The region $W$ is the cone shown below (see image).
The angle at the vertex is $π/3$, and the top is flat and at a height of $7\sqrt{3}$.
Write the limits of integration for $\int_W dV$ in the following coordinates (do not reduce the domain of integration by taking advantage of symmetry):
Progress
As seen in the image, the ones highlighted in red are those I cannot figure out. I'm not sure if I have some fundamental misconceptions that allows me to derive some of the boundaries but not others, but if someone can kindly explain how to find those values in red, I would be grateful. Thanks in advance.
Best Answer
The cross-sectional radius of the cone increases linearly with $z$, so the lower limit of $z$ also increases linearly. So $1/7$th of the way up, the lower $z$ limit needs to be $\sqrt{3}$. From your bounds on $x$ and $y$, $\sqrt{x^2 + y^2} = \sqrt{2}$ at $z=1$, so the limits of $z$ are $\left(\sqrt{3/2}\sqrt{x^2+y^2}\right)$ to $7\sqrt{3}.$
For the cylindrical coordinates, it's much the same. Replace $\sqrt{x^2+y^2}$ with $r$. (You should have $r$ as the integrand instead of $\rho$.)
For the spherical coordinates, the $\rho$ dependence is with $\phi$ only. At $\phi=0$, $\rho = 7\sqrt{3}$, while at $\phi=\pi/6$, $\rho=7\sqrt{3}/\cos(\pi/6).$ So your upper limit is $7\sqrt{3}/\cos\phi$.
The integrand is determined from the spherical volume element, which is
$$dV = \rho^2 \sin \theta\; d\rho\; d\theta\; d\phi.$$