Hi people! I came across this problem while writing an entrance test to CMI (Chennai Mathematical Institute) today. The question is to find the area between the two curves

\( \displaystyle x^2 + y^2 = 144 \) and \( \displaystyle \sin(2x + 3y) \le 0 \).

Well, it can be done, definitely, by "brute force" method - what we usually do to find area between two curves. But that would be insanely complicated! So is there a "nice" ("nice" is subjective - I know) way to solve this problem?

Just to share what I started off with:

\( \displaystyle \sin(2x + 3y) \le 0 \) \( \displaystyle \implies (2n+1) \pi \le 2x + 3y \le (2n+2) \pi \) where \( n \) is obviously an integer. Also, the tangents to the circle with the slope \( -\dfrac{2}{3} \) are \( 2x + 3y = \pm 12\sqrt{13} \). We are interested in the region between these two tangents. So, to obtain the lower value of \( n \), we have:

\( \displaystyle (2n_{min} + 1) \pi) \ge -12\sqrt{13} \implies n = -7 \)

To obtain the upper value of \(n\), we have: \( \displaystyle (2n_{max} + 2) \pi) \le 12\sqrt{13} \implies n = +5 \)

So, the area we have to find is that of the intersection of the circle and the "strips" given by the region \( \displaystyle \implies (2n+1) \pi \le 2x + 3y \le (2n+2) \pi \), where \( n = -7, -6, \ldots, 5 \). Let's just expand this to get a better idea:

The "strips" are:

- \( \displaystyle \implies -13 \pi \le 2x + 3y \le -12 \pi \)
- \( \displaystyle \implies -11 \pi \le 2x + 3y \le -10 \pi \)
- \( \displaystyle \implies -9 \pi \le 2x + 3y \le -8 \pi \)
- \( \displaystyle \implies -7 \pi \le 2x + 3y \le -6 \pi \)
- \( \displaystyle \implies -5 \pi \le 2x + 3y \le -4 \pi \)
- \( \displaystyle \implies -3 \pi \le 2x + 3y \le -2 \pi \)
- \( \displaystyle \implies -1 \pi \le 2x + 3y \le 0 \pi \)
- \( \displaystyle \implies 1 \pi \le 2x + 3y \le 2 \pi \)
- \( \displaystyle \implies 3 \pi \le 2x + 3y \le 4 \pi \)
- \( \displaystyle \implies 5 \pi \le 2x + 3y \le 6 \pi \)
- \( \displaystyle \implies 7 \pi \le 2x + 3y \le 8 \pi \)
- \( \displaystyle \implies 9 \pi \le 2x + 3y \le 10 \pi \)
- \( \displaystyle \implies 11 \pi \le 2x + 3y \le 12 \pi \)

I think (note: here I don't have any solid justification), that, the area between the circle and strip number #13 (that is, \( \displaystyle \implies 11 \pi \le 2x + 3y \le 12 \pi \) ) is the same as the area between the circle and \( \displaystyle \implies -12 \pi \le 2x + 3y \le -11 \pi \). My reason to think so: The strips are kind of "symmetrical" about the \(6.5^{\text{th}}\) strip. I mean, the half of the \(6^{\text{th}}\) strip. So, we can move the strips "above" it, and "place" them in the "gaps" below the strip of symmetry. So by "moving" all the strips above the strip of symmetry, we get a continuous, "thicker" strip, which is easier (relatively) to integrate.

In the end ( if you have read till here :P ), I have to say, I couldn't actually find the area. That's why the question is here!

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## Comments

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TopNewestWell look at this argument ( I am not writing the complete solution) :

For every point in the circle \( (x, y) \) satisfying the given equations, there exists a point \((-x,-y)\) that lies in the circle and doesn't satisfy the equations. And the vice-versa is also true.

So the answer is simply half the circle's area, I.e., \( 72 \pi\)

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This seems too simple (not saying incorrect) to believe! Seriously! I spent so much time over it, typing it here, and writing it in the paper, and here you're dusting it off in one line. Man this is awesome!

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So how did the entrance test go, Parth?

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And you might not worry about the case when \(sin (2x+3y) =0\), because it resembles sets of lines, after all, and wouldn't affect the area. :D

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What were the other questions in that you found interesting?

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Will surely be adding them soon!

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Here's one: A fun problem - Find the formula of number of functions from a power set to another set

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Here's another: And you thought limits were always easy

I'll post others sometime later. Thanks for showing interest :D

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I couldn't stop myself from posting this one: Polynomials? That sounds familiar

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hey buddy i also wrote the same examination. in the question sin(2x+3y)<=0 you can write it as 2x+3y<=0.since if you take sin inverse on both sides you get the above equation and hence can find the area easily

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Consider the point \( (\pi/2, 0) \). Does this satisfy \( \sin(2x+3y) \le 0 \)? Yes. Does it satisfy \( 2x + 3y \le 0 \)? No. You can't write that way.

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you can write it that way.if you are that unsure you can check it out in that new pattern iit by arihant. it has the same question but with different data.

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yup buddy the ans is 72pie, yhe same question with a diff version came in ISI , its just a straight line passing thru (0,0) also the centre.

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I Cleared bothe Cmi & ISI

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