perl
  • perl
region bounded by polar curves. inside r = 2 + 2 cos t , outside r = 6 cos t. Im a bit stumped
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
do we know what the graphs look like?
perl
  • perl
yes we do
perl
  • perl
i graphed it , one sec

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perl
  • perl
|dw:1333337337946:dw|
perl
  • perl
the curve to the left is r = 6 cos t, the smaller heart shaped curve is r = 2 + 2 cos t
amistre64
  • amistre64
|dw:1333337385158:dw| these areas right?
perl
  • perl
right
amistre64
  • amistre64
since the top is just a mirror of the bottom all we need is the point of intersection in the q1
perl
  • perl
theta = pi/3 , theta = 5pi/3
perl
  • perl
|dw:1333337578255:dw|
amistre64
  • amistre64
pi/3 is good enough, then we will just double our results
perl
  • perl
ok , so...
amistre64
  • amistre64
r = pi/3 to pi this sweeps out the area that we are concerned with right?
perl
  • perl
im not 100 % sure the areas are symmetric, but whatever
perl
  • perl
no pi/3 to pi.2
perl
  • perl
thats q1, and q2 is pi/2 to 5pi/3
perl
  • perl
thats why i dont think the regions are symmetric
amistre64
  • amistre64
the angle that we need to sweep out to include the top portion is from the intersection all the way to 180 degrees
amistre64
  • amistre64
they are symmetric
perl
  • perl
no i drew it symmetric, but the graph looks a bit different
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=polar+plot+r%3D2%2B2cos%28t%29+and+r%3D6cos%28t%29
perl
  • perl
it is 30 degrees sweeped in q1 and 60 degrees sweeped in q2
perl
  • perl
in any case, what do you get for q1?
amistre64
  • amistre64
same shape and countour and mirror image across the xaxis
amistre64
  • amistre64
|dw:1333338046085:dw|
amistre64
  • amistre64
so \[\int_{pi/3}^{pi}\int_{0}^{2+2cos(t)}r\ dr.dt\]
perl
  • perl
can you do it with single integral?
amistre64
  • amistre64
i could, but this is more realistic
amistre64
  • amistre64
since r ints up to r^2/2 and 0 is pointless; its goes to\[\int_{pi/3}^{pi}\frac{1}{2}(2+2cos(t))^2\ dt\]
amistre64
  • amistre64
the angle sweep of the 6cos(t) for that portion is different tho
perl
  • perl
i dont think that is right
perl
  • perl
so for q1 , the first shaded region that is what you got?
amistre64
  • amistre64
thats q1+q2 regions
perl
  • perl
wat is q2?
perl
  • perl
the smaller curve, lets call it the heart
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=polar+plot+r%3D2%2B2cos%28t%29%2C+t+%3D+pi%2F3+to+pi
perl
  • perl
but you are integrating too much
amistre64
  • amistre64
r=6cos(t) makes the deletion when the angle ssweeps from pi/3 to pi/2
amistre64
  • amistre64
you keep saying that but you really havent explained to me why you think that is in error
perl
  • perl
youre not using the second curve
perl
  • perl
are you saying integral 1/2 [( 2 + 2 cos t)^2 -( 6 cos t)^2]
perl
  • perl
i dont know what you mean by deletion
amistre64
  • amistre64
that is definantly not what i am saying
perl
  • perl
these are polar integrals http://tutorial.math.lamar.edu/Classes/CalcII/PolarArea.aspx
perl
  • perl
check it out :)
amistre64
  • amistre64
i am saying that the area to be considered is contained in the curce 2+2cos(t) from pi/3 to pi; and next we trim that
perl
  • perl
we want the area outside the curve 6 cost t
perl
  • perl
ok and how do you trim that?
perl
  • perl
can you write out the whole expression,
amistre64
  • amistre64
correct; so we take the heart shape as a striaght line, and trim out the 6cos(t) that abuts it afterwards
perl
  • perl
i was using the formula 1/2 (r^2 - r^2) outer radius - inner radius
perl
  • perl
so you are doing 1/2 int ( 2 + 2 cos t ) - 1/2 int (6 cos t ) , from pi/3 to 5pi/3 ?
perl
  • perl
forgot to square
amistre64
  • amistre64
the 6cos(t) affects our area from t = pi/3 to pi/2 such that:\[2\left(\frac{1}{2}\int_{pi/3}^{pi}(2+2cos(t))^2\ dt-\frac{1}{2}\int_{pi/3}^{pi/2}(6cos(t))^2\ dt\right)\] defines the area of concern
amistre64
  • amistre64
heart - circle = area shaded
perl
  • perl
ok so dont use what i did , integral 1/2 (r^2 - r^2)
amistre64
  • amistre64
that might work with some extra work thrown in to account for the missing portion of q2
perl
  • perl
and can we check the double integral, to make sure
amistre64
  • amistre64
i could have just as well split my integral of the heart into 2 parts; but i dont see the pooint really
amistre64
  • amistre64
the double just comes about to the single after integrating for r
perl
  • perl
I got for q1 8 - 2pi
amistre64
  • amistre64
\[\int_{0}^{f(t)} r\ dr=\left.\frac{1}{2}r^2\right|_{0}^{f(t)}\to\ \frac{1}{2}(f(t))^2\]
perl
  • perl
what is that?
