How Do You Calculate the Length of the Polar Curve \( r = 3 \sin \theta \)?

[Exeneva]

Homework Statement

r = 3 sin $$\vartheta$$

0 $$\leq$$ $$\vartheta$$ $$\leq$$ $$\pi$$/3

Homework Equations

Arc Length: $$\int$$ $$\sqrt{r^{2} + (dr/d\vartheta)^{2}}d\vartheta$$

The Attempt at a Solution

$$r^{2} = 9 (sin \vartheta)^{2} = 9 (1/2 - cos 2\vartheta/2)$$

$$r^{2} = 9/2 - 9/2 cos 2\vartheta$$

$$dr/d\vartheta = 3 cos \vartheta$$

$$\int (9/2 - 9/2 cos 2\vartheta + 3 cos \vartheta)^{1/2} d\vartheta$$
from 0 to \pi/3

Not sure how to integrate this.

 

[Mark44]

Homework Statement

r = 3 sin $$\vartheta$$

0 $$\leq$$ $$\vartheta$$ $$\leq$$ $$\pi$$/3

Homework Equations

Arc Length: $$\int$$ $$\sqrt{r^{2} + (dr/d\vartheta)^{2}}d\vartheta$$

The Attempt at a Solution

$$r^{2} = 9 (sin \vartheta)^{2} = 9 (1/2 - cos 2\vartheta/2)$$

$$r^{2} = 9/2 - 9/2 cos 2\vartheta$$

$$dr/d\vartheta = 3 cos \vartheta$$

$$\int (9/2 - 9/2 cos 2\vartheta + 3 cos \vartheta)^{1/2} d\vartheta$$
from 0 to \pi/3

Not sure how to integrate this.

You really made it hard on yourself.

$$r^{2} = 9 sin^2( \theta)$$
$$(dr/d\theta)^2 = (3 cos(\theta))^{2} = 9cos^2(\theta)$$