Mathematical Immaturity

14.19 Exercises

• Sketch the curve whose polar equation is $r = \sin^2 \theta,$ $0 \leq \theta \leq 2\pi,$ and show that it consists of two loops. (a) Find the area of region enclosed by one loop of the curve. (b) Compute the length of one loop of the curve.
• The first loop is drawn out by the radius from $\theta = 0$ to $\theta = \pi$ where $r = 0.$ The second loop is drawn likewise from $\theta = \pi$ to $\theta = 2\pi.$ (a) The area of the radial set from $\theta = a$ to $\theta = b$ is: $$ \begin{align*} \\ a(R) &= \frac{1}{2}\int_a^b f^2(\theta)\,d\theta. \end{align*} $$ (b) The arc length $s$ of the curve described by the polar equation $r = f(\theta)$ from $\theta = a$ to $\theta = b,$ where $a \leq \theta \leq b \leq a + 2\pi$ is: $$ \begin{align*} \\ s &= \int_a^b \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta. \end{align*} $$ Then, use the double-angle identities of the sine and cosine $$ \begin{align*} \\ \sin 2\theta &= 2\sin\theta\cos\theta \\ \\ \cos 2\theta &= 1 - 2\sin^2\theta \end{align*} $$ to get the integrand into the form $\sqrt{u^2 + 1}\,du$ from which point you can apply the tangent substitution $u = \tan t$ and $du = \sec^2 t\,dt$ as shown in Section 6.24 Example 2.
• (a) The area $a(R)$ swept out by the radius $r = f(\theta)$ from $\theta = a$ to $\theta = b$ is $$ \begin{align*} \\ a(R) &= \frac{1}{2}\int_a^b f^2(\theta)\,d\theta. \end{align*} $$ Setting $a = 0$ and $b = \pi,$ using the double-angle identities for the sine and cosine $$ \begin{align*} \\ \sin 2\theta &= 2\sin\theta\cos\theta \\ \\ \cos 2\theta &= 1 - 2\sin^2\theta \end{align*} $$ we get: $$ \begin{align*} \\ a(R) &= \frac{1}{2}\int_0^\pi \sin^4\theta\,d\theta \\ \\ &= \frac{1}{8}\int_0^\pi (1 - \cos 2\theta)^2\,d\theta \\ \\ &= \frac{\pi}{8} + \frac{1}{8}\int_0^\pi \cos^2 2\theta - 2\cos 2\theta\,d\theta \\ \\ &= \frac{\pi}{8} + \frac{1}{8}\int_0^\pi \cos^2 2\theta\,d\theta \end{align*} $$ Integration by parts with $$ \begin{align*} \\ u &= \cos2\theta, \quad dv = \cos2\theta \\ v &= \frac{1}{2}\sin2\theta, \quad du = -2\sin2\theta \end{align*} $$ gives us: $$ \begin{align*} \\ \int_0^\pi \cos^2 2\theta\,d\theta &= \frac{1}{2}\sin2\theta\cos2\theta\,\Biggr|_0^{\pi} + \int_0^{\pi}\sin^2 2\theta \\ \\ &= \frac{1}{2}\int_0^{\pi}(1 - \cos4\theta)\,d\theta \\ \\ &= \frac{\pi}{2} - \frac{1}{8}\sin4\theta\,\Biggr|_0^{\pi} \\ \\ &= \frac{\pi}{2}. \end{align*} $$ Plugging this back into our previous equation for $a(R)$ gives us: $$ \begin{align*} \\ a(R) &= \frac{\pi}{8} + \frac{1}{8}\int_0^\pi \cos^2 2\theta\,d\theta \\ \\ &= \frac{\pi}{8} + \frac{\pi}{16} \\ \\ &= \frac{3\pi}{16} \quad \blacksquare \end{align*} $$ (b) We know from Exercise 4 that the arc length $s$ of the curve given by the polar equation $r = f(\theta)$ from $\theta = a$ to $\theta = b,$ where $a \leq \theta \leq b \leq a + 2\pi$ is: $$ \begin{align*} \\ s &= \int_a^b \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta. \end{align*} $$ Evaluating for $r = \sin^2\theta$ over the interval $[0, \pi]$ we get: $$ \begin{align*} \\ s &= \int_0^\pi \sqrt{\sin^4\theta + \left(2\sin\theta \cos\theta\right)^2}\,d\theta \\ \\ &= \int_0^\pi \sin\theta \sqrt{\sin^2\theta + 4\cos^2\theta}\,d\theta \\ \\ &= \int_0^\pi \sin\theta \sqrt{1 + 3\cos^2\theta}\,d\theta. \end{align*} $$ Using the substitution $u = -\sqrt{3}\cos\theta$ and $du = \sqrt{3}\sin\theta,$ noting that cosine is an even function $($ie, the integral from -$\sqrt{3}$ to $0$ is equal to that from $0$ to $\sqrt{3})$ the above becomes: $$ \begin{align*} \\ s &= \frac{1}{\sqrt{3}} \int_{-\sqrt{3}}^{\sqrt{3}} \sqrt{u^2 + 1}\,du \\ \\ &= \frac{2}{\sqrt{3}} \int_{0}^{\sqrt{3}} \sqrt{u^2 + 1}\,du. \end{align*} $$ Now, recall from Section 6.25 that integrals of the form $$ \begin{align*} \\ \int R\left(x, \sqrt{a^2 + (cx + d)^2}\right)\,dx \end{align*} $$ can be handled using the substitution $cx + d = a\tan t$ and $dx = a\sec^2 t\,dt.$ Accordingly, we perform the substitution $u = \tan t$ and $du = \sec^2 t\,dt$ to give us: $$ \begin{align*} \\ s &= \frac{2}{\sqrt{3}} \int_{0}^{\pi/3} \sec^2 t\sqrt{1 + \tan^2 t}\,dt \\ \\ &= \frac{2}{\sqrt{3}} \int_{0}^{\pi/3} \sec^3 t\,dt. \end{align*} $$ Using integration by parts, with $$ \begin{align*} \\ x &= \sec t, \quad dy = \sec^2 t\,dt \\ y &= \tan t, \quad dx = \tan t \sec t\,dt \end{align*} $$ the integral becomes: $$ \begin{align*} \\ \frac{2}{\sqrt{3}} \int_{0}^{\pi/3} \sec^3 t\,dt &= \frac{2}{\sqrt{3}} \sec t \tan t\,\Biggr|_{0}^{\pi/3} - \frac{2}{\sqrt{3}} \int_{0}^{\pi/3} \tan^2 t \sec t\,dt \\ \\ &= 4 - \frac{2}{\sqrt{3}}\int_{0}^{\pi/3} \sec^3 t - \sec t\,dt. \end{align*} $$ Then, noting that we have the integral term $\int_{0}^{\pi/3} \sec^3 t\,dt$ on both sides, we can rearrange the above to get $$ \begin{align*} \\ \frac{2}{\sqrt{3}} \int_{0}^{\pi/3} \sec^3 t\,dt &= 2 + \frac{1}{\sqrt{3}}\int_{0}^{\pi/3} \sec t\,dt. \end{align*} $$ Recall from Section 14.13, Exercise 8 that $\int \sec t\,dt = \log(\sec t + \tan t) + C.$ As such, we obtain $$ \begin{align*} \\ s &= \frac{2}{\sqrt{3}} \int_{0}^{\pi/3} \sec^3 t\,dt \\ \\ &= 2 + \frac{1}{\sqrt{3}}\log(\sec t + \tan t )\,\Biggr|_0^{\pi/3} \\ \\ &= 2 + \frac{1}{3}\sqrt{3}\log(2 + \sqrt{3}) \quad \blacksquare \end{align*} $$