$\int_{0}^{1}\int_{0}^{\sqrt{1-x^2}} \sqrt{1-x^2-y^2} dy dx$

Question

I need $\int_{0}^{1}\int_{0}^{\sqrt{1-x^2}} \sqrt{1-x^2-y^2}dy\;dx$

the first integral is very difficult, I plan to change the variable to obtain $r\sqrt{1-r^2}\;dr\;d\theta$

but the new limits of integration confuse me

Answer 1

Note that the region you are integrating over is the part of the unit circle in the first quadrant. Thus, the new limits of integration are:

$$\int_{0}^{\frac{\pi}{2}}\int_{0}^{1}r\sqrt{1 - r^{2}}\ dr\ d\theta=\int_{0}^{\frac{\pi}{2}}-\frac{(1-r^{2})^{\frac{3}{2}}}{3}\bigg\vert_{0}^{1}\ d\theta = \int_{0}^{\frac{\pi}{2}}\frac{1}{3}\ d\theta = \frac{\theta}{3}\bigg\vert_{0}^{\frac{\pi}{2}} = \boxed{\frac{\pi}{6}}$$