Finding the Arclength of an Astroid Curve

  • Thread starter ILoveBaseball
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In summary: I also used another math program on my computer to verify it. i integrated from 0 to pi/2 and got 9.558*4 = 38.232 but it's incorrect and i don't understand why. Also tried to integrate from 0 to 2pi and got 50.253217, but it won't take that either. Simplify your integrand. It becomes \int_{0}^{2\pi}18|\sin\theta\cos\theta|d\theta
  • #1
ILoveBaseball
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If [tex]f(\theta)[/tex] is given by:[tex]f(\theta) = 6cos^3(\theta)[/tex] and [tex]g(\theta)[/tex] is given by:[tex]g(\theta) = 6sin^3(\theta)[/tex]
Find the total length of the astroid described by [tex]f(\theta)[/tex] and [tex]g(\theta)[/tex].
(The astroid is the curve swept out by ([tex]f(\theta)[/tex],[tex]g(\theta)[/tex]) as [tex]\theta[/tex] ranges from 0 to 2pi )

[tex]f/d(\theta) = -18*cos(x)^2*sin(x)[/tex]
[tex]g/d(\theta) = 18*sin(x)^2*cos(x)[/tex]

this is asking for arclength right?
my integral:

[tex]\int_{0}^{2\pi}\sqrt{(-18*cos(\theta)^2*sin(\theta))^2+(18*sin(\theta)^2*cos(\theta))^2}[/tex]

anyone what's wrong with my integral? cause i keep getting the wrong answer.
 
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  • #2
I suppose it's all right with your integral. What a result do You receive? In which way You integrated it?
 
  • #3
ILoveBaseball said:
my integral:

[tex]\int_{0}^{2\pi}\sqrt{(-18*cos(\theta)^2*sin(\theta))^2+(18*sin(\theta)^2*cos(\theta))^2}[/tex]

anyone what's wrong with my integral? cause i keep getting the wrong answer.

Do not forget that

[tex]\sqrt{(\sin\theta)^2(\cos\theta)^2}=|\sin\theta\cos\theta|[/tex]

Integral from 0 to pi/2 and multiple the result by 4.

ehild
 
  • #4
i used my calculator to integrate my function. I also used another math program on my computer to verify it. i integrated from 0 to pi/2 and got 9.558*4 = 38.232 but it's incorrect and i don't understand why. Also tried to integrate from 0 to 2pi and got 50.253217, but it won't take that either.
 
  • #5
From 0 to pi/2 i and "Mathematica" got 9. Thus the total length is 9*4=36. Agree?
 
  • #6
awesome thanks
 
  • #7
ILoveBaseball said:
i used my calculator to integrate my function. I also used another math program on my computer to verify it. i integrated from 0 to pi/2 and got 9.558*4 = 38.232 but it's incorrect and i don't understand why. Also tried to integrate from 0 to 2pi and got 50.253217, but it won't take that either.

Simplify your integrand. It becomes

[tex]\int_{0}^{2\pi}18|\sin\theta\cos\theta|d\theta[/tex]

Integral from 0 to pi/2, multiply by 4. The result should be 36.

You could have made the mistake with your programs that you did not set to radians and the program calculated with degrees.

ehild
 

FAQ: Finding the Arclength of an Astroid Curve

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