Finding the Arclength of an Astroid Curve

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The discussion focuses on finding the arclength of an astroid curve defined by the parametric equations f(θ) = 6cos³(θ) and g(θ) = 6sin³(θ). Participants are troubleshooting an integral for the arclength, which is expressed as ∫ from 0 to 2π of the square root of the sum of the squares of the derivatives of f and g. One user suggests simplifying the integrand to 18|sin(θ)cos(θ)| and integrating from 0 to π/2, then multiplying the result by 4 to obtain the total length. There is a caution about ensuring the integration is performed in radians, as using degrees could lead to incorrect results. The correct total length of the astroid is ultimately confirmed to be 36.
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If f(\theta) is given by:f(\theta) = 6cos^3(\theta) and g(\theta) is given by:g(\theta) = 6sin^3(\theta)
Find the total length of the astroid described by f(\theta) and g(\theta).
(The astroid is the curve swept out by (f(\theta),g(\theta)) as \theta ranges from 0 to 2pi )

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

this is asking for arclength right?
my integral:

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

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

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

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

Do not forget that

\sqrt{(\sin\theta)^2(\cos\theta)^2}=|\sin\theta\cos\theta|

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

ehild
 
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.
 
From 0 to pi/2 i and "Mathematica" got 9. Thus the total length is 9*4=36. Agree?
 
awesome thanks
 
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

\int_{0}^{2\pi}18|\sin\theta\cos\theta|d\theta

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
 
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