Triple integration with limits

Question

$$\iiint_T\sqrt{x^2+y^2+z^2}\,dx\, dy\, dz\\T:x^2+y^2+z^2\leqslant y$$ I cannot find the limits...please help

Answer 1

The integration range is a ball centered at $\left(0;\frac{1}{2};0\right)$ with radius $\frac{1}{2}$. For any $y\in[0,1]$ the $y$-section is a circle with radius $\sqrt{y-y^2}$ and the integral over the section of the distance from the origin equals $$ 2\pi \int_{0}^{\sqrt{y-y^2}}\rho\sqrt{\rho^2+y^2}\,d\rho =\frac{2\pi}{3}(y^{3/2}-y^3)$$ so the wanted integral equals $$ \frac{2\pi}{3}\int_{0}^{1} y^{3/2}-y^{3}\,dy = \frac{2\pi}{3}\left(\frac{2}{5}-\frac{1}{4}\right)=\color{red}{\frac{\pi}{10}}.$$