perl
  • perl
general formula?
perl
  • perl
so the answer should be 16 - 4pi
amistre64
  • amistre64
thats how the formula for the polar area stuff is determined yes; they teach you the "formula" to use in a single integral since they think you cant handle integrationg more than once
amistre64
  • amistre64
the total area I calcuate, if the wolf hasnt lied to me; is pi
amistre64
  • amistre64
after pulling out the halfs and multiplying them by 2 they cancel to 1 so i get:http://www.wolframalpha.com/input/?i=integrate+%282%2B2cos%28t%29%29%5E2+dt+from+pi%2F3+to+pi+-+integrate+%286cos%28t%29%29%5E2+dt+from+pi%2F3+to+pi%2F2
perl
  • perl
let me check
perl
  • perl
hmmm, that looks right, and you multiplied by 2
amistre64
  • amistre64
yes; the 1/2 pull out and when i double it up (*2) they simple go to one and cause no trouble
perl
  • perl
how come i cant integrate 1/2 ( 2 + 2 cos t)^2 - 1/2 ( 6cost))^2 from pi/3 to 5pi/3 ,
perl
  • perl
im just wondering, i agree with what you did though
perl
  • perl
, does the area become negative>
amistre64
  • amistre64
you might be able to pull that off with the heart part; but the 6cos(t) doesnt play fair in that interval; it actually doubles back around again so that you have to adjust it
perl
  • perl
they are going in opposite directions , true
perl
  • perl
when i do it this way, i get -8pi as the solution
amistre64
  • amistre64
at t= 0 we get 6cos(0) = 6 and 2+2cos(0) = 4 and they travel in the same direction; but the issue is that 2+2cos(t) is moving slower than the other one
amistre64
  • amistre64
so by taking that area of 6cos(t) from pi/3 to 5pi/3 your actaully subtracting the whole area 1 and a half times over just about
amistre64
  • amistre64
|dw:1333340310037:dw| this is what we get if we do both from pi/3 to pi
perl
  • perl
ohhhh
amistre64
  • amistre64
|dw:1333340375323:dw| and from pi to 5pi/3 results in this
amistre64
  • amistre64
so, we had to reevaluate 6cos(t) to fit the part
perl
  • perl
yeah it sweeps 1 and 1/2 times around
perl
  • perl
well let me adjust the second integral
perl
  • perl
these radians are confusing me
perl
  • perl
so on the curve r = 6 cos t , where does it interest, we have theta= pi/3 and ...
perl
  • perl
theta = 2pi/3
perl
  • perl
yes it worked
amistre64
  • amistre64
6cos(t) will sweep the area from ppi/3 to 5pi/3 just fine, but the heart wont catch up by then
perl
  • perl
yay!!
perl
  • perl
right, so we have to use different limits for the 6 cos t
amistre64
  • amistre64
right; 60 to 120
perl
  • perl
http://www.wolframalpha.com/input/?i=integral+%282%2B2cos%28t%29%29^2+dt%29+from+pi%2F3+to+5pi%2F3+-+integral+%286cos%28t%29%29^2+dt+from+pi%2F3+to+2pi%2F3
amistre64
  • amistre64
heart goes from 60 to 240; circle from 60 to 120 to fit the area on the graph
perl
  • perl
just divide that by 2 , because of the 1/2 in front
perl
  • perl
right!!
perl
  • perl
so polar integration we use sectors of circles to integral instead of rectangles
amistre64
  • amistre64
i think you areaed the total of it; while i was focused on half of it
amistre64
  • amistre64
correct
amistre64
  • amistre64
yes; you should have 2pi; then /2 = pi
perl
  • perl
ok , i was confused , this problem had tricky limits
perl
  • perl
because of the different speeds of the curve made me very mad
amistre64
  • amistre64
that tends to be why i dont try to combine them into one integral; i like to subtract one from the other
perl
  • perl
i assumed they both moved at the same speed, so the limits kept busting me
amistre64
  • amistre64
:)
perl
  • perl
right, but i thought it wasnt symmetric because the limits were not symmetric then i realised it isnt the graph that is not symmetric, but the speed
perl
  • perl
i thought the graph was not symmetric, oh well... now i know it must be my eyes
perl
  • perl
there is a way to do this with double integrals? sorry, i didnt follow what you tried to do at first

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