Area of curve in parametric equation

Question

Given the curve defined by the parametric equations:

$$ x=7\cos{3t}\\ y=7\sin{3t}\\ 0\le t\le2\pi $$

What is the area of the region bounded by this curve?

Clearly,

$$ x^2+y^2=(7\cos{3t})^2+(7\sin{3t})^2=7^2 $$

which is a circle centered at the origin and radius 7. Therefore, the area is $\pi\cdot7^2=49\pi$.

However, it is also true that the area is

$$ A=\int{ydx}=\int_{0}^{2\pi}{y(t)x'(t)dt}=\int_{0}^{2\pi}{7\sin{3t}(-7\sin3t)3dt}=-49\int_{0}^{2\pi}{\sin^2{3t}\;3dt}=-49\cdot(3\pi)=-3(49\pi) $$

which is the same area as before, but multiplied by $-3$.

What is answer is correct? The $3t$ in the $\sin$ and $\cos$ terms means that as $t$ goes from $0$ to $2\pi$, the circle is traversed $3$ times, right? Which would explain why the area is multiplied by 3. But what about the minus sign?

Answer 1

Either you change to polar coordinates or else you use cartesian ones. If it is as you wrote it then the parametrization is

$$-7\le x\le 7\;,\;\;-\sqrt{49-x^2}\le y\le\sqrt{49-x^2}$$

and thus

$$\oint_Cy\,dx=\int_{-7}^72\sqrt{49-x^2}\,dx=28\int_0^7\sqrt{1-\left(\frac x7\right)^2}=196\int_0^{\pi/2}\cos^2t\,dt=$$

$$=\left.98(t+\cos t\sin t)\right|_0^{\pi/2}=49\pi$$

Answer 2

In your second way of computing it, you have gone around the circle three times instead of one; thus you have (erroneously) tripled the answer. Also, there is a difference between clockwise and counterclockwise and you clearly got unlucky there, to get a negative area.

Answer 3

Looks like a circle to me: centre $(0,0)$ and radius $7$ so area $49\pi$.

But the path winds around it three times. So, do you count the area of the circle once, or three times? You pays your money and you takes your choice. But since you are going around anticlockwise, the traditional positive direction, you must demand a positive area.

Answer 4

The main cause of the difficulty starts here:

$$ A \stackrel?= \int y\,dx. $$

This doesn't become the area of a circle until you set the appropriate bounds on the integral (making it a definite integral). An indefinite integral is not a number, it's a family of functions. Is the area of a circle of radius $7$ a family of functions?

If you express $y$ as a function of $x,$ that is, $y = \sqrt{49 - x^2},$ then it is legitimate to write something like this:

$$ A = 2 \int_{-7}^7 y\,dx. $$

This supplies the bounds of integration, so we have a definite integral that will evaluate to some particular number. The factor $2$ comes in because the integral only measures the area between the $x$-axis and the curve $y = \sqrt{49 - x^2},$ that is, it only measures the half of the disk above the $x$-axis. To get the full area of the disk you have to double the integral. (Alternatively, you could integrate $2y,$ which is the vertical distance between the lower arc of the circle and the upper arc, and get the same answer.)

Now, observing that $x = -7$ when $t = \frac\pi3,$ that $x = 7$ when $t = 0,$ and that $x$ increases continuously from $-7$ to $7$ as $t$ decreases from $\frac\pi3$ to $0,$ we can make the substitution as follows:

\begin{align} A &= 2 \int_{-7}^{7} y\,dx \\ &= 2 \int_{\pi/3}^{0} (7 \sin (3t)) (-7 \sin(3t)) (3\,dt) \\ &= 6 \cdot 7^2 \int_0^{\pi/3} \sin^2 (3t) \,dt \\ &= 6 \cdot 7^2 \cdot \frac\pi6 \\ &= 49 \pi. \end{align}

Notice that in order to make a correct substitution, the first integral over must "start" at the value of $t$ that corresponds with the "starting" value of $x$ in the integral over $x$; and when we swap the upper and lower bounds of the integral in order to integrate in the usual direction (increasing $t$), the swap negates the sign of the integral (and in this case, it cancels the negative sign).

Answer 5

Parametric curves are just screaming out to be solved in the complex plane. Consider that

$$ z=x+iy=7(\cos 3t+i\sin 3t)=7e^{i3t}\\ A=\frac{1}{2}\int\mathfrak{Im}\{z^* \dot z\}\ dt $$ $$ z^*=7e^{-i3t}\\ \dot z=21ie^{-i3t}\\ \mathfrak{Im}\{z^* \dot z\}=147 $$

Therefore

$$A=\frac{147}{2}\int_0^{2\pi}dt=147\pi$$

Or, if we wish to avoid the ambiguity of going around the circle three times,

$$A=\frac{147}{2}\int_0^{2\pi/3}dt=49\pi$$

Easy as $\pi$.