Question

Find the area of the interior of r = 12sin(theta)

Answer 1

sounds like an integral haha

Answer 2

:O? :) I like integrals.

Answer 3

idk...i just said sounds like...usually when i see area i think of integrals =))

Answer 4

i dont know about integrals of polar coordinates though :C

Answer 5

first find the limits for theta
solve
12sin(theta)=0

Answer 6

so it is an integral @sami-21 ?

Answer 7

yes

Answer 8

@lgbasallote is is just to find the area of circle whose radius is 6

Answer 9

:O hmm :)

Answer 10

here is helpful graph.

Answer 11

;) yeah I saw wolfram. So woulod it be arctan(12) = theta?

Answer 12

arcsin(0) :P

Answer 13

12 disappears?

Answer 14

I figured it swapped, but wasn't sure where the 12 goes.

Answer 15

so theta = 0 anyways....

Answer 16

yes
0=12sin(theta)
sin(theta)=0/12
theta=arcsin(0) !

Answer 17

Ah okay I thought about 0/12 too :P.

Answer 18

theta = 0.

Answer 19

and where else sin is zero in the range of 0..2pi :P

Answer 20

pi :)

Answer 21

yes !!!!

Answer 22

:D

Answer 23

so theta = pi?

Answer 24

you know the integral for area in polar coordinates is.
\[\Large Area=\int\limits_{\theta_{1}}^{\theta_{2}}\frac{1}{2}r^2d \theta\]
so here
r=12sin(theta)
and you have found the limits for theta.
so integral becomes
\[\Large Area=\frac{1}{2}\int\limits_{0}^{\pi} (12\sin(\theta))^2d \theta\]

Answer 25

so we should change it to 144/2 = 72 int sin^2(theta) d theta

Answer 26

so for sin^2 should we change it using the double angel/half angel formulas?

Answer 27

yes !!

Answer 28

This wolfram thing has ruined the student to do integration by themselves :P
Even simple integral cannot be solved by students themselves who are addicted to wolfram :(.
When I was studying integrals there was no wolfram thing :P

Answer 29

:)

Answer 30

sorry just got back.

Answer 31

wolfram is good to check your answers or to ask basic questions.

Answer 32

dit you get it now ?

Answer 33

yessir.. It asks me to graph it tho so Idk what to do with that...

Answer 34

"Graph and then find the area r = 12sin(theta)"

Answer 35

It gives me a circle from -6 to 6 for both x,y...

Answer 36

which is what the circle is.....

Answer 37

or should be from 0 to 12 for y.

Answer 38

should be from zero to 12 on y. because sin(90)=1 so r=12 !!

Answer 39

let me see if I can post it...

Answer 40

@sami-21

Answer 41

what you want me to do ?
to check the graph ?

Answer 42

im not sure how to graph it...
it should be 0-12

Answer 43

also I gpt 36x - 18sin(2x)

Answer 44

yes it should be
just try
o,pi/2,pi,3pi/2 ,2pi in the given equation r=12sin(theta) and plot the values of r yo get from above angles . then connect those point.

Answer 45

should I do the formula sin(2u) = 2sin u cos u?

Answer 46

or can I leave it as that?

Answer 47

it should come out the same with the limits...

Answer 48

nooooooooooooooooooooooooooooooooooooooooooo :P
it is sin^2(x) not sin(2x)
use
\[\Large \sin^2(x)=\frac{(1-\cos(2x)}{2}\]

Answer 49

yeah that's up top... But I end up with 36x-18sin(2x)

Answer 50

I ended up geting 36pi.

Answer 51

good that's correct !!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 52

the reason I ask is when I checked with the evil wolfram it said it switch it over...

http://www.wolframalpha.com/input/?i=integral+1%2F2+%2812sin%28x%29%29%5E2

Answer 53

Just to check your answer !
I told you it is to find the area of circle whose radius is 6 you know formula for area is

\[\Large Area=\pi r^2\]
So
\[\Large Area= \pi (6(^2)=36 \pi \]

Answer 54

oh cooool!

Answer 55

so it is ok now :)

Answer 56

now to graph this stupid thing :(

Answer 57

so you said to plug in 0 pi etc for theta and plot huh....

Answer 58

http://www.wolframalpha.com/input/?i=plot+%2812sin%280%29%29%2C%2812sin%282%2F3pi%29%29%2C%2812sin%28pi%29%29%2C%2812sin%282pi%29%29

Answer 59

:O?

Answer 60

it is the equation of circle centered on y axis with radius r=6
the standard equation of circle centered at y axis with radius a is given by.
\[\Large r=2asin(\theta)\]

Answer 61

hmmm...

Answer 62

this means it will not comes down below y axis.
study about polar graphs in your text.

Answer 63

so how do I graph this bs grrr!! :(

Answer 64

it should be all the wa to the top, do I stop at 6 like wtf do i DPOOO :(

Answer 65

maybe I just weont graph it LOL