Calculate the volume determined by the cone $4z^2=x^2+y^2$ and the planes $z=1$ and $z= \dfrac{1}{2}$. Also $y≤x$.
To calculate the volume I use the following triple integral:
$$ \iint_D\int_{1/2}^1 1\,dz\,dy dx, $$
D is the projection of the cone in the xy-plane
My problem is that I'm not sure about the integral limits:
$\theta$ run from $\frac{5\pi}4$ to $\frac\pi4$? and $r$ run from $1$ to $2$? I thougt this because the radius $r$ is determinate in this case by $z$. So when I replacing whit $z= \dfrac{1}{2}$ in the cone ecuation: $4z^2=x^2+y^2$ becomes $1=x^2+y^2$ then $r=1$. In the same way when $z=1$ I have $r=2$. Is this correct? Are these the limits of the integral for $r$? or $r$ runs from $0$ to $2$ because $r=2$ is the largest radio and the integral must start at the origin. If this is correct why should the integral start at the origin and not in the smaller radius?
I appreciate all help and correction/s!
Note that the volume is the volume difference of the two cones with circular bases, one with the base radius 2 and height 1 and the other with the base radius 1 and height 1/2. The condition $y\le x$ accounts for half of the volume.
Thus, the volume is
$$\frac{1}{2}\cdot \frac{\pi}{3}\left( 2^2\cdot 1-1^2\cdot \frac{1}{2} \right) =\frac{7\pi}{12} $$
Edit:
The triple integral for the volume can be set up in cylindrical coordinates, $z = \frac12r$, as follows, $$\int_{\pi /4}^{5\pi /4}{\int_0^2{\left( 1-\frac{1}{2}r \right)}}rdrd\theta -\int_{\pi /4}^{5\pi /4}{\int_0^1{\left( \frac{1}{2}-\frac{1}{2}r \right)}}rdrd\theta =\frac{2\pi}{3}-\frac{\pi}{12}=\frac{7\pi}{12} $$