Question

Help. I dont know how to figure out the bounds

Answer 1

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Answer 2

It has been awhile since I have done this
but I think we need to find when r is 0

Answer 3

Yeah. Im confused how to go about it though

Answer 4

replace r with 0 and solve for theta

Answer 5

you can use unit circle

Answer 6

is pi/6 one?

Answer 7

yep

Answer 8

what about the other?

Answer 9

hint quadrant 2

Answer 10

second quadrant would be 3pi/2 but i dont think that would be right

Answer 11

im not the most comfortable with trig functions

Answer 12

3pi/2 isn't in the second quadrant

Answer 13

3pi/2 is between the 3rd and 4th quadrant

Answer 14

you are looking for where the y-coordinate is 1/2

Answer 15

you seen this at pi/6 which was in the 1st quadrant

Answer 16

and you should see this at 5pi/6 which is in the 2nd quadrant

Answer 17

Ok. I was alos thinking because of pi/6 just adding pi/2 to it to get where it would be in the 4th quadrant but that would give me 4pi/6

Answer 18

so it would be1/2 the integral from5pi/6 to pi/6 of (1-2sin(theta))^2

Answer 19

pi/6+pi/2 is 4pi/6 or reducing 2pi/3 which is in the 2nd quadrant not 4th

Answer 20

from pi/6 to 5pi/6

Answer 21

yeah i was thinking about that after you said my first guess was wrong

Answer 22

thank you :)