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FP2 Polar Graph Area

Screenshot_2020-03-13-00-57-46.png

Shouldn't the enclosed area be multiplied by 2 because there are two petals of sin4θ
that are enclosed by sin2θ in 0<θ<π/2
Original post by Roha125


Shouldn't the enclosed area be multiplied by 2 because there are two petals of sin4θ
that are enclosed by sin2θ in 0<θ<π/2


No because the second petal arises from θ∈(5π4,6π4)\theta \in (\frac{5\pi}{4},\frac{6\pi}{4}) which is when sin⁡(4θ)<0\sin(4\theta) < 0 hence rr goes pointing into the opposite direction, hence the appearance of a 'second' petal in the first quadrant instead.
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Roha125
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Original post by RDKGames
No because the second petal arises from θ∈(5π4,6π4)\theta \in (\frac{5\pi}{4},\frac{6\pi}{4}) which is when sin⁡(4θ)<0\sin(4\theta) < 0 hence rr goes pointing into the opposite direction, hence the appearance of a 'second' petal in the first quadrant instead.

Sorry I am not getting it
Sin4t has 8 petals and 2 are in first quadrant so two enclosed area
Screenshot_2020-03-13-01-30-16.png187.7KB
Original post by Roha125
Sorry I am not getting it
Sin4t has 8 petals and 2 are in first quadrant so two enclosed area


Restrict θ\theta be between 00 and π2\dfrac{\pi}{2} for your polar curves ... right now they are between 0 and 2pi !
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Roha125
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Original post by RDKGames
Restrict θ\theta be between 00 and π2\dfrac{\pi}{2} for your polar curves ... right now they are between 0 and 2pi !

Oh,thank you so much
I got it now

Can you tell how me how should I know which petal appears first when I increase range from pi/4 to pi/2 to pi
Is there a sequence or should i memories
Can I know my putting values of theta in sin
Original post by Roha125
Oh,thank you so much
I got it now

Can you tell how me how should I know which petal appears first when I increase range from pi/4 to pi/2 to pi
Is there a sequence or should i memories
Can I know my putting values of theta in sin


When theta is between pi/4 and pi/2 (first quadrant), we have r = sin(4theta) < 0 [the fact that it's negative is important here] so the petal will be drawn in the third quadrant instead.

When theta is between pi/2 and 3pi/4 (second quadrant), we have r = sin(4theta) > 0. So the petal will be in the same quadrant.

When theta is between 3pi/4 and pi, we have r < 0. So the petal will be drawn in the opposite (fourth) quadrant.

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