# Differential area for the lateral surface of frustum of a cone

I am studying Fluid Mechanics and I needed a differential area element of the side or lateral surface of a frustum. This frustum is cut from a cone. In solution manual of the book I study, differential area is given for side surfaces by $$dA=2\pi r dz$$

The geometry and the coordinate axes are given in the picture: https://imgur.com/TQMeAQs

Please ignore SAE 10W oil and remaining parts other than the frustum. They are something to do with mechanical engineering

Can you give me a hint on how to write the equation above? Do I need to consider the unfolded frustum and draw a differential curved strip element or something?

For a rotational surface around the $$z$$-axis you need the function $$z\mapsto \rho(z)$$ that gives the radius of the latitude circles at height $$z$$. The area element then is $${\rm d}A=2\pi\>\rho(z)\>\sqrt{1+\rho'^2(z)}\>dz\ .$$ In the case at hand the function $$\rho(z)$$ increases linearly from $$d$$ to $$D$$, hence $$\rho'(z)$$ is constant. In fact $$\sqrt{1+\rho'^2(z)}={1\over\cos\alpha}\ ,$$ where $$\alpha$$ is the angle between the frustum generators and the $$z$$-axis.
• For $\rho (z)$, z is the independent variable and $\rho$ is the dependent variable, but how that arc length $\sqrt{1+\rho(z)'^2}$ is represented in $dA=2\pi rdz$? – Ali Kıral May 18 '19 at 21:37
• The formula $dA=2\pi r dz$ is just for a cylinder with vertical walls. In this case $\rho'(z)\equiv0$. – Christian Blatter May 19 '19 at 8:25