Vtyjkyuk

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 $7sqrt3$.
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.

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 $sqrt3$. From your bounds on $x$ and $y$, $sqrtx^2 + y^2 = sqrt2$ at $z=1$, so the limits of $z$ are $left(sqrt3/2sqrtx^2+y^2right)$ to $7sqrt3.$

For the cylindrical coordinates, it's much the same. Replace $sqrtx^2+y^2$ with $r$. (You should have $r$ as the integrand instead of $rho$.)

For the spherical coordinates, the $rho$ dependence is with $phi$ only. At $phi=0$, $rho = 7sqrt3$, while at $phi=pi/6$, $rho=7sqrt3/cos(pi/6).$ So your upper limit is $7sqrt3/cosphi$.

The integrand is determined from the spherical volume element, which is

$$dV = rho^2 sin theta; drho; dtheta; dphi.$$