How do you find the area between the loops of #r=2(1+2sintheta)#?

Question

Evelyn Agard · Accepted Answer

#r = sqrt( x^2 + y^2 )>= 0. So, there is only one loop.

See graph.

#0 <= r = 2 ( 1 + 2 sin theta ) rArr sin theta >= - 1/2# #rArr theta notin (7/6pi, 11/6pi )# . graph{x^2+y^2-2sqrt(x^2+y^2)-4y=0[-14 14 -7 7]} Of course, just as an exercise, r-negative loop can be inserted, and the area in between can be evaluated..

Charles Arocha · Answer

undefined To find the area between the loops of the polar curve $ r = 2(1 + 2\sin(\theta)) $, you need to determine the values of $ \theta $ where the loops intersect and then integrate the difference between the outer and inner curves with respect to $ \theta $.

First, find the intersection points by solving $ r = 2(1 + 2\sin(\theta)) = 0 $. This yields $ \sin(\theta) = -\frac{1}{2} $, which occurs at $ \theta = \frac{7\pi}{6} $ and $ \theta = \frac{11\pi}{6} $.

Then, integrate the difference between the outer curve $ r_1 = 2(1 + 2\sin(\theta)) $ and the inner curve $ r_2 = 0 $ from $ \theta = \frac{7\pi}{6} $ to $ \theta = \frac{11\pi}{6} $:

$$ A = \frac{1}{2} \int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}} (r_1^2 - r_2^2) \, d\theta $$

$$ = \frac{1}{2} \int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}} \left( 4(1 + 2\sin(\theta))^2 - 0 \right) \, d\theta $$

$$ = \frac{1}{2} \int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}} (4 + 16\sin(\theta) + 16\sin^2(\theta)) \, d\theta $$

$$ = \frac{1}{2} \int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}} (4 + 16\sin(\theta) + 16(1 - \cos^2(\theta))) \, d\theta $$

$$ = \frac{1}{2} \int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}} (20 + 16\sin(\theta) - 16\cos^2(\theta)) \, d\theta $$

Now, integrate term by term and evaluate the integral from $ \frac{7\pi}{6} $ to $ \frac{11\pi}{6} $ to find the area between the loops.

HIX Tutor · Answer

undefined To find the area between the loops of the polar curve $ r = 2(1 + 2\sin\theta) $, follow these steps:

1. **Find Points of Intersection**:
   - Set $ r = 0 $ to find when the curve crosses the pole:  
     $ 2(1 + 2\sin\theta) = 0 $  
     Solve for $ \sin\theta $:  
     $ 1 + 2\sin\theta = 0 $  
     $ \sin\theta = -\frac{1}{2} $  
   - The angles for $ \sin\theta = -\frac{1}{2} $ are:  
     $ \theta = \frac{7\pi}{6}, \frac{11\pi}{6} $.

2. **Set Up the Area Integral**:  
   Use the formula for area in polar coordinates:  
   $$ A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta $$  
   Here, $ r = 2(1 + 2\sin\theta) $ and $ \alpha = \frac{7\pi}{6}, \beta = \frac{11\pi}{6} $.

3. **Square $ r $**:  
   $$ r^2 = \left(2(1 + 2\sin\theta)\right)^2 = 4(1 + 2\sin\theta)^2 $$  
   Expand it:  
   $$ = 4(1 + 4\sin\theta + 4\sin^2\theta) $$  
   $$ = 4 + 16\sin\theta + 16\sin^2\theta $$

4. **Integrate**:  
   Set up the integral:  
   $$ A = \frac{1}{2} \int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}} \left(4 + 16\sin\theta + 16\sin^2\theta\right) \, d\theta $$

5. **Compute the Integral**:  
   - Break it into parts:  
     $$ = \frac{1}{2} \left( \int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}} 4 \, d\theta + \int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}} 16\sin\theta \, d\theta + \int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}} 16\sin^2\theta \, d\theta \right) $$
   - Compute each part. The first part is simple:  
     $$ \int 4 \, d\theta = 4\theta \Big|_{\frac{7\pi}{6}}^{\frac{11\pi}{6}} = 4 \left( \frac{11\pi}{6} - \frac{7\pi}{6} \right) = 4 \cdot \frac{4\pi}{6} = \frac{8\pi}{3} $$
   - The sine integrals can be computed using known values or trigonometric identities.

6. **Combine results**:  
   Sum the results of each part and multiply by $ \frac{1}{2} $.

7. **Final Result**:  
   Calculate to find the final area.

This will give you the area between the loops of the curve!