(a) \(\displaystyle\int {\sqrt {1 - {x^2}} } \,{\rm{d}}x\)\(x = \sin {\rm{\theta }}\;\;\; \Rightarrow \;\;\;{\rm{d}}x = \cos {\rm{\theta }}\,{\rm{d\theta }}\)substituting: \(\displaystyle\int {\sqrt {1 - {x^2}} } \,{\rm{d}}x = \int {\sqrt {1 - {{\sin }^2}{\rm{\theta }}} } \cos {\rm{\theta }}\,{\rm{d\theta }} = \int {\cos {\rm{\theta }}\cos {\rm{\theta }}} \,{\rm{d\theta }} = \int {{{\cos }^2}{\rm{\theta }}} \,{\rm{d\theta }}\)applying trig identity: \(\displaystyle\cos 2{\rm{\theta }} = 2{\cos ^2}{\rm{\theta }} - 1\;\;\; \Rightarrow \;\;\;{\cos ^2}{\rm{\theta }} = \frac{1}{2} + \frac{1}{2}\cos 2{\rm{\theta }}\)substituting: \(\displaystyle\int {{{\cos }^2}{\rm{\theta }}} \,{\rm{d\theta }} = \int {\left( {\frac{1}{2} + \frac{1}{2}\cos 2{\rm{\theta }}} \right)} \,{\rm{d\theta }} = \frac{1}{2}\int {{\rm{d\theta }}} + \frac{1}{2}\int {\cos 2{\rm{\theta }}} \,{\rm{d\theta }}\)\( =\displaystyle \frac{1}{2}\int {{\rm{d\theta }}} + \frac{1}{2}\int {\cos 2{\rm{\theta }}} \,{\rm{d\theta }} = \frac{1}{2}{\rm{\theta }} + \frac{1}{2}\left( {\frac{1}{2}\sin 2{\rm{\theta }}} \right) = \frac{1}{2}{\rm{\theta }} + \frac{1}{4}\sin 2{\rm{\theta }} = \frac{1}{2}{\rm{\theta }} + \frac{1}{4}\left( {2\sin {\rm{\theta }}\cos {\rm{\theta }}} \right)\)from the right triangle above: \({\rm{\theta }} = \arcsin x\), \(\sin {\rm{\theta }}=x\) and \(\cos {\rm{\theta }} = \sqrt {1 - {x^2}} \)\(\displaystyle\int {\sqrt {1 - {x^2}} } \,{\rm{d}}x = \frac{1}{2}{\rm{\theta }} + \frac{1}{2}\sin {\rm{\theta }}\cos {\rm{\theta }}\)substituting: \(\displaystyle\int {\sqrt {1 - {x^2}} } \,{\rm{d}}x = \frac{1}{2}\arcsin x + \frac{1}{2}x\sqrt {1 - {x^2}} \)Thus, \(\displaystyle\int {\sqrt {1 - {x^2}} } \,{\rm{d}}x = \frac{{\arcsin x + x\sqrt {1 - {x^2}} }}{2}\) QED(b) The result from (a) can be used to find the volume of the donut (torus) by setting up a definite integral that sums washers created by rotating the right half of the circle about the x-axis. So, we initially...