Help with area between polar curves: √2 sin2θ = 1

[chahud]

On a quiz my professor gave me the problem of finding the area between the two polar curves r=√2sin2θ and r=1 (if you graph it, we were looking for the area outside of under the polar sine function). I got a 9/10 on the problem but dont know what I did wrong cause he just crossed out the answer and took a point off. Here's my solution, I'd appreciate it if you can tell me what I did wrong. First I solved for the interception points by setting them equal to eachother.
√2sin2θ=1
2sin2θ=1
sin2θ=½
2θ=π/6 and 5π/6
θ=π/12 and 5π/12
Then I set up and solved the integral
∫½(2sin2θ - 1)dθ
∫sin2θ-½dθ
-(cos2θ)/2-½θ
Then I plugged in the limits of integration
(-cos(π/6)/2-½·π/12)-(-cos(5π/6)/2-½·5π/12)
and everything simplified down to
(-√(3)/4-π/24)-(-√(3)/4-5π/24)
which equals π/6

Where did I go wrong? Thanks!

 

[HallsofIvy]

The first graph is a "four leafed rose". The second is, of course, a circle so the problem is to find the area outside the circle and inside the rose. Using symmetry, I would calculate the area inside one leaf of the rose, outside the circle, then multiply by 4. It looks to me like you calculated only the area inside one leaf, outside the circle, but did not multiply by 4.

 

[Steven G]

On a quiz my professor gave me the problem of finding the area between the two polar curves r=√2sin2θ and r=1 (if you graph it, we were looking for the area outside of under the polar sine function). I got a 9/10 on the problem but dont know what I did wrong cause he just crossed out the answer and took a point off. Here's my solution, I'd appreciate it if you can tell me what I did wrong. First I solved for the interception points by setting them equal to eachother.
√2sin2θ=1
2sin2θ=1
sin2θ=½
2θ=π/6 and 5π/6
θ=π/12 and 5π/12
Then I set up and solved the integral
∫½(2sin2θ - 1)dθ
∫sin2θ-½dθ
-(cos2θ)/2-½θ
Then I plugged in the limits of integration
(-cos(π/6)/2-½·π/12)-(-cos(5π/6)/2-½·5π/12)
and everything simplified down to
(-√(3)/4-π/24)-(-√(3)/4-5π/24)
which equals π/6

Where did I go wrong? Thanks!

When you square a product you just don't square one factor, you square both. Think about it, how do you know which factor to square? In particular (see red above), [√2sin2θ]² \(\displaystyle \neq\) 2sin2θ, in fact, [√2sin2θ]² = 2sin²2θ

 

[chahud]

The first graph is a "four leafed rose". The second is, of course, a circle so the problem is to find the area outside the circle and inside the rose. Using symmetry, I would calculate the area inside one leaf of the rose, outside the circle, then multiply by 4. It looks to me like you calculated only the area inside one leaf, outside the circle, but did not multiply by 4.

The picture we got on the quiz looked more like a "two leafed rose" on an angle, with one "leaf" in the first quadrant, and one in the third. our goal was to calculate the area in the first quadrant under the polar graph and over the circle. Idk if that was a sufficient explanation but hopefully it got the point across, so that's why it looks like I calculated the area inside one leaf outside the circle. Because that's what i was trying to do, hah.

 

[chahud]

When you square a product you just don't square one factor, you square both. Think about it, how do you know which factor to square? In particular (see red above), [√2sin2θ]² \(\displaystyle \neq\) 2sin2θ, in fact, [√2sin2θ]² = 2sin²2θ

I don't think you understood the question right, although I could just be totally wrong as well. The function we were given was √(2sin2θ) where the whole thing is under the root. In that case I do believe squaring it would just eliminate the radical. I was trying to find a decent program to write out the equations as they appeared on the quiz but the best I could find was that keyboard, and it makes things confusing. I should've been more clear with this, sorry about that! If I'm still wrong let me know.

 

[chahud]

Heres a picture of the problem and my work to clear anything problems with the question up.

 

[chahud]

Bump

Bump

 

[Dr.Peterson]

Heres a picture of the problem and my work to clear anything problems with the question up.View attachment 9399

The picture helps a lot, though it could be much easier to read. I see that you originally stated the equation without using necessary parentheses, which delayed the process of getting an answer. You wrote r=√2sin2θ when you meant r=√(2sin2θ).

But did you check your work line by line? You made at least two mistakes that I can see: you evaluated the definite integral in the wrong order, taking F(pi/6) - F(5pi/6) instead of F(5pi/6) - F(pi/6); and you made a sign error later so that the two cosines cancel out when they shouldn't.

Just go through your work carefully, asking yourself everywhere whether what you wrote agrees with the line above, and you should find the errors.