Area of polar graphs

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rousie313

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Need help setting up the integral to find the area of a singular "petal". The graph looks as those it only has 3 petals but my professor said it had 6. Maybe because of the way one petal is bisected by the x-axis? Anyway are the 3 petals that are formed equal? so they can be used to find the total area enclosed?
 
rousie313 said:
Need help setting up the integral to find the area of a singular "petal". The graph looks as those it only has 3 petals but my professor said it had 6. Maybe because of the way one petal is bisected by the x-axis? Anyway are the 3 petals that are formed equal? so they can be used to find the total area enclosed?
Click to expand...

What was your function?
 
Hello, rousie313!

While waiting for your clarification, I'll make some observations.

Need help setting up the integral to find the area of a singular "petal".

The graph looks as those it only has 3 petals . . The function would be: .\(\displaystyle r \:=\:a\cos3\theta\;(\text{or }a\sin3\theta)\)
. . but my professor said it had 6. . . The function would be: .\(\displaystyle r^2 \:=\:a\cos3\theta\)

Anyway, are the 3 petals that are formed equal? . Yes!
So they can be used to find the total area enclosed? . Yes!
Click to expand...
 
rousie313 said:
Sorry i forgot to include the function. It is r= cos(3?/2)
Click to expand...

As far as I can see - that will have 3 petals

To find the limits - decide on the values of ? where cos(3?/2) = 0 and where cos(3?/2) = ±1
 
Thank you that help but if i now look at r=sin(3?/2), two questions: 1.would the same method of finding the the limits work(setting the equation equal to 0 and 1)
2. by making it r^2=sin(3?/2), does it double the amount of petals via reflecting it over the y-axis?
 
Here's a graph of \(\displaystyle r^{2}=sin(\frac{3\theta}{2}) \;\ and \;\ sin(\frac{3\theta}{2})\)

The number of petals are the same.
 
That plot is confusing to me.

For example - why is it showing r=1 for ? = ?? What am I missing?
 
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