\,dx = 2\cos \theta d\theta \cr
& \,\,\,\,\theta = {\sin ^{ - 1}}\left( {\frac{{x - 2}}{2}} \right) \cr
& \,\,\,\,x = 4 \to \theta = \pi /2 \cr
& \,\,\,\,x = 0 \to \theta = - \pi /2 \cr
& \int_0^4 {\sqrt {4 - {{\left( {x - 2} \right)}^2}} } dx = \int_{ - \pi /2}^{\pi /2} {\sqrt {4 - 4{{\sin }^2}\theta } \left( {2\cos \theta } \right)d\theta } \cr
& = \int_{ - \pi /2}^{\pi /2} {2\sqrt {1 - {{\sin }^2}\theta } \left( {2\cos \theta } \right)d\theta } \cr
& = 4\int_{ - \pi /2}^{\pi /2} {\sqrt {{{\cos }^2}\theta } \cos \theta d\theta } \cr
& = 4\int_{ - \pi /2}^{\pi /2} {{{\cos }^2}\theta d\theta } \cr
& = 2\int_{ - \pi /2}^{\pi /2} {\left( {1 + \cos 2\theta } \right)} d\theta \cr
& \cr
& {\text{Integrate}} \cr
& = 2\left( {\theta + \frac{1}{2}\sin 2\theta } \right)_{ - \pi /2}^{\pi /2} \cr
& {\text{Evaluate the limits}} \cr
& = 2\left( {\frac{\pi }{2} + \frac{1}{2}\sin 2\left( {\frac{\pi }{2}} \right)} \right) - 2\left( { - \frac{\pi }{2} + \frac{1}{2}\sin 2\left( { - \frac{\pi }{2}} \right)} \right) \cr
& = 2\left( {\frac{\pi }{2} + 0} \right) - 2\left( { - \frac{\pi }{2} + 0} \right) \cr
& = \pi + \pi \cr
& = 2\pi \cr} $$
