Question

Consider a polar graph for the equation \( r = 2cos2\theta \). The range of \( \theta \) is \(0 \le \theta < 2\pi \).
The question asks me to calculate the exact value of area for one loop.

My confusions are as follows.
1. What is meant by loop here?
2. If I integrate the equation, using the limits \( 0\) and \( \frac{\pi}{2}\), I get 0 as answer where as I get a correct answer by using limits \( \frac{\pi}{4} \) and \( \frac{3\pi}{4}\). Why is the first limit incorrect?

Answer 1

@Irishboy123 @ganeshie8 @Kainui

Answer 2

Woah, I'm sorry I can't help with this, it looks like a foreign language to me :/

Answer 3

Lol, no problem

Answer 4

graph it
it will form a loop
find the area inside it

Answer 5

\(\color{#0cbb34}{\text{Originally Posted by}}\) @spongebob.
graph it
it will form a loop
find the area inside it
\(\color{#0cbb34}{\text{End of Quote}}\)
Don't make explanation looks like I have already did it?

Answer 6

\(\color{#0cbb34}{\text{Originally Posted by}}\) @spongebob.
graph it
it will form a loop
find the area inside it
\(\color{#0cbb34}{\text{End of Quote}}\)
Doesn't my explanation looks like I have already did it?

Answer 7

@zepdrix

Answer 8

well the loop according to the acc is this ig http://prntscr.com/cae9mx

but the thing is -that loop is not a part of our graph hmm

Answer 9

\(\color{#0cbb34}{\text{Originally Posted by}}\) @spongebob.
well the loop according to the acc is this ig http://prntscr.com/cae9mx

but the thing is -that loop is not a part of our graph hmm
\(\color{#0cbb34}{\text{End of Quote}}\)
I meant what loop is meant here..

Answer 10

For graphs of polar curves, you can think of a loop as referring to a part of the curve \(r(\theta)\) for which \(r(\theta_1)=r(\theta_2)\) with \(\theta_1\lt\theta_2\) such that \(|\theta_1-\theta_2|\) is minimized. This would explain why integrating over \(\left[0,\dfrac{\pi}{2}\right]\) doesn't give the answer you expect, because \(2\cos0=2\) while \(2\cos\dfrac{\pi}{2}=0\). Meanwhile, \(\left[\dfrac{\pi}{4},\dfrac{3\pi}{4}\right]\) does work because \(2\cos\dfrac{\pi}{2}=2\cos\dfrac{3\pi}{2}=0\).

You can try finding all points that *potentially* constitute a loop by solving \(r(\theta_1)=r(\theta_2)\):
\[\begin{align*}
2\cos2\theta_1&=2\cos2\theta_2\\[1ex]
\cos2\theta_1&=\cos2\theta_2
\end{align*}\]which has infinitely many solutions for either variable, e.g. \(\theta_1=0\) and \(\theta_2=2\pi\).

Given the shape of this particular curve, a loop would refer to any one "petal" of the "rose". (See http://mathworld.wolfram.com/Rose.html for more info.)

Since the curve passes through the origin several times, a natural choice for \(\theta_1\) would be any angle that returns \(2\cos2\theta_1=0\), and \(\theta_1=\dfrac{\pi}{4}\) fits the bill. Then \(\theta_2\) can have infinitely many values that also return \(2\cos2\theta_2=0\), but only one that minimizes the distance between it and \(\theta_1\).
\[\cos2\theta_2=\cos\dfrac{2\pi}{4}=\cos\dfrac{\pi}{2}=0\implies\theta_2=\frac{1}{2}\left(n\pi+\frac{\pi}{2}\right)\quad\color{lightgray}{\text{where }n\in\mathbb{Z}}\]Setting \(n=1\) satisfies our requirements, so \(\theta_2=\dfrac{3\pi}{4}\) as the upper limit will give you the correct area.

Answer 11

I assume you plotted the it?
From wolfram
for theta from 0 to 2 pi gives a clover leaf pattern

but for theta from 0 to pi/2 , we don't get a closed loop

Answer 12

Think this is my bad, i was thinking 8 of these:

so 8 of these:

https://www.wolframalpha.com/input/?i=%5Cint_(0)%5E(pi%2F4)++1%2F2+*+4+cos%5E2+(2t)++dt

......as i was guessing that a loop means all the way around the entire clover.

maybe it means one leaf of the clover in which case it is only 2 of those.

read holster and phi.