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

- anonymous

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

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- roadjester

let me guess, third year calculus?

- anonymous

yeah why?

- roadjester

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

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## More answers

- anonymous

ok thanks man its so tough

- roadjester

i can relate, gimmie a sec

- anonymous

ok

- roadjester

have you done double integrals?

- anonymous

Not yet

- roadjester

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

- anonymous

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

- roadjester

give me as much info as you can then

- anonymous

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

- anonymous

we have done polar coordinates, taylor series , maclaurin series

- roadjester

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

- 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

- roadjester

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

- roadjester

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.

- anonymous

ok I see

- roadjester

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.

- anonymous

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

- roadjester

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

- roadjester

Yes

- anonymous

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

- roadjester

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

- anonymous

Ok so now what?

- roadjester

what are your values for theta?

- anonymous

0 and 1/2?

- roadjester

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.

- anonymous

I thought you meant those values though

- roadjester

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?

- anonymous

The circle, so wouldn't it be 30 degrees

- roadjester

|dw:1351461552476:dw|

- anonymous

I barely remember the unit circle

- roadjester

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|

- anonymous

I thought it was 30 degrees?

- roadjester

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.

- anonymous

sqrt(3)/2?

- anonymous

That is the best I could remember

- roadjester

no, no, the DEFINITION of cosine

- anonymous

like which degree or just the definition?

- roadjester

forget all numbers

- anonymous

A/H

- roadjester

forget angles, forget numbers, just the textbook definition

- roadjester

GOOD.

- roadjester

Okay, so what is A/H?

- roadjester

what does it stand for?

- anonymous

sqrt(3)/2

- anonymous

adjacent over hypotenuse

- roadjester

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

- anonymous

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

- roadjester

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

- anonymous

Oh yeah so 1/2

- roadjester

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

- anonymous

yes

- roadjester

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

- anonymous

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

- roadjester

|dw:1351462522741:dw|

- anonymous

60 degrees will give us 1/2

- roadjester

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

- anonymous

ok yes cause they are equivalent

- roadjester

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

- roadjester

Hint: you can't do this using a triangle

- anonymous

its just 0

- roadjester

bingo

- roadjester

those are your limits of integration

- anonymous

so 0 to pi/3?

- roadjester

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

- roadjester

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

- anonymous

like Big R ^2 - Little r ^2?

- roadjester

YES

- anonymous

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

- roadjester

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

- anonymous

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

- roadjester

Did you guess that?

- anonymous

No I graphed it

- roadjester

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

- anonymous

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

- roadjester

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

- anonymous

so whats our final integral going to look like then?

- roadjester

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

- roadjester

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

- anonymous

You realize this is not a double intgreal question right?

- roadjester

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

- anonymous

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

- roadjester

|dw:1351463619332:dw|

- roadjester

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

- roadjester

It's the same idea here, but using polar

- anonymous

it is top-bottom though

- anonymous

Can you draw out what the ending integral will look like

- roadjester

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

- roadjester

|dw:1351463857221:dw|

- anonymous

Yeah I have that

- roadjester

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.

- anonymous

Ok cool thanks I think I got it from here

- roadjester

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

- anonymous

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

- roadjester

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

- roadjester

Back of the book I mean

- anonymous

yes thats my answer, oh no its webwork

- anonymous

webwork didnt take my answer

- roadjester

Uh, I got an 8...

- roadjester

and no idea what webwork is

- anonymous

what did you get 8 what?

- roadjester

8 as the final answer

- roadjester

Is that the answer in your text?

- anonymous

ok but our answers are both wrong

- roadjester

OUGH

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