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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
adjacent over hypotenuse
 2 years 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?
 2 years ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

roadjester Group TitleBest ResponseYou've already chosen the best response.1
dw:1351463619332:dw
 2 years 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?
 2 years ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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