# Another parametric question

1. Apr 6, 2005

### ILoveBaseball

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.

Last edited: Apr 6, 2005
2. Apr 6, 2005

### Yegor

I suppose it's all right with your integral. What a result do You receive? In which way You integrated it?

3. Apr 6, 2005

### ehild

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

4. Apr 6, 2005

### ILoveBaseball

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 dont understand why. Also tried to integrate from 0 to 2pi and got 50.253217, but it wont take that either.

5. Apr 6, 2005

### Yegor

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

6. Apr 6, 2005

### ILoveBaseball

awesome thanks

7. Apr 6, 2005

### ehild

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