Change of variables?

The region we're integrating over is the quarter of the unit circle in the first quadrant of the Cartesian coordinate plane:

This integral is fairly difficult to evaluate using Cartesian coordinates, but changing to polar coordinates makes it much easier: $0 \le x \le 1,\quad 0 \le y \le \sqrt{1-x^2}\quad\Rightarrow\quad 0 \le r \le 1,\quad 0 \le \theta \le \frac{\pi}{2}$ $1 - x^2 - y^2\quad\Rightarrow\quad 1 - r^2$ $dy\: dx\quad\Rightarrow\quad r\: dr\: d\theta$

Now we can evaluate: $\int_0^{\frac{\pi}{2}} \int_0^1 (1 - r^2)r\: dr\: d\theta$ $=\int_0^{\frac{\pi}{2}} d\theta \int_0^1 (r - r^3)\: dr$ $=\frac{\pi}{2} \left.\left(\frac{r^2}{2} - \frac{r^4}{4}\right)\right|_0^1$ $=\frac{\pi}{2} \left(\frac{1}{2} - \frac{1}{4}\right) = \boxed{\dfrac{\pi}{8}}$

$Let\quad I=\int _{ 0 }^{ 1 }{ \int _{ 0 }^{ \sqrt { 1-{ x }^{ 2 } } }{ (1-{ x }^{ 2 }-{ y }^{ 2 } } ) } dy\quad dx\\ \quad \quad =\int _{ 0 }^{ 1 }{ { I }_{ 0 } } dx\quad \quad \quad \quad [where\quad { I }_{ 0 }=\int _{ 0 }^{ \sqrt { 1-{ x }^{ 2 } } }{ 1-{ x }^{ 2 } } -{ y }^{ 2 }dy]\\ Now\quad in\quad integral\quad { I }_{ 0 }\quad we\quad are\quad integrating\quad with\quad respect\quad to\quad y.\\ So,\quad x\quad is\quad constant\quad in\quad { I }_{ 0 }$

${ I }_{ 0 }=\int _{ 0 }^{ \sqrt { 1-{ x }^{ 2 } } }{ 1-{ x }^{ 2 } } -{ y }^{ 2 }dy\\ \quad =y-{ yx }^{ 2 }-\frac { y^{ 3 } }{ 3 } \quad from\quad 0\quad to\sqrt { 1-{ x }^{ 2 } } \\ On\quad simplifying\quad we\quad get\quad \\ { I }_{ 0 }=\frac { 2 }{ 3 } { \left( 1-{ x }^{ 2 } \right) }^{ \frac { 3 }{ 2 } }$

$So,\quad I=\int _{ 0 }^{ 1 }{ { I }_{ 0 } } dx\\ \quad \quad \quad \quad =\frac { 2 }{ 3 } \int _{ 0 }^{ 1 }{ { \left( 1-{ x }^{ 2 } \right) }^{ \frac { 3 }{ 2 } } } dx\\ Let\quad x=sinT\\ \quad \quad dx=cosT\quad dT\\ when\quad x\rightarrow 0\\ \quad \quad T\rightarrow 0\\ and\quad x\rightarrow 1\\ \quad T\rightarrow \frac { \pi }{ 2 }$

$\therefore \quad I=\frac { 2 }{ 3 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ { cos }^{ 3 } } T.cosT\quad dT\\ \quad \quad =\frac { 2 }{ 3 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ { cos }^{ 4 } } T\quad dT\\ \quad \quad =\frac { 2 }{ 3 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ { \left( \frac { 1+cos2T }{ 2 } \right) }^{ 2 } } dT\\ \quad \quad =\frac { 1 }{ 6 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ 1+{ cos }^{ 2 } } 2T+2cos2T\quad dT\\ \quad \quad =\frac { 1 }{ 6 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ 1+\left( \frac { 1+cos4T }{ 2 } \right) } +2cos2T\quad dT\\ \quad \quad =\frac { 1 }{ 6 } \left( \frac { 3T }{ 2 } +\frac { sin4T }{ 8 } +sin2T \right) \quad \quad from\quad T=0\quad to\quad \frac { \pi }{ 2 } \\ \quad \quad =\frac { 1 }{ 6 } \left( \frac { 3\pi }{ 4 } \right) \\ \quad \quad =\frac { \pi }{ 8 }$