ilikephysics2
HELP Find the area of the region that lies inside both curves r = 8sin(2theta), r = 8sin(theta)
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roadjester
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let me guess, third year calculus?
ilikephysics2
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yeah why?
roadjester
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It was torture for me :) let me see if I can help, I need some scratch paper though
ilikephysics2
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ok thanks man its so tough
roadjester
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i can relate, gimmie a sec
ilikephysics2
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ok
roadjester
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have you done double integrals?
ilikephysics2
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Not yet
roadjester
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dang, okay gotta think of how to do this using single...
ilikephysics2
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we haven't gotten that far we are in chapter 9 with areas of regions and in curves
roadjester
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give me as much info as you can then
ilikephysics2
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well the question is just right there lol, idk what else you want
ilikephysics2
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we have done polar coordinates, taylor series , maclaurin series
roadjester
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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
ilikephysics2
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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
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\[8\sin(2\theta)=8\sin(\theta)\]
\[\sin(2\theta)=\sin(\theta)\]
\[2\sin(\theta)\cos(\theta)=\sin(\theta)\]
\[2\sin(\theta)\cos(\theta)-\sin(\theta)=0\]
\[\sin(\theta)(2\cos(\theta)-1)=0\]
\[\sin(\theta)=0\] and \[2\cos(\theta)-1=0\]
Solve for theta and these are your limits of integration.
roadjester
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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.
ilikephysics2
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ok I see
roadjester
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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.
ilikephysics2
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so its 2 cos(theta) = 1,
so cos(theta) = 1/2?
roadjester
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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
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Yes
ilikephysics2
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OK so i have sin(theta) = 0 and cos(theta) = 1/2
roadjester
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yep, and notice how both functions are even( you're in polar not rectangular) so you need to take into account the negative answer
ilikephysics2
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Ok so now what?
roadjester
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what are your values for theta?
ilikephysics2
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0 and 1/2?
roadjester
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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.
ilikephysics2
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I thought you meant those values though
roadjester
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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?
ilikephysics2
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The circle, so wouldn't it be 30 degrees
roadjester
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|dw:1351461552476:dw|
ilikephysics2
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I barely remember the unit circle
roadjester
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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|
ilikephysics2
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I thought it was 30 degrees?
roadjester
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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.
ilikephysics2
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sqrt(3)/2?
ilikephysics2
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That is the best I could remember
roadjester
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no, no, the DEFINITION of cosine
ilikephysics2
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like which degree or just the definition?
roadjester
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forget all numbers
ilikephysics2
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A/H
roadjester
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forget angles, forget numbers, just the textbook definition
roadjester
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GOOD.
roadjester
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Okay, so what is A/H?
roadjester
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what does it stand for?
ilikephysics2
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sqrt(3)/2
ilikephysics2
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adjacent over hypotenuse
roadjester
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good, now when we had \[2\cos(\theta)-1=0\]
how did you solve it?
ilikephysics2
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I got cos(theta) = 1/2 by adding then dividing
roadjester
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good, now look at that triangle I drew. Do you see a 1 for the adjacent, and 2 for the hypotenuse?
ilikephysics2
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Oh yeah so 1/2
roadjester
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right, when you have a fraction, that's what that fraction means
ilikephysics2
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yes
roadjester
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so what angle, 30, 60, or 90, will give you cos(theta)=1/2?
ilikephysics2
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Nope 60 degrees since its a/h which is 1/2
roadjester
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|dw:1351462522741:dw|
ilikephysics2
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60 degrees will give us 1/2
roadjester
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got it! so theta is 60 degrees or pi/3
ilikephysics2
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ok yes cause they are equivalent
roadjester
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ok, so can you do the same for\[\sin(\theta)=0\]
roadjester
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Hint: you can't do this using a triangle
ilikephysics2
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its just 0
roadjester
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bingo
roadjester
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those are your limits of integration
ilikephysics2
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so 0 to pi/3?
roadjester
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yes, but do you remember the area(not under a curve) between two curves?
roadjester
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how would you integrate if I said y=x^2 and y=3-4x^2?
ilikephysics2
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like Big R ^2 - Little r ^2?
roadjester
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YES
ilikephysics2
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so we have 0 to pi/3 of our 2 functions?
roadjester
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now you have your limits of integration (in radians). You need to figure out which is the top function and which is the bottom
ilikephysics2
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top is 8 sin(theta) and bottom is 8sin(2theta)
roadjester
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Did you guess that?
ilikephysics2
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No I graphed it
roadjester
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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
ilikephysics2
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so then we have to go from -pi/3 to 0?
roadjester
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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
ilikephysics2
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so whats our final integral going to look like then?
roadjester
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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
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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
ilikephysics2
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You realize this is not a double intgreal question right?
roadjester
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yup, you're going to add the two integrals, just like if you had a break in the middle of a graph
ilikephysics2
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So integral from 0 to 2pi (8sin(theta)-8sin(2theta)) + 0 to pi/3 ((8sin(theta)-8sin(2theta))
roadjester
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|dw:1351463619332:dw|
roadjester
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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?
roadjester
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It's the same idea here, but using polar
ilikephysics2
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it is top-bottom though
ilikephysics2
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Can you draw out what the ending integral will look like
roadjester
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true, you're right there, just that the limits(BOTH LIMITS) will change
roadjester
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|dw:1351463857221:dw|
ilikephysics2
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Yeah I have that
roadjester
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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.
ilikephysics2
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Ok cool thanks I think I got it from here
roadjester
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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
ilikephysics2
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it was good thanks, I still have to solve for the answer though, what was your result? Mine was 12.75516082
roadjester
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uh...do you by any chance have the answer? Just a yes or no.
roadjester
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Back of the book I mean
ilikephysics2
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yes thats my answer, oh no its webwork
ilikephysics2
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webwork didnt take my answer
roadjester
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Uh, I got an 8...
roadjester
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and no idea what webwork is
ilikephysics2
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what did you get 8 what?
roadjester
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8 as the final answer
roadjester
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Is that the answer in your text?
ilikephysics2
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ok but our answers are both wrong
roadjester
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OUGH