Question

find the area of the commmon interior of the polar eq r = 4(1+sin(theta)) and r = 4(1-sin(theta)) Show the integration.

Answer 1

0 = 4 - 4sin(theta)
0 = 4+4sin(theta)

0/4 = -4sin(theta)
0/4 = 4sin(theta)

4 = sin(theta)
-4 = sin(theta)

theta = arcsin(4)
theta = arcsin(-4)

Answer 2

@sami-21 :)

Answer 3

@satellite73 do you know about this? :)

Answer 4

libniz is answering =P

Answer 5

We can hope :).

Answer 6

i know the general idea of what you need to do

Answer 7

alright, this is what I am going by

http://www.wolframalpha.com/input/?i=polar+%7Br+%3D+4%281%2Bsin%28theta%29%29+and+r+%3D+4%281-sin%28theta%29%29%7D

\[\int_{0}^{\pi/2}\int_{0}^{4-4sin(\theta)}rdrd\theta\]
this takes care of top half

top half
\[\int_{\pi/2}^{\pi}\int_{0}^{4+4sin(\theta)}rdrd\theta\]

Answer 8

all I need to know is if I need to do 2 integrals or where I go from here...

Answer 9

first graph to see what the common area of both eqautions are

Answer 10

and libniz if the areas are the same because it being symetrical you don't have to do ... you can do one

Answer 11

and multiply by 2

Answer 12

good idea @Outkast3r09

Answer 13

my graph is weird.... It goes to 8, this only goes to 6...?

Answer 14

so take either integral and double it

Answer 15

The last question went to 12 like WTF?

Answer 16

use agraphing calc

Answer 17

look at my link

Answer 18

I did it goes from 0 to 8.

Answer 19

0? to 8 it should go from thetas

Answer 20

ok, well I'm not sure what to do then..... How do I graph it...?

Answer 21

r is like the radius of the line to the point

Answer 22

theta is the angle and t is the time

Answer 23

I thought I was supposed to graph what was from wolfram...

Answer 24

that's the graph

Answer 25

or any graphing utility.

Answer 26

Yeah, so y = 8 at the top point, what I'm saying is the graph I'm provided goes from -6 to 6 for both x,y.

Answer 27

that's what I'm graphing right?

Answer 28

The integral you said I can just do 1 right?

Answer 29

\[2\int\limits_{0}^{\pi/2} 4\sin(\theta)+4 d \theta\]

Answer 30

so at the top theta=pi/2

r=4+4 sin(pi/2)
r=4+4(1)=8

Answer 31

mhmhmhm

Answer 32

yeah , just do one integral and double the answer

Answer 33

I'm going to jujst redo the graph screw it... :)

Answer 34

ok so I set the integral up correctly? :)

Answer 35

\[2\int_{0}^{\pi/2}\int_{0}^{4+4sin(\theta)}rdrd\theta\]
take the inner integral

integral of r is r^2/2 , plug in limit value

Answer 36

I've never seen integrals set up like that...

Answer 37

that's double integral

Answer 38

hmm.....? lol

Answer 39

so it's r dr and then that dtheta?

Answer 40

yep

Answer 41

so we found r to be 8 then? from what you said? so we do the integral of 8?

Answer 42

no, that's just graphing to see the graph;

Answer 43

ok haha.

Answer 44

of 4(sin(theta)+1)?

Answer 45

:(

Answer 46

yes

Answer 47

:)

Answer 48

and we do that from 0 to pi/2 right? or 0 to 4+4sin(θ)...?

Answer 49

you intergrate the inner part first

Answer 50

I got 4 θ-4 cos(θ)?

Answer 51

\[\int_{0}^{4+4sin(\theta)}rdr\]

do this first

Answer 52

from 0 to 4+4sin(theta) hmm..

Answer 53

http://www.wolframalpha.com/input/?i=integral+4%28sin%28theta%29%2B1%29+from+0+to+4%2B4sin%28theta%29 ???? Wtf LOL

Answer 54

I got something more complicated
http://www.wolframalpha.com/input/?i=int%5Br%2C%7Br%2C0%2C4%2B4sin%28theta%29%7D%5D

Answer 55

wtf.....

Answer 56

loooks better than what I got....

Answer 57

let me do the whole thing

Answer 58

http://www.wolframalpha.com/input/?i=int%5Bint%5Br%2C%7Br%2C0%2C4%2B4sin%28theta%29%7D%5D%2C%7Btheta%2C0%2Cpi%2F2%7D%5D

2*(16+6pi)

Answer 59

thanks :).