What is the area enclosed by #r=2cos((2theta)/3+(5pi)/3)+4sin(theta/2+pi/4) -theta# between #theta in [0,pi]#?

Question

Emily Ammerman · Accepted Answer

See the answer below:

Christopher Allers · Answer

undefined To find the area enclosed by the curve $ r = 2\cos\left(\frac{2\theta}{3}+\frac{5\pi}{3}\right) + 4\sin\left(\frac{\theta}{2}+\frac{\pi}{4}\right) $ and the line $ \theta $ between $ 0 $ and $ \pi $, you can follow these steps:

1. Determine the points of intersection between the curve and the line $ \theta = \pi $ and $ \theta = 0 $.

2. Use the formula for the area enclosed by a polar curve:
$$ A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta $$

3. Integrate $ r^2 $ with respect to $ \theta $ from $ \alpha $ to $ \beta $, where $ \alpha $ and $ \beta $ are the angles of intersection.

Here are the detailed steps:

1. Find the points of intersection:
   - At $ \theta = 0 $, $ r = 2\cos\left(\frac{5\pi}{3}\right) + 4\sin\left(\frac{\pi}{4}\right) $.
   - At $ \theta = \pi $, $ r = 2\cos\left(\frac{2\pi}{3}+\frac{5\pi}{3}\right) + 4\sin\left(\frac{\pi}{2}+\frac{\pi}{4}\right) $.

2. Calculate the values of $ r $ at $ \theta = 0 $ and $ \theta = \pi $.

3. Determine the range of integration: $ \theta $ varies from $ 0 $ to $ \pi $.

4. Integrate $ r^2 $ with respect to $ \theta $ over the given range.

5. Calculate the area using the formula $ A = \frac{1}{2} \int_{0}^{\pi} r^2 \, d\theta $.

6. Evaluate the integral and find the area enclosed by the curve and the line.

7. The result will be the area enclosed by the curve and the line within the specified range of $ \theta $.