# Surface area of solid rotated about a slant

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

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

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

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

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