Calculate the volume determined by the paraboloid $4z=x^2+y^2$ and the planes $z=1$ and $z= \dfrac{1}{2}$. Also $y≤x$.
I made a variable change: $x=r\cos \theta $ and $y=r\sin \theta$ then $z=\dfrac{r}{2}$ with $\dfrac{1}{4}≤r≤\dfrac{1}{2}$ and here is my first doubt: is $\theta$ going from $\dfrac{- \pi}{4}$ to $\dfrac{\pi}{4}$? because if I have two circles in the $xy$ plane centered on $(0,0)$ and I draw the line $y = x$ then the values of $x$ less than or equal to $y$ are swept by the angle that goes from $\dfrac{- \pi}{4}$ to $\dfrac{\pi}{4}$.
Finally, I would like to consult what intergral I should use to obtain this volume and if it can be obtained only by a triple integral. Because I have seen cases in which a volume is obtained through a double integral and I don't understand how this is possible and when I can do it.
I appreciate all help!
The change to polar coordinates gives you $z=\frac{r^2}4$, but that's not what you need. The volume can be calculated by $$ \int_{1/2}^1\iint_D 1\,dA\,dz, $$ where $D$ is the disk of radius $2\sqrt z$. The restriction $y\leq x$ can be achieved on $\theta$; it's the half plane below the line $y=x$: so it is given by making $\theta$ run from $\frac{3\pi}2-\frac\pi4=\frac{5\pi}4=-\frac{3\pi}4$ to $\frac\pi4$. So your volume is $$ \int_{1/2}^1\iint_D 1\,dA\,dz =\pi\,\int_{1/2}^1\int_0^{2\sqrt z}r\,dr\,dz=\pi\,\int_{1/2}^12z\,dz=\frac{3\pi}4. $$ A volume can be calculated using a triple integral if you integrate the function $1$, or by a double integral if you integrate the different between the "roof" and the "floor". The same one that a one-variable integral can be used (trivially) to calculate a length if you integrate the function $1$, or to calculate the area between two curves if you integrate their difference.