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I have two functions $$r=2$$ $$r= 3+2sin\theta$$

and I want to find the area of the yellow region in the picture below.

The limits of the integral solving the equation must be $\theta=-\pi/6,7\pi/6$.

Can someone explain how to do this?

enter image description here

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    I found the answer as 11π/3-11√3 ? whats wrong ?2012-03-04

3 Answers 3

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I do not think you can express the area using only the limits of integration that you specify.

However, if

$\ \ A=$ the area of the region outside the circle and inside the cardioid (the pink region below),

$\ \ B=$ the area of the region inside the cardioid (pink and brown below)

and

$\ \ C=$ the area of the yellow region,

then $C=B-A$.

You can evaluate both $B$ and $A$ in polar coordinates.

The Cardioid is generated once, in the counterclockwise direction, as $\theta$ takes values from $0$ to $2\pi$. For any particular value of $\theta$ $r$ takes values from $0$ to $3+2\sin \theta$. So $$ B=\int_0^{2\pi}\int_0^{3+2\sin\theta} r dr\,d\theta $$

Note that the region $A$ is described by the set $$ \{ (r,\theta) : -\pi/6\le\theta\le 7\pi/6,\ 2\le r\le 3+2\sin\theta \}; $$ in particular the points of intersection of the circle with the cardioid are given when $\theta=-\pi/6$ and $\theta=7\pi/6$. (The maroon line segment below represents the typical $r$ range over the ray with angle $\theta$.)

We have: $$ A=\int_{-\pi/6}^{7\pi/6}\int_2^{3+2\sin\theta} r dr\,d\theta $$


Alternatively (and more simply), you can take the area of the circle, $4\pi$, and subtract the area of the green region below: $$ \int_{7\pi/6}^{11\pi/6}\int_{3+2\sin\theta}^2 r dr\,d\theta $$
enter image description here

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    Thank you. I had the same idea but I couldn't make the integral.2012-01-22
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    Did you calculate it? Is the result $18\pi$?2012-01-22
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    @tamakisnen According to Wolfram, the answer is ${19\pi\over3}-{11\sqrt3\over2}$.2012-01-22
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    I did again and I found $5\pi$.2012-01-22
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Hint :

  • Search the intersection of the two curves in polar coordinates (you should find your angles)
  • Draw the two lines joining the intersection points to the origine
  • Compute the surface at the upper part (first equation) and lower part (second equation) delimited by these two lines in polar coordinates (you get two 'pie charts)'.
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I suspect you will be using polar coordinates to solve and use an iterated integral. There are two different boundary curves: the circle and the whatchamacallit (lemniscate? one of those). You need to find out where the circle and the whatchamacallit intersect. So do that.

You think $-\pi/6$ is one of those spots. Let's see: $\sin \pi/6 = 1/2$, so $3 + 2 \sin \pi/6 = 2$. Great! That's where these bounds come in.

So you now set up your integral. For the arc from $-\pi/6$ to $7 \pi / 6$, you integrate the radius up to the whatchamacallit. For the rest, just $r$ from $0$ to $2$.

EDITED to respond to OP in comments

The OP states: Well, looking at the picture a possible solution might be $2\int_0^2\int_{-\pi/6}^0 2\sin\theta + \int_0^\pi r$

This is a very intriguing guess. I want to consider only the second term for a moment, $\int_0^\pi r \mathrm{d} \theta$. This evaluates to $\pi r$. Why is that $r$ there? We know the final answer should be a number, as there is no uncertainty here. So perhaps you should explain how you got at this (incomplete) answer.

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    I think it's "Cardioid" or "Limaçon".2012-01-22
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    I have problem constructing the correct integral. It is a Cardioid.2012-01-22
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    Oh, way to make me look bad... although to be honest, I think cardiod's have cusps. I don't know if this has a name other than a dimpled limacon. Or maybe I'm wrong? Of course, I mean whatchamacallit as a term of endearment. And my spellchecker doesn't mark it wrong! Whoa.2012-01-22
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    @tamakisnen: so what do your integrals look like?2012-01-22
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    They don't look like. The problem is that I am trying to do it having venn diagrams in mind.2012-01-22
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    @tamakisnen: I don't understand. How about you edit your question to include your work, and I'll work with you from there?2012-01-22
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    Well, looking at the picture a possible solution might be $2\int_0^2\int_{-\pi/6}^0 2sin\theta + \int_0^\pi r$2012-01-22
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    That was embarrassing. I didn't even write what I intended to write. Problem solved after the explanation from DavidMitra.2012-01-22