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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?

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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.

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  • $\begingroup$ 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 $dA=2\pi rdz$? $\endgroup$ – Ali Kıral May 18 at 21:37
  • $\begingroup$ The formula $dA=2\pi r dz$ is just for a cylinder with vertical walls. In this case $\rho'(z)\equiv0$. $\endgroup$ – Christian Blatter May 19 at 8:25

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