What is the area of this closed curve $\alpha(t) = (\sin(2t), 2\sin t)$ on the plane?

Question

The closed curve $\alpha(t) = (\sin(2t), 2\sin t)$ with $0 \leq t \leq \pi$. Find the area of the $A$ enclosed by the $\alpha$ on the $\mathbb{R}$.

I took the formula $\int_{0}^{\pi} \frac{1}{2}{f(\theta)}^2 d\theta$ for the $r = f(\theta)$.

In my thought, $\alpha(t) =(x(t), y(t)) = (\sin(2t), 2\sin t)$. So the $r^2(t) = x^2(t) + y^2(t) = \sin^2(2t) + 4\sin^2t$

My answer is $\int_{0}^{\pi} \frac{1}{2}{f(t)}^2 dt = \int_{0}^{\pi} \frac{1}{2} r^2 dt = \int_{0}^{\pi} \frac{1}{2} (\sin^2(2t) + 4\sin^2t) dt = \frac{5\pi}{4}$

But the answer was not my thing. Actually the answer was $\frac{8}{3}$. So my question is What the point did I have a mistake? Why is the formula $\int_{0}^{\pi} \frac{1}{2}{f(\theta)}^2 d\theta$ not hold for this case?

Answer 1

In polar coordinates, $x = r \cos\theta, y = r\sin\theta$ where $\theta$ is the angle with the positive x coordinate axis. That is not the case here and hence the formula you write cannot be applied as is. You can check it yourself. For the $r$ you obtained, does $r \cos\theta ~$ give you $x$?

$y = 2\sin t$

$x = \sin2t \implies x^2 = 4 \sin^2 t (1 - \sin^2 t)$

$ ~ \displaystyle x^2 = y^2 - \frac{y^4}{4} ~ $ is the equation of the curve in cartesian coordinates.

That leads to $\displaystyle - \sqrt{y^2 - \frac{y^4}{4}} \leq x \leq \sqrt{y^2 - \frac{y^4}{4}}$

So the integral to find area can be written as,

$ 2 \displaystyle \int_0^2 \sqrt{y^2 - \frac{y^4}{4}} ~ dy$

Answer 2

The angle $\theta$ you need for the area is the geometric angle on the graph. For this angle, $\cot\theta=x/y=\cos t$. Use that to write $r$ as a function of $\theta$ instead.

Answer 3

Your formula for the computation of the area in polar coordinates if fine, but you did not express your curve in such coordinates. Note that\begin{align}\bigl(\sin(2t),2\sin(t)\bigr)&=2\sin(t)\bigl(\cos(t),1\bigr)\\&=2\sin(t)\sqrt{1+\cos^2(t)}\left(\frac{\cos(t)}{\sqrt{1+\cos^2(t)}},\frac1{\sqrt{1+\cos^2(t)}}\right).\end{align}Note also that each point of your curve belongs to the upper half-plane, and that therefore if you express the curve in polar coordinates, you will be able to express it as $r(\theta)\bigl(\cos(\theta),\sin(\theta)\bigr)$ with $\theta\in[0,\pi]$. And if $\theta\in[0,\pi]$ is such that$$\tan(\theta)=\frac{\frac1{\sqrt{1+\cos^2(t)}}}{\frac{\cos(t)}{\sqrt{1+\cos^2(t)}}}=\sec(t),$$you actually have $\theta\in\left[\frac\pi4,\frac{3\pi}4\right]$ (since $\tan(\theta)=\sec(t)\in(-\infty,-1]\cup[1,\infty)$). So, for such a $\theta$, since$$\left(\frac{\cos(t)}{\sqrt{1+\cos^2(t)}},\frac1{\sqrt{1+\cos^2(t)}}.\right)=\pm\bigl(\cos(\theta),\sin(\theta)\bigr)\tag2$$and since $\frac1{\sqrt{1+\cos^2(t)}},\sin(\theta)>0$, we must have the $+$ sign in $(2)$. So,$$\bigl(\sin(2t),2\sin(t)\bigr)=2\sin(t)\sqrt{1+\cos^2(t)}\bigl(\cos(\theta),\sin(\theta)\bigr).\tag1$$Now, since$$\sqrt{1+\cos^2(t)}=\frac1{\sin(\theta)},$$and since\begin{align}\sin(t)&=\sqrt{\sin^2(t)}\\&=\sqrt{1-\cos^2(t)}\\&=\sqrt{1-\frac1{\tan^2(\theta)}}\\&=\sqrt{1-\frac{\cos^2(\theta)}{\sin^2(\theta)}}\\&=\frac{\sqrt{\sin^2(\theta)-\cos^2(\theta)}}{\sin(\theta)}\\&=\frac{\sqrt{-\cos(2\theta)}}{\sin(\theta)},\end{align}then $(1)$ becomes$$\bigl(\sin(2t),2\sin(t)\bigr)=2\frac{\sqrt{-\cos(2\theta)}}{\sin^2(\theta)}\bigl(\cos(\theta),\sin(\theta)\bigr),$$and therefore the function $r$ is$$\begin{array}{rccc}r\colon&\left[\frac\pi4,\frac{3\pi}4\right]&\longrightarrow&\Bbb R\\&\theta&\mapsto&2\frac{\sqrt{-\cos(2\theta)}}{\sin^2(\theta)}.\end{array}$$And it turns out that, indeed,$$\int_{\pi/4}^{3\pi/4}\frac12r^2(\theta)\,\mathrm d\theta=\frac83.$$

Answer 4

Your mistake was already pointed out in other answers. You could also have used Green's theorem: $$ \oint_C L dx + M dy = \iint_D(M'_x-L'_y) dx dy. $$

If you take, in particular, $L = -y, M=0$, you know that $$ \underbrace{\iint_D 1 dx\,dy}_{\textrm{= area}}=\oint_C-y\,dx $$

Now,

\begin{align*} -\oint_C y dx = -\int_0^{\pi} 2 \cos(2t) \cdot 2 \sin t dt= \cdots = \frac 83 \end{align*}