Answer
$I = \mathop \smallint \limits_{x = - 1}^1 \mathop \smallint \limits_{y = 0}^{\sqrt {1 - {x^2}} } \mathop \smallint \limits_{z = 0}^1 \left( {x + y + z} \right){\rm{d}}z{\rm{d}}y{\rm{d}}x = \frac{2}{3} + \frac{\pi }{4}$
Work Step by Step
Evaluate:
$I = \mathop \smallint \limits_{x = - 1}^1 \mathop \smallint \limits_{y = 0}^{\sqrt {1 - {x^2}} } \mathop \smallint \limits_{z = 0}^1 \left( {x + y + z} \right){\rm{d}}z{\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{x = - 1}^1 \mathop \smallint \limits_{y = 0}^{\sqrt {1 - {x^2}} } \left( {\left( {xz + yz + \frac{1}{2}{z^2}} \right)|_0^1} \right){\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{x = - 1}^1 \mathop \smallint \limits_{y = 0}^{\sqrt {1 - {x^2}} } \left( {x + y + \frac{1}{2}} \right){\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{x = - 1}^1 \left( {\left( {xy + \frac{1}{2}{y^2} + \frac{1}{2}y} \right)|_0^{\sqrt {1 - {x^2}} }} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = - 1}^1 \left( {x\sqrt {1 - {x^2}} + \frac{1}{2} - \frac{1}{2}{x^2} + \frac{1}{2}\sqrt {1 - {x^2}} } \right){\rm{d}}x$
(1) ${\ \ \ }$ $I = \mathop \smallint \limits_{x = - 1}^1 x\sqrt {1 - {x^2}} {\rm{d}}x + \frac{1}{2}\mathop \smallint \limits_{x = - 1}^1 {\rm{d}}x - \frac{1}{2}\mathop \smallint \limits_{x = - 1}^1 {x^2}{\rm{d}}x + \frac{1}{2}\mathop \smallint \limits_{x = - 1}^1 \sqrt {1 - {x^2}} {\rm{d}}x$
1. The antiderivative of $x\sqrt {1 - {x^2}} $ is $ - \frac{1}{3}{\left( {1 - {x^2}} \right)^{3/2}}$ because
$\frac{d}{{dx}}\left( { - \frac{1}{3}{{\left( {1 - {x^2}} \right)}^{3/2}}} \right) = x\sqrt {1 - {x^2}} $
2. The antiderivative of $\sqrt {1 - {x^2}} $ can be found by changing of variable: $x = \sin \theta $.
So, $dx = \cos \theta d\theta $ and $\sqrt {1 - {x^2}} = \cos \theta $. Thus,
$\mathop \smallint \limits_{x = - 1}^1 \sqrt {1 - {x^2}} {\rm{d}}x = \mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} {\cos ^2}{\rm{d}}\theta $
Using the double-angle formulas in Section 1.4:
${\cos ^2}\theta = \frac{1}{2}\left( {1 + \cos 2\theta } \right)$
we get
$\mathop \smallint \limits_{x = - 1}^1 \sqrt {1 - {x^2}} {\rm{d}}x = \mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} {\cos ^2}\theta {\rm{d}}\theta $
$ = \frac{1}{2}\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} \left( {1 + \cos 2\theta } \right){\rm{d}}\theta $
$ = \frac{1}{2}\left( {\theta + \frac{1}{2}\sin 2\theta } \right)|_{ - \pi /2}^{\pi /2} = \frac{1}{2}\left( {\frac{\pi }{2} + \frac{\pi }{2}} \right) = \frac{\pi }{2}$
Substituting these results back in equation (1):
$I = \mathop \smallint \limits_{x = - 1}^1 x\sqrt {1 - {x^2}} {\rm{d}}x + \frac{1}{2}\mathop \smallint \limits_{x = - 1}^1 {\rm{d}}x - \frac{1}{2}\mathop \smallint \limits_{x = - 1}^1 {x^2}{\rm{d}}x + \frac{1}{2}\mathop \smallint \limits_{x = - 1}^1 \sqrt {1 - {x^2}} {\rm{d}}x$
gives
$I = - \frac{1}{3}\left( {{{\left( {1 - {x^2}} \right)}^{3/2}}|_{ - 1}^1} \right) + 1 - \frac{1}{2}\left( {\frac{1}{3}{x^3}|_{ - 1}^1} \right) + \frac{\pi }{4}$
$ = 1 - \frac{1}{2}\left( {\frac{2}{3}} \right) + \frac{\pi }{4} = \frac{2}{3} + \frac{\pi }{4}$
So, $I = \mathop \smallint \limits_{x = - 1}^1 \mathop \smallint \limits_{y = 0}^{\sqrt {1 - {x^2}} } \mathop \smallint \limits_{z = 0}^1 \left( {x + y + z} \right){\rm{d}}z{\rm{d}}y{\rm{d}}x = \frac{2}{3} + \frac{\pi }{4}$.
