How to recompose a triple integral over a domain into a single integral Given:
$$
V_1 = \left\lbrace\;\;
             (x, \, y,\, z) \;\;
         \left| \;\; \rule{0pt}{12pt}
             x^2 +y^2  \leq 1, \;\;
             0  \leq z \leq 3 \sqrt{x^2 + y^2}, \;\;
             x \geq 0 \;
         \right.
         \right\rbrace\
$$
how can I rewrite this:
$$\displaystyle \iiint_{V_1} f(z)\, dxdydz$$
as an expression of this form:
$$\int_{0}^{3}\displaystyle K*f(z)\, dz$$ - what should be the value of K? why?
The excersize is supposed to convey that sometimes we need to pick a different order of integration than the "natural" one.
my current attempt:
I personally think the answer should be $\pi (\frac{z^2}{8})^\left(0.5\right)$ since if we take a "slice" we can see that z is line in a triangle with a side length of r (in the end), a hypotenuse of 3r so we can maybe extract it - not sure about it at all though and wether or not this was the intent.
 note: K doesn't have to be a constant - in fact I'm fairly sure it's a function
 A: Please note that we can take area of the cross section of the region parallel to xy-plane and then integrate wrt $dz$.
As $ \displaystyle z \leq 3 \sqrt{x^2 + y^2} \implies r \geq \frac{z}{3} ~$ but as we are inside the cylinder $x^2 + y^2 \leq 1 \implies r \leq 1$. Also note that $x \geq 0 \implies -\pi/2 \leq \theta \leq \pi/2$, which is half of the area of the cross section between cone and the cylinder.
So the area of the cross section is  $A(z) = \displaystyle \frac{1}{2} \cdot \pi \cdot 1^2 - \frac{1}{2} \cdot \frac {\pi z^2}{9}$.
At $r = 0, z = 0$ and at $r = 1, z = 3$
So the integral can be expressed as,
$ \displaystyle \int_0^3 \left(\frac{\pi}{2} - \frac {\pi z^2}{18}\right) \cdot f(z) ~ dz$
So, $ \displaystyle K = \frac{\pi}{2} - \frac {\pi z^2}{18}$ where $K$ is a function of $z$.
A: If you compute that triple integral in cylindrical coordinates, you get that\begin{align}\iiint_{V_1}f(z)\,\mathrm dz&=\int_{-\pi/2}^{\pi/2}\int_0^3\int_{z/3}^1f(z)\rho\,\mathrm d\rho\,\mathrm dz\,\mathrm d\theta\\&=\pi\int_0^3f(z)\left[\frac{\rho^2}2\right]_{\rho=z/3}^{\rho=1}\,\mathrm dz\\&=\int_0^3\pi\left(\frac12-\frac{z^2}{18}\right)f(z)\,\mathrm dz.\end{align}So, take $K(z)=\pi\left(\frac12-\frac{z^2}{18}\right)$.
