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let me guess, third year calculus?

yeah why?

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

ok thanks man its so tough

i can relate, gimmie a sec

ok

have you done double integrals?

Not yet

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

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

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

we have done polar coordinates, taylor series , maclaurin series

ok I see

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

Yes

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

Ok so now what?

what are your values for theta?

0 and 1/2?

I thought you meant those values though

The circle, so wouldn't it be 30 degrees

|dw:1351461552476:dw|

I barely remember the unit circle

I thought it was 30 degrees?

sqrt(3)/2?

That is the best I could remember

no, no, the DEFINITION of cosine

like which degree or just the definition?

forget all numbers

A/H

forget angles, forget numbers, just the textbook definition

GOOD.

Okay, so what is A/H?

what does it stand for?

sqrt(3)/2

adjacent over hypotenuse

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

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?

Oh yeah so 1/2

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

yes

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

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

|dw:1351462522741:dw|

60 degrees will give us 1/2

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

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

its just 0

bingo

those are your limits of integration

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?

like Big R ^2 - Little r ^2?

YES

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

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

Did you guess that?

No I graphed it

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

so whats our final integral going to look like then?

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

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

|dw:1351463619332:dw|

It's the same idea here, but using polar

it is top-bottom though

Can you draw out what the ending integral will look like

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

|dw:1351463857221:dw|

Yeah I have that

Ok cool thanks I think I got it from here

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

Back of the book I mean

yes thats my answer, oh no its webwork

webwork didnt take my answer

Uh, I got an 8...

and no idea what webwork is

what did you get 8 what?

8 as the final answer

Is that the answer in your text?

ok but our answers are both wrong

OUGH