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I just want to find out if my thought process and reasoning of my derivation of this formula is correct

Let $C$ be the arc of the curve $y=f(x)$ between the points $P(p,f(p))$ and $Q(q,f(q))$ where $p\lt q$.

If $\forall x\in\{x\in\mathbb{R}\,|\,p\le x\le q\}$, $y=mx+b\le f(x)$, derive an integral expression for the surface area of the solid obtained when $R$, the region bounded by $C$, $f(x)$, and the two lines perpendicular to $y=mx+b$ that intersect $P$ and $Q$, is rotated about the line $y=mx+b$.

Let $k\in\{k\in\mathbb{Z}\,|\,0\le k\le n\}$ for some $n\in \mathbb{N}$, where $x_0=p$ and $x_n=q$

Let $L_k$ be the length between the points $P_{k-1}(x_{k-1}, f(x_{k-1}))$ and $P_k(x_k, f(x_k))$, where $1\le k\le n$

So $L_k=|P_{k-1}P_k|=\sqrt{{\Delta{x}}^2+\big[f(x_k)-f(x_{k-1})\big]^2}$

Since the mean value theorem states that

$$\exists c\in \{x\in\mathbb{R}\,|\,a\lt x\lt b\}, \frac{f(b)-f(a)}{b-a}=f'(c)$$

provided that $f(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$, $f(x_k)-f(x_{k-1})$ can be written as $f'(x_k^*)\Delta{x}$ for some $x_k^*\in \{x\in\mathbb{Z}\,|\,x_{k-1}\lt x\lt x_k\}$

So $L_k$ becomes $\sqrt{\Delta{x}^2+\big[f'(x_k^*)\Delta{x}\big]^2}=\sqrt{1+\big[f'(x_k^*)\big]^2}\Delta{x}$

The surface area of a frustum is $\pi(r_1+r_2)L$, which in this case turns out to be

$$\pi\big(f(x_{k-1})+f(x_k)\big)\sqrt{1+\big[f'(x_k^*)\big]^2}\Delta{x}=2\pi\left(\frac{f(x_{k-1})+f(x_k)}{2}\right)\sqrt{1+\big[f'(x_k^*)\big]^2}\Delta{x}$$

Since the intermediate value theorem states that

$$\forall c\in (f(a),f(b)), \exists r\in(a,b), f(r)=c$$

provided $f(x)$ is continuous on $[a,b]$, $\frac{1}{2}\big(f(x_{k-1})+f(x_k)\big)$ can be written as $f(x_k^*)$ for some $x_k^*\in \{x\in\mathbb{Z}\,|\,x_{k-1}\lt x\lt x_k\}$

So the surface area becomes $2\pi f(x_k^*)\sqrt{1+\big[f'(x_k^*)\big]^2}\Delta{x}$ if rotated about the x-axis

Now, we determine relationships that expresses $\Delta{u}$, the increments along $y=mx+b$, and the distance between $f(x_k^*)$ and $y=mx+b$ in terms of $\Delta{x}$, the increments along the x-axis.

Let $\alpha$ be the angle between the x-axis and the secant line of $f(x)$ formed by the points $P_{k-1}$ and $P_k$, and $\beta$ be the angle between $y=mx+b$ and the x-axis, where $\tan(\beta)=m$

$$1)\cos(\alpha)=\frac{\Delta{x}}{L_k}\implies L_k=\frac{\Delta{x}}{\cos(\alpha)}$$

$$2)\sin(\alpha+90-\beta)=\frac{\Delta{u}}{L_k}\implies L_k=\frac{\Delta{u}}{\cos(\alpha-\beta)}$$

$$\frac{\Delta{x}}{\cos(\alpha)}=\frac{\Delta{u}}{\cos(\alpha-\beta)}$$

$$\Delta{u}=\frac{\Delta{x}}{\cos(\alpha)}(\cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta))=\Delta{x}(\cos(\beta)+\tan(\alpha)\sin(\beta))$$

By the trigonometric identities $\sec^2(\theta)=1+\tan^2(\theta)$ and $\csc^2(\theta)=1+\cot^2(\theta)$, we obtain

$$1)\cos(\theta)=\sqrt{\frac{1}{1+\tan^2(\theta)}}\implies\cos(\beta)=\sqrt{\frac{1}{m^2+1}}$$

$$2)\sin(\theta)=\sqrt{\frac{1}{1+\cot^2(\theta)}}\implies\sin(\beta)=\sqrt{\frac{1}{1+\frac{1}{m^2}}}=\frac{m}{\sqrt{m^2+1}}$$

Also

$$\tan(\alpha)=\frac{f(x_k)-f(x_{k-1})}{\Delta{x}}=\frac{f'(x_k^*)\Delta{x}}{\Delta{x}}=f'(x_k^*)$$

Therefore

$$\Delta{u}=\Delta{x}\left(\sqrt{\frac{1}{m^2+1}}+f'(x_k^*)\frac{m}{\sqrt{m^2+1}}\right)=\left(\frac{1+mf'(x_k^*)}{\sqrt{m^2+1}}\right)\Delta{x}$$

Let $\overline{D_k}$ be the line segment between $f(x_k^*)$ and $y=mx+b$ where $\overline{D_k}$ makes a $90^{\circ}$ angle with $y=mx+b$

$$\cos(\beta)=\frac{D_k}{f(x_k^*)-(mx_k^*+b)}\implies D_k=\big(f(x_k^*)-mx_k^*-b\big)\cos(\beta)=\frac{f(x_k^*)-mx_k^*-b}{\sqrt{m^2+1}}$$

So

$$SA=\lim_{n\to\infty}\sum_{k=1}^n 2\pi D_k\sqrt{1+f'({x_k^*})^2}\Delta{u}$$

$$=2\pi\lim_{n\to\infty}\sum_{k=1}^n \left(\frac{f(x_k^*)-mx_k^*-b}{\sqrt{m^2+1}}\right)\sqrt{1+\left[f'(x_k*)\right]^2}\left(\frac{1+mf'(x_k^*)}{\sqrt{m^2+1}}\right)\Delta{x}$$

$$=\frac{2\pi}{m^2+1}\lim_{n\to\infty}\sum_{k=1}^n (f(x_k^*)-mx_k^*-b)(1+mf'(x_k^*))\sqrt{1+\left[f'(x_k^*)\right]^2}\Delta{x}$$

$$=\frac{2\pi}{m^2+1}\int_p^q (f(x)-mx-b)(1+mf'(x))\sqrt{1+\left[f'(x)\right]^2}dx$$

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