# Area of polar curve

1. Apr 1, 2013

### steel1

1. The problem statement, all variables and given/known data
Find the area enclosed by the graph r=2+sin(4θ)

2. Relevant equations
Area = .5∫r^2

3. The attempt at a solution
Area = .5∫(2+sin(4θ))^2
=.5(4.5θ-1/16sin(8θ)-cos(4θ))

I can do the integration and all, but I am having trouble finding the limits of integration

I set sin(4θ)=0 and solve, and get 0 and pi/4. but when i use these limits, i get the wrong answer.

2. Apr 1, 2013

### CompuChip

Have you tried plotting the curve (either using a calculator or computer, or by hand)?
That should give you a clue (hint: the correct answer is related to the period of r(theta)).

3. Apr 2, 2013

### haruspex

It is? i don't see that. Since r > 0 for all theta, the curve cannot join up until it has gone all the way around 2π.

4. Apr 2, 2013

### CompuChip

Exactly. So why the pi / 4?

5. Apr 3, 2013

### haruspex

I agree the pi/4 is wrong, but the period of r would be pi/2, which is also wrong.

6. Apr 3, 2013

### CompuChip

Ah, you are right, the problem was in my wording. Thanks!