# Homework Help: Area of polar curves

1. Jul 14, 2012

### knv

1. Given the curves r = 2sin(θ) and r = 2sin(2θ), 0≤θ≤π/2, find the area of the region outside the first curve and inside the second curve

2.not sure which equations to use

3. I got 1 and 1/2 as the area and they were wrong. I do not really know how to work this problem. A detailed step by step explanation would be really helpful!

2. Jul 14, 2012

### tiny-tim

hi knv! welcome to pf!

show us your full calculations, and then we'll see what went wrong, and we'll know how to help!

3. Jul 14, 2012

### knv

Well I dont exactly know how to even start this problem but lets see.

∫2sin(2θ) - ∫2 sin(θ)

do the bounds change for the first one to π/4? Any hints on how to even start this problem. Im not looking for an answer. I do not understand my books explanation so Im really lost

4. Jul 14, 2012

### LCKurtz

Generally the area inside a polar graph is $\int_a^b \frac 1 2 r^2\, d\theta$ between appropriate limits $a$ and $b$. If you want the area between two graphs the integrand would be $\frac 1 2( r_{outer}^2-r_{inner}^2)$. So you should sketch the two curves in the first quadrant and determine where they intersect and what the requestion region looks like. Then use the formula.

5. Jul 14, 2012

### knv

∫sin2(θ) - ∫ sin2(2θ)

since sin2 = 1/2(1-cos 2θ) we plug it in and bring the 1/2 to the front of each integral giving us..

1/2∫(1-cos 2θ) - 1/2∫(1-cos 4θ)

am I heading in the right direction? how do I find the bounds?

6. Jul 14, 2012

### LCKurtz

You follow my suggestion and plot the two curves in the first quadrant. How else will you know which is the inner and outer curve and what the area looks like? To find where they intersect set the $r$'s equal and find out which $\theta$ in the first quadrant does that.

7. Jul 15, 2012

### knv

Im still having trouble with this problem. Can anyone help? Please Im so lost. Just need some help on setting it up and getting it started

8. Jul 15, 2012

### Staff: Mentor

You've been given good advice about sketching the graphs of the two functions. Have you done that yet?

9. Jul 15, 2012

### knv

Yeah I did that. And the inner one is 2 sin θ and the outer one is 2 sin 2θ

They cross at 0 and π/4

Correct?

10. Jul 15, 2012

### knv

can I solve for half of the area using only one of the functions and then doubling it?

11. Jul 15, 2012

### knv

I got the answer 1/4. Does anyone know if that is correct before I use my last attempt?

12. Jul 15, 2012

### LCKurtz

No, why would you think that unless you are just guessing?

If you know the outer curve and inner curve and where they cross, you know everything you need to know to use the formula I gave you in post #4. Try it.

13. Jul 15, 2012

### knv

No i had watched a video where a guy did that so I wasn't sure if it would work in this case.

I worked it out doing what you told me to do and came up with the answer 1/4.

Can you tell me if that is correct?

∫0->π/4 = ∫1/2 (4sin2 2θ) dθ -∫1/2(2 sin 2θ) dθ

2∫(1/2)(1-cos4θ)dθ - ∫(1/2)(1-cos 2θ) dθ

[(θ-sin4θ)/4] - 1/2[(θ-sin 2θ)/2]

(π/4-0)/4) - 1/2[(π/4 - 4/4)/2]

so it would end up being

π/16 - π/16 +1/4

14. Jul 15, 2012

### LCKurtz

You are close. The answer should be 1/2, so you are off by a factor of 2.
Are you sure you got everything squared that is supposed to be in your setup?

15. Jul 15, 2012

### knv

I had found 1/2 earlier and it was counted wrong.

16. Jul 15, 2012

### LCKurtz

I didn't notice you had the intersection point wrong. Please show me the steps how you got $\pi/4$.

17. Jul 16, 2012

### SammyS

Staff Emeritus
No.

They intersect at θ = 0 and θ = π/3 .