Mouting triple integrals to calculate circular cone

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Good night, I know too little about Calculus III. I'm entering now in the triple integrals world. I started with the Sphere volume and Cylinder. I could calculate it as well. But to the Circular Cone I just can't mount the equation to calculate the volume. My Circular cone have 5cm of height and 3cm as radius. And I'm totally lost with these papers on internet, I'm using cylinders coordinates to calculate. I'm really lost, how do I calculate the volume of this circular cone with triple integrals?? I just need somebody to help me to construct the triple integrals standard to calculate. I'm not at school, I don't have a teacher. I just want to learn it by myself!

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Consider the following figure

enter image description here

The equation of the thick red line in an $r,z$ coordinate system is $$z=\frac53 r.$$ The vertical black line intersects the cone at $\frac53 r$ and at $z=5$.

So we can get the volume of the cone as the following integral

$$\int_0^{2\pi}\int_0^3\int_{\frac53r}^5\ dz\ dr \ d\varphi=\int_0^{2\pi}\int_0^3\ 5-\frac53 r\ dr\ d\varphi=$$ $$=\int_0^{2\pi}\frac{15}2\ d\varphi=15\pi.$$