Question

Find the Area. Details inside

Answer 1

if\[r=\sin \theta \] and \[(\frac{ \pi }{ 3 })\le \theta \le (\frac{ 2\pi }{ 3 })\]
use \[A= \frac{ 1 }{ 2 }\int\limits_{\frac{ \pi }{ 2 }}^{\frac{ 2\pi }{ 3 }}( \sin^2 \theta) d \theta\]

Answer 