## anonymous 4 years ago HELP Find the area of the region that lies inside both curves r = 8sin(2theta), r = 8sin(theta)

let me guess, third year calculus?

2. anonymous

yeah why?

It was torture for me :) let me see if I can help, I need some scratch paper though

4. anonymous

ok thanks man its so tough

i can relate, gimmie a sec

6. anonymous

ok

have you done double integrals?

8. anonymous

Not yet

dang, okay gotta think of how to do this using single...

10. anonymous

we haven't gotten that far we are in chapter 9 with areas of regions and in curves

give me as much info as you can then

12. anonymous

well the question is just right there lol, idk what else you want

13. anonymous

we have done polar coordinates, taylor series , maclaurin series

okay good, you can't do this in rectangular, polar is your best, no only shot unless you want roots. it's also cleaner. I just need to remember how to do a single integral sadly. I haven't done Calc 3 in a long time

15. anonymous

Yeah I hear you, I'm looking at this problem and I don't even know where to start, if you can show me step by step what you do, that would be so helpful

$8\sin(2\theta)=8\sin(\theta)$ $\sin(2\theta)=\sin(\theta)$ $2\sin(\theta)\cos(\theta)=\sin(\theta)$ $2\sin(\theta)\cos(\theta)-\sin(\theta)=0$ $\sin(\theta)(2\cos(\theta)-1)=0$ $\sin(\theta)=0$ and $2\cos(\theta)-1=0$ Solve for theta and these are your limits of integration.

Notice how I cancelled the 8 but didn't cancel the sin(theta). You never want to cancel out anything that isn't a constant since it might be a potential solution.

18. anonymous

ok I see

Can you pick it up from there since you've done polar coordinates? I don't know if you've done integration using polar yet or not.

20. anonymous

so its 2 cos(theta) = 1, so cos(theta) = 1/2?

But believe me, you'll need a strong grip on polar since double and triple integrals build on that. Cylindrical is like a second layer to polar and spherical...well just be strong in polar and double and triple will be easier for you

Yes

23. anonymous

OK so i have sin(theta) = 0 and cos(theta) = 1/2

yep, and notice how both functions are even( you're in polar not rectangular) so you need to take into account the negative answer

25. anonymous

Ok so now what?

what are your values for theta?

27. anonymous

0 and 1/2?

come on, your values for theta, not sin(theta) and cos(theta). Think trig or geometry. This is calculus so if you're willing to take calc, especially calc 3, you should know how to take the inverse of a trig function.

29. anonymous

I thought you meant those values though

theta refers to the angle. What are the angles? Stop and don't think calculus right now. Go back to your geometry or trig. How did you find an angle using trig functions like sine and cosine?

31. anonymous

The circle, so wouldn't it be 30 degrees

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33. anonymous

I barely remember the unit circle

That is a right triangle. Forget the unit circle, focus on the triangle. Cosine of which angle will give you 1/2? What two sides do you use for cosine? |dw:1351461655567:dw|

35. anonymous

I thought it was 30 degrees?

Oh and use that graph for sine. It's crappy but it was the best I had. Ok, what is cosine? Think back to your geometry.

37. anonymous

sqrt(3)/2?

38. anonymous

That is the best I could remember

no, no, the DEFINITION of cosine

40. anonymous

like which degree or just the definition?

forget all numbers

42. anonymous

A/H

forget angles, forget numbers, just the textbook definition

GOOD.

Okay, so what is A/H?

what does it stand for?

47. anonymous

sqrt(3)/2

48. anonymous

good, now when we had $2\cos(\theta)-1=0$ how did you solve it?

50. anonymous

I got cos(theta) = 1/2 by adding then dividing

good, now look at that triangle I drew. Do you see a 1 for the adjacent, and 2 for the hypotenuse?

52. anonymous

Oh yeah so 1/2

right, when you have a fraction, that's what that fraction means

54. anonymous

yes

so what angle, 30, 60, or 90, will give you cos(theta)=1/2?

56. anonymous

Nope 60 degrees since its a/h which is 1/2

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58. anonymous

60 degrees will give us 1/2

got it! so theta is 60 degrees or pi/3

60. anonymous

ok yes cause they are equivalent

ok, so can you do the same for$\sin(\theta)=0$

Hint: you can't do this using a triangle

63. anonymous

its just 0

bingo

those are your limits of integration

66. anonymous

so 0 to pi/3?

yes, but do you remember the area(not under a curve) between two curves?

how would you integrate if I said y=x^2 and y=3-4x^2?

69. anonymous

like Big R ^2 - Little r ^2?

YES

71. anonymous

so we have 0 to pi/3 of our 2 functions?

now you have your limits of integration (in radians). You need to figure out which is the top function and which is the bottom

73. anonymous

top is 8 sin(theta) and bottom is 8sin(2theta)

Did you guess that?

75. anonymous

No I graphed it

Ok, so then you know we're missing the left part of the graph then correct? since we are only going from 0 to pi/3

77. anonymous

so then we have to go from -pi/3 to 0?

Nope, don't go into negative. Use the trace on your calculator and you'll see that the graph keeps going from 0 to 2pi. A lot of graphs are like that but not all

79. anonymous

so whats our final integral going to look like then?

Remember, the independent variable here isn't x like in rectangular coordinates, the angle theta (in radians) is the independent variable. Both x and y are dependent

Good question. There are actually two limits I didn't want to get into yet, but since you brought it up, there are TWO solutions to both sin(theta)=0 and cos(theta)=1/2 (within the restriction that theta is from 0 to 2pi.) Without that restriction, theta would have an infinite number of solutions

82. anonymous

You realize this is not a double intgreal question right?

yup, you're going to add the two integrals, just like if you had a break in the middle of a graph

84. anonymous

So integral from 0 to 2pi (8sin(theta)-8sin(2theta)) + 0 to pi/3 ((8sin(theta)-8sin(2theta))

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notice how if you did top minus bottom( which is the x axis) you need to change your limits of integration?

It's the same idea here, but using polar

88. anonymous

it is top-bottom though

89. anonymous

Can you draw out what the ending integral will look like

true, you're right there, just that the limits(BOTH LIMITS) will change

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92. anonymous

Yeah I have that

I focused on three points, but one of those points is actually a different limit of integration. For the graph of sine, what TWO values will give you 0? That's a big hint on your limits.

94. anonymous

Ok cool thanks I think I got it from here

I hope I didn't hold back too much, but I wanted you to think and not just give you the answers. I hope I did okay

96. anonymous

it was good thanks, I still have to solve for the answer though, what was your result? Mine was 12.75516082

uh...do you by any chance have the answer? Just a yes or no.

Back of the book I mean

99. anonymous

yes thats my answer, oh no its webwork

100. anonymous

Uh, I got an 8...

and no idea what webwork is

103. anonymous

what did you get 8 what?

106. anonymous

ok but our answers are both wrong