region bounded by polar curves. inside r = 2 + 2 cos t , outside r = 6 cos t. Im a bit stumped

- perl

region bounded by polar curves. inside r = 2 + 2 cos t , outside r = 6 cos t. Im a bit stumped

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- amistre64

do we know what the graphs look like?

- perl

yes we do

- perl

i graphed it , one sec

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## More answers

- perl

|dw:1333337337946:dw|

- perl

the curve to the left is r = 6 cos t, the smaller heart shaped curve is r = 2 + 2 cos t

- amistre64

|dw:1333337385158:dw|
these areas right?

- perl

right

- amistre64

since the top is just a mirror of the bottom all we need is the point of intersection in the q1

- perl

theta = pi/3 , theta = 5pi/3

- perl

|dw:1333337578255:dw|

- amistre64

pi/3 is good enough, then we will just double our results

- perl

ok , so...

- amistre64

r = pi/3 to pi this sweeps out the area that we are concerned with right?

- perl

im not 100 % sure the areas are symmetric, but whatever

- perl

no pi/3 to pi.2

- perl

thats q1, and q2 is pi/2 to 5pi/3

- perl

thats why i dont think the regions are symmetric

- 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

they are symmetric

- perl

no i drew it symmetric, but the graph looks a bit different

- amistre64

http://www.wolframalpha.com/input/?i=polar+plot+r%3D2%2B2cos%28t%29+and+r%3D6cos%28t%29

- perl

it is 30 degrees sweeped in q1 and 60 degrees sweeped in q2

- perl

in any case, what do you get for q1?

- amistre64

same shape and countour and mirror image across the xaxis

- amistre64

|dw:1333338046085:dw|

- amistre64

so
\[\int_{pi/3}^{pi}\int_{0}^{2+2cos(t)}r\ dr.dt\]

- perl

can you do it with single integral?

- amistre64

i could, but this is more realistic

- 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

the angle sweep of the 6cos(t) for that portion is different tho

- perl

i dont think that is right

- perl

so for q1 , the first shaded region that is what you got?

- amistre64

thats q1+q2 regions

- perl

wat is q2?

- perl

the smaller curve, lets call it the heart

- amistre64

http://www.wolframalpha.com/input/?i=polar+plot+r%3D2%2B2cos%28t%29%2C+t+%3D+pi%2F3+to+pi

- perl

but you are integrating too much

- amistre64

r=6cos(t) makes the deletion when the angle ssweeps from pi/3 to pi/2

- amistre64

you keep saying that but you really havent explained to me why you think that is in error

- perl

youre not using the second curve

- perl

are you saying integral 1/2 [( 2 + 2 cos t)^2 -( 6 cos t)^2]

- perl

i dont know what you mean by deletion

- amistre64

that is definantly not what i am saying

- perl

these are polar integrals http://tutorial.math.lamar.edu/Classes/CalcII/PolarArea.aspx

- perl

check it out :)

- 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

we want the area outside the curve 6 cost t

- perl

ok and how do you trim that?

- perl

can you write out the whole expression,

- amistre64

correct; so we take the heart shape as a striaght line, and trim out the 6cos(t) that abuts it afterwards

- perl

i was using the formula 1/2 (r^2 - r^2) outer radius - inner radius

- 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

forgot to square

- 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

heart - circle = area shaded

- perl

ok so dont use what i did , integral 1/2 (r^2 - r^2)

- amistre64

that might work with some extra work thrown in to account for the missing portion of q2

- perl

and can we check the double integral, to make sure

- amistre64

i could have just as well split my integral of the heart into 2 parts; but i dont see the pooint really

- amistre64

the double just comes about to the single after integrating for r

- perl

I got for q1 8 - 2pi

- 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

what is that?

- perl

general formula?

- perl

so the answer should be 16 - 4pi

- 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

the total area I calcuate, if the wolf hasnt lied to me; is pi

- 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

let me check

- perl

hmmm, that looks right, and you multiplied by 2

- amistre64

yes; the 1/2 pull out and when i double it up (*2) they simple go to one and cause no trouble

- perl

how come i cant integrate
1/2 ( 2 + 2 cos t)^2 - 1/2 ( 6cost))^2 from pi/3 to 5pi/3 ,

- perl

im just wondering, i agree with what you did though

- perl

, does the area become negative>

- 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

they are going in opposite directions , true

- perl

when i do it this way, i get -8pi as the solution

- 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

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

|dw:1333340310037:dw|
this is what we get if we do both from pi/3 to pi

- perl

ohhhh

- amistre64

|dw:1333340375323:dw|
and from pi to 5pi/3 results in this

- amistre64

so, we had to reevaluate 6cos(t) to fit the part

- perl

yeah it sweeps 1 and 1/2 times around

- perl

well let me adjust the second integral

- perl

these radians are confusing me

- perl

so on the curve r = 6 cos t , where does it interest, we have theta= pi/3 and ...

- perl

theta = 2pi/3

- perl

yes it worked

- amistre64

6cos(t) will sweep the area from ppi/3 to 5pi/3 just fine, but the heart wont catch up by then

- perl

yay!!

- perl

right, so we have to use different limits for the 6 cos t

- amistre64

right; 60 to 120

- 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

heart goes from 60 to 240; circle from 60 to 120 to fit the area on the graph

- perl

just divide that by 2 , because of the 1/2 in front

- perl

right!!

- perl

so polar integration we use sectors of circles to integral instead of rectangles

- amistre64

i think you areaed the total of it; while i was focused on half of it

- amistre64

correct

- amistre64

yes; you should have 2pi; then /2 = pi

- perl

ok , i was confused , this problem had tricky limits

- perl

because of the different speeds of the curve
made me very mad

- amistre64

that tends to be why i dont try to combine them into one integral; i like to subtract one from the other

- perl

i assumed they both moved at the same speed, so the limits kept busting me

- amistre64

:)

- 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

i thought the graph was not symmetric, oh well... now i know it must be my eyes

- 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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