On known formulae

1. Evaluate the area of the disk of center $(0, 0)$ and radius $R>0$
2. Let $f$ be a continuous function such that $f(z) \geq 0 , \; z \in [0, R]$. Prove , using the previous question as well as Cavalieri's principal that the volume $\mathcal{M}=\{(x,y, z) \in \mathbb{R}^3 : z \in [0, R] \mid x^2+y^2 \leq f^2(z) \}$ produced by an entire rotation by the graph of $f$ (which is a curve on the $xy$ plane) around the $z$ axis is equal to:
   
   $$\mathcal{V}\left ( \mathcal{M} \right )= \pi \int_{0}^{R}f^2(z) \, {\rm d}z$$


3.  If the function of the previous question is continuously differentiable , then prove that the area of the surface:
   
   $$\mathbb{S}=\left \{ \left ( f(z)\cos \varphi, f(z) \sin \varphi, z \right ) \in \mathbb{R}^3 : z \in [0, R], \varphi \in [0, 2\pi] \right \}$$
   
   produced by an entire rotation by the graph of $f$ around the $z$ axis is equal to:
   
   $$\mathcal{A}\left ( \mathbb{S} \right ) = 2\pi \int_{0}^{R} f(z) \sqrt{1+\big(f'(z)\big)^2} \, {\rm d}z$$

Solution

The exercise comes from an examination paper of Integral Multivariable Calculus. There were two more questions / applications regarding calculation of certain solids but for the purposes of this post these questions are omitted. 

1. This particular disk can be described as:
   
   $$\mathbb{D}=\big\{(x,y)\in\mathbb{R}^2\; |\; x^2+y^2\leqslant R^2\big\}$$
   
   Thus its area is equal to:
   
   \begin{align*} \mathcal{A}(\mathbb{D})&=\mathop{\iint}\limits_{\mathbb{D}}\; {\rm d}(x , y)\\
   &\stackrel{(*)}{=}\int_{0}^{2\pi}\int_{0}^{R}\rho\,{\rm d}\rho\,{\rm d} \varphi\\
   &=\frac{R^2}{2}\int_{0}^{2\pi}\,{\rm d}\varphi\\
   &=\pi\,R^2
   \end{align*}
   
   $(*)$ we converted to polar coordinates.


2. The solid
   
   $$\mathcal{M}=\big\{(x,y,z) \in \mathbb{R}^3 \;|\; z \in [0, R] \,, \; x^2+y^2 \leqslant f^2(z) \big\}$$
   
   where $f(z) \geq 0 , \; z \in [0, R]$ is a continuous function, is Jordan countable and lies between the plane $z=0$ and $z=R$. Also for every $\xi \in [0, R]$ the intersection of $\mathcal{M}$ with the plane $z=\xi$ is the disk
   
   $$\mathbb{D}_{f(\xi)}=\{(x,y)\in\mathbb{R}^2\; |\; x^2+y^2\leqslant f^2(\xi)\}$$
   
   of area $\pi\,f^2(\xi)$ . Using Cavalieri's principal we conclude that the function $g(\xi)=\pi\,f^2(\xi)\,, \; \xi\in[0, R]$ is integrable and that the volume of $\mathcal{M}$ is:
   
   $$\mathcal{V}( \mathcal{M})= \displaystyle\int_{0}^{R}g(\xi) \,{\rm d} \xi=\pi \int_{0}^{R}f^2(\xi) \, {\rm d}\xi$$


3.  A perpendicular vector to the surface
   
   $$\overline{\mathbb{S}}(z,\varphi)=\left({ \begin{array}{c} f(z)\cos \varphi\\ f(z) \sin \varphi\\ z \end{array} }\right)\,,\quad z \in [0, R],\; \varphi \in [0, 2\pi]$$
   
   is:
   
   \begin{align*} \overline{N}(z,\varphi)&=\displaystyle\frac{\partial\overline{\mathbb{S}} }{\partial z}\times\frac{\partial\overline{\mathbb{S}} }{\partial \varphi}\\
   &=\left|{\begin{array}{ccc} \overline{e}_1 & \overline{e}_2 & \overline{e}_3\\ f'(z)\cos \varphi & f'(z)\sin\varphi & 1\\ -f(z)\sin\varphi & f(z)\cos\varphi & 0 \end{array}}\right|\\\\
   &=-f(z)\sin\varphi\,\overline{e}_1-f(z)\cos\varphi\,\overline{e}_2+f(z)\,f'(z)\,\overline{e}_3 \end{align*}
   
   Thus the area of the surface is equal to:
   
   \begin{align*} \mathcal{A}(\mathbb{S})&=\iiint \limits_{\mathbb{S}}{\big\|{\overline{N}(z,\varphi)}\big\|\,{\rm d}(z,\varphi)}\\
   &=\int_0^R\int_{0}^{2\pi}{\sqrt{(-f(z)\sin\varphi)^2+(-f(z)\cos\varphi)^2+(f(z)\,f'(z))^2}\;{\rm d} \varphi\,{\rm d}z}\\
   &=\int_0^R\int_{0}^{2\pi}{\rm d} \varphi{\sqrt{f^2(z)+f^2(z)\,(f'(z))^2}\;\,{\rm d}z}\\
   &=\int_0^R{2\pi\,|f(z)|\sqrt{1+(f'(z))^2}\;\,{\rm d}z}\\
   &=2\pi \int_{0}^{R} f(z) \sqrt{1+(f'(z))^2} \, {\rm d}z
   \end{align*}

This post was migrated from mathimatikoi.org