Problem source

The   curve $P$ has the parametric equations
$$
x= \sin\theta, \quad y=\cos2\theta
\qquad\hbox{ for }-\pi/2 \le \theta \le \pi/2.
$$
Show that $P$ is part of the parabola $y=1-2x^2$ and sketch $P$.
Show that the length of $P$ is $\surd (17) + {1\over 4} \sinh^{-1}4$.
Obtain the volume of the solid enclosed when  $P$  is rotated through $2\pi$ radians about the line $y=-1$.

Solution source

First notice that $y = \cos 2 \theta = 1 - 2\sin^2 \theta = 1- 2x^2$, therefore $P$ is lies on that parabola.

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){2*(#1)*((#1)^2 - 5)/((#1)^2-4)};
    \def\xl{-1.5};
    \def\xu{1.5};
    \def\yl{-1.5};
    \def\yu{1.5};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{10/\xrange}
    \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the styles for the axes and grid
    \tikzset{
        axis/.style={very thick, ->},
        grid/.style={thin, gray!30},
        x=\xscale cm,
        y=\yscale cm
    }
    
    % Define the bounding region with clip
    \begin{scope}
        % You can modify these values to change your plotting region
        \clip (\xl,\yl) rectangle (\xu,\yu);
        
        % Draw a grid (optional)
        % \draw[grid] (-5,-3) grid (5,3);
        
        \draw[thick, blue, smooth, domain={-pi/2}:{pi/2}, samples=100] 
            plot ({sin(180/pi*\x)}, {cos(2*180/pi*\x)});

        \node[left] at (0, 1) {$1$};
        \node[left] at (0, -1) {$-1$};
        \node[below] at ({sqrt(1/2)}, 0) {$\frac{1}{\sqrt{2}}$};
        \node[below] at ({-sqrt(1/2)}, 0) {$-\frac{1}{\sqrt{2}}$};
    \end{scope}
    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
    
    \end{tikzpicture}
\end{center}

The arc length is

\begin{align*}
&& s &= \int_{-\pi/2}^{\pi/2} \sqrt{\left ( \frac{\d x}{\d \theta} \right)^2+\left ( \frac{\d y}{\d \theta} \right)^2} \d \theta\\
&& &= \int_{-\pi/2}^{\pi/2} \sqrt{\cos^2 \theta+16 \sin^2 \theta \cos^2 \theta } \d \theta\\
&& &= \int_{-\pi/2}^{\pi/2} \cos \theta\sqrt{1+16 \sin^2 \theta} \d \theta\\
u = \sin \theta, \d u = \cos \theta \d \theta && &= \int_{u=-1}^{u=1} \sqrt{1+16 u^2} \d u\\
4u = \sinh v, 4\d u = \cosh v: && &= \int_{v=-\sinh^{-1} 4}^{v=\sinh^{-1} 4} \sqrt{1+\sinh^2 v} \tfrac14\cosh v \d v\\
&& &= \frac14 \int_{-\sinh^{-1} 4}^{\sinh^{-1} 4}  \cosh^2 v \d v\\
&& &= \frac18 \int_{-\sinh^{-1} 4}^{\sinh^{-1} 4}  (1 + \cosh 2v) \d v\\
&& &= \frac14 \sinh^{-1} 4 + \frac18\left [ \frac12\sinh 2v \right]_{-\sinh^{-1} 4}^{\sinh^{-1} 4}\\
&& &= \frac14 \sinh^{-1} 4 + \frac18\left [ \sinh v \sqrt{1 + \sinh^2 v} \right]_{-\sinh^{-1} 4}^{\sinh^{-1} 4}\\
&& &= \frac14 \sinh^{-1} 4 + \left (\frac18 \cdot 4 \sqrt{17} \right) - \left (\frac18 \cdot (-4) \sqrt{17} \right)\\
&& &= \frac14 \sinh^{-1} 4 + \sqrt{17}\\
\end{align*}

The volume of revolution is

\begin{align*}
&& V &=\pi \int_{-1}^1 (2-2x^2)^2 \d x \\
&&&= \pi \left [4x-\frac83x^3+\frac45x^5 \right]_{-1}^1 \\
&&&= \pi \left ( 8-\frac{16}3+\frac85 \right) \\
&&&= \frac{64}{15}\pi
\end{align*}