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ilikephysics2
Group Title
HELP Find the area of the region that lies inside both curves r = 8sin(2theta), r = 8sin(theta)
 one year ago
 one year ago
ilikephysics2 Group Title
HELP Find the area of the region that lies inside both curves r = 8sin(2theta), r = 8sin(theta)
 one year ago
 one year ago

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roadjester Group TitleBest ResponseYou've already chosen the best response.1
let me guess, third year calculus?
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
yeah why?
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
It was torture for me :) let me see if I can help, I need some scratch paper though
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
ok thanks man its so tough
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
i can relate, gimmie a sec
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
have you done double integrals?
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
Not yet
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
dang, okay gotta think of how to do this using single...
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
we haven't gotten that far we are in chapter 9 with areas of regions and in curves
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
give me as much info as you can then
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
well the question is just right there lol, idk what else you want
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
we have done polar coordinates, taylor series , maclaurin series
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
\[8\sin(2\theta)=8\sin(\theta)\] \[\sin(2\theta)=\sin(\theta)\] \[2\sin(\theta)\cos(\theta)=\sin(\theta)\] \[2\sin(\theta)\cos(\theta)\sin(\theta)=0\] \[\sin(\theta)(2\cos(\theta)1)=0\] \[\sin(\theta)=0\] and \[2\cos(\theta)1=0\] Solve for theta and these are your limits of integration.
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
ok I see
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
so its 2 cos(theta) = 1, so cos(theta) = 1/2?
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
OK so i have sin(theta) = 0 and cos(theta) = 1/2
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
yep, and notice how both functions are even( you're in polar not rectangular) so you need to take into account the negative answer
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
Ok so now what?
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
what are your values for theta?
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
0 and 1/2?
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
I thought you meant those values though
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
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?
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
The circle, so wouldn't it be 30 degrees
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
dw:1351461552476:dw
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
I barely remember the unit circle
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
I thought it was 30 degrees?
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
sqrt(3)/2?
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
That is the best I could remember
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
no, no, the DEFINITION of cosine
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
like which degree or just the definition?
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
forget all numbers
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
forget angles, forget numbers, just the textbook definition
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
Okay, so what is A/H?
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
what does it stand for?
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
sqrt(3)/2
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
adjacent over hypotenuse
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
good, now when we had \[2\cos(\theta)1=0\] how did you solve it?
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
I got cos(theta) = 1/2 by adding then dividing
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
good, now look at that triangle I drew. Do you see a 1 for the adjacent, and 2 for the hypotenuse?
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
Oh yeah so 1/2
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
right, when you have a fraction, that's what that fraction means
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
so what angle, 30, 60, or 90, will give you cos(theta)=1/2?
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
Nope 60 degrees since its a/h which is 1/2
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
dw:1351462522741:dw
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
60 degrees will give us 1/2
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
got it! so theta is 60 degrees or pi/3
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
ok yes cause they are equivalent
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
ok, so can you do the same for\[\sin(\theta)=0\]
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
Hint: you can't do this using a triangle
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
its just 0
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
those are your limits of integration
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
so 0 to pi/3?
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
yes, but do you remember the area(not under a curve) between two curves?
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
how would you integrate if I said y=x^2 and y=34x^2?
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
like Big R ^2  Little r ^2?
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
so we have 0 to pi/3 of our 2 functions?
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
now you have your limits of integration (in radians). You need to figure out which is the top function and which is the bottom
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
top is 8 sin(theta) and bottom is 8sin(2theta)
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
Did you guess that?
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
No I graphed it
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
so then we have to go from pi/3 to 0?
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
so whats our final integral going to look like then?
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
You realize this is not a double intgreal question right?
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
yup, you're going to add the two integrals, just like if you had a break in the middle of a graph
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
So integral from 0 to 2pi (8sin(theta)8sin(2theta)) + 0 to pi/3 ((8sin(theta)8sin(2theta))
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
dw:1351463619332:dw
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
notice how if you did top minus bottom( which is the x axis) you need to change your limits of integration?
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
It's the same idea here, but using polar
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
it is topbottom though
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
Can you draw out what the ending integral will look like
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
true, you're right there, just that the limits(BOTH LIMITS) will change
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
dw:1351463857221:dw
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
Yeah I have that
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
Ok cool thanks I think I got it from here
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
it was good thanks, I still have to solve for the answer though, what was your result? Mine was 12.75516082
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
uh...do you by any chance have the answer? Just a yes or no.
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
Back of the book I mean
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
yes thats my answer, oh no its webwork
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
webwork didnt take my answer
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
Uh, I got an 8...
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
and no idea what webwork is
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
what did you get 8 what?
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
8 as the final answer
 one year ago

roadjester Group TitleBest ResponseYou've already chosen the best response.1
Is that the answer in your text?
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
ok but our answers are both wrong
 one year ago
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