anonymous
  • anonymous
HELP Find the area of the region that lies inside both curves r = 8sin(2theta), r = 8sin(theta)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
roadjester
  • roadjester
let me guess, third year calculus?
anonymous
  • anonymous
yeah why?
roadjester
  • roadjester
It was torture for me :) let me see if I can help, I need some scratch paper though

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

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