How to find maximum and minimum volumes of solid obtained by rotating $y=\sin x$ around $y=c$

Question

I would appreciate if somebody could help me with the following problem:

Q: Let $S$ be the region bounded by the curves $y=\sin x \ (0 \leq x \leq \pi)$ and $y=0$. Let $V(c)$ be the volume of the solid of obtained by rotating the region $S$ around the line $y=c \ (0 \leq c \leq 1)$.

1. Find $c_1$ such that $V(c_1)$ is the minimum of $V(c)$.

2. Find $c_2$ such that $V(c_2)$ is the maximum of $V(c)$.

Answer 1

Here is the answer. If $\frac{1}{2}\le c\le 1$, then $$ V(c)=2c^2\pi^2-2\int_0^{\arcsin c}\pi(\sin x-c)^2dx. $$ If $0<c<\frac{1}{2}$, then $$ V(c)=\int_{\arcsin c}^{\pi-\arcsin c}\pi(\sin x-c)^2dx+2c^2\pi\arcsin (2c)-2\int_{\arcsin c}^{\arcsin(2c)}\pi(c-\sin x)^2dx-2\int_0^{\arcsin c}\pi(c-\sin x)^2dx. $$ I will come back to finish this.

Answer 2

$\mathbf{A)}$ If $\frac{1}{2}\le c\le1$, we have that $$V(c)=\int_0^{c}2\pi(c-y)(\pi-2\sin^{-1}y)dy=$$

$$2\pi\bigg[-\frac{\pi}{2}(c-y)^2-2cy\sin^{-1}y-2c\sqrt{1-y^2}+y^2\sin^{-1}y-\frac{1}{2}\sin^{-1}y+\frac{1}{2}y\sqrt{1-y^2}\bigg]_{0}^{c},$$

so$\;\;\;$ $V(c)=2\pi\big(-c^2\sin^{-1}c-\frac{3}{2}c\sqrt{1-c^2}+\frac{\pi}{2}c^2+2c-\frac{1}{2}\sin^{-1}c\big)$

and $\;$ $V^{\prime}(c)=2\pi\big(\pi c+2-2c\sin^{-1}c-2\sqrt{1-c^2}\big)$.

Since $V^{\prime}(\frac{1}{2})=2\pi(2-\sqrt{3}+\frac{\pi}{3})>0$ and $V^{\prime\prime}(c)=2\pi(\pi-2\sin^{-1}c)>0$ for $c\in(\frac{1}{2},1)$,

$\;\;\;$V is increasing on $[\frac{1}{2},1]$.

$\mathbf{B)}$ If $0\le c\le\frac{1}{2}$, we have that

$\displaystyle V(c)=\int_0^{c}2\pi(c-y)(\pi-2\sin^{-1}y)dy+\int_{2c}^{1}2\pi(y-c)(\pi-2\sin^{-1}y)dy=$

$\;\;\;\;\;$ $2\pi\bigg[-\frac{\pi}{2}(c-y)^2-2cy\sin^{-1}y-2c\sqrt{1-y^2}+y^2\sin^{-1}y-\frac{1}{2}\sin^{-1}y+\frac{1}{2}y\sqrt{1-y^2}\bigg]_{0}^{c}-2\pi\bigg[-\frac{\pi}{2}(y-c)^2-2cy\sin^{-1}y-2c\sqrt{1-y^2}+y^2\sin^{-1}y- \frac{1}{2}\sin^{-1}y+\frac{1}{2}y\sqrt{1-y^2}\bigg]_{2c}^{1}$

$\;\;\;\;$$=2\pi\big(-c^2\sin^{-1}c-\frac{3}{2}c\sqrt{1-c^2}-\frac{1}{2}\sin^{-1}c+2c+\frac{\pi}{2}(1-c)^2+\pi c$

$\;\;\;\;\;\;\;\;$ $-\frac{\pi}{4}-c\sqrt{1-4c^2}-\frac{1}{2}\sin^{-1}2c\big)$.

Then $V^{\prime}(c)=2\pi\big(\pi c+2-2c\sin^{-1}c-2\sqrt{1-c^2}-2\sqrt{1-4c^2}\big)$

so $V^{\prime}(0)=-4\pi$ and $V^{\prime\prime}(c)=2\pi\big(\pi+\frac{8c}{\sqrt{1-4c^2}}-2\sin^{-1}c\big)>0$ for $0<c<\frac{1}{2}$ since $\pi>2\sin^{-1}c$.

Since $V^{\prime}(\frac{1}{2})=2\pi\big(2-\sqrt{3}+\frac{\pi}{3}\big)>0$,$\;\;$ $V(c)$ has exactly one local minimum on $(0,\frac{1}{2})$

at the solution of the equation

$V^{\prime}(c)=2\pi\big(\pi c+2-2c\sin^{-1}c-2\sqrt{1-c^2}-2\sqrt{1-4c^2}\big)=0$,

which is approximately given by $c\approx.4133665$.

$\mathbf{C)}$ Since $$V(1)=\frac{8\pi-\pi^{2}}{2}>\frac{\pi^2}{2}=V(0),$$

$\:\:$ we can conclude that $V(c)$ has a minimum for $c\approx.4133665$ and a maximum for $c=1$.

. . . . . . . . . . . . . . . . .

Notice that the integrand in both integrals could have been simplified using the equation $\pi-2\sin^{-1}y=2\cos^{-1}y$.

Answer 3

We will use the disk (or annulus) method to compute the volume of revolution.

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For $c\in\left[0,\frac12\right]$, the points of intersection are $x=\sin^{-1}(c)$ and $x=\sin^{-1}(2c)$

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So, by symmetry, the volume is $$ \begin{align} &2\pi\left(\int_0^{\sin^{-1}(c)}\left(c^2-(c-\sin(x))^2\right)\,\mathrm{d}x +\int_{\sin^{-1}(c)}^{\sin^{-1}(2c)}c^2\,\mathrm{d}x +\int_{\sin^{-1}(2c)}^{\pi/2}(\sin(x)-c)^2\,\mathrm{d}x\right)\\[6pt] &=\pi^2c^2+4\pi c-3\pi c\sqrt{1-c^2}-\pi(2c^2+1)\sin^{-1}(c) -2\pi c\sqrt{1-4c^2}+\pi\cos^{-1}(2c) \end{align} $$ whose derivative is $$ 4\pi\left(1-\sqrt{1-c^2}+c\cos^{-1}(c)-\sqrt{1-4c^2}\right) $$ which is increasing on $\left[0,\frac12\right]$ and $0$ when $c\doteq0.4133664985118219064957589$.

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For $c\in\left[\frac12,1\right]$, the intersection at $\sin^{-1}(2c)$ goes away

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So, by symmetry, the volume is $$ \begin{align} &2\pi\left(\int_0^{\sin^{-1}(c)}\left(c^2-(c-\sin(x))^2\right)\,\mathrm{d}x +\int_{\sin^{-1}(c)}^{\pi/2}c^2\,\mathrm{d}x\right)\\[6pt] &=\pi^2c^2+4\pi c-3\pi c\sqrt{1-c^2}-\pi(2c^2+1)\sin^{-1}(c) \end{align} $$ whose derivative is $$ 4\pi\left(1-\sqrt{1-c^2}+c\cos^{-1}(c)\right) $$ which is positive for $c\in\left[\frac12,1\right]$.

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Plotting volume vs $c$:

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We get the extremes to be $$ \begin{array}{} &c&\text{volume}\\ &0.0000000000000000&4.9348022005446793\\ c_1=&0.4133664985118219&1.9531551992492437\\ c_2=&1.0000000000000000&7.6315684138144936 \end{array} $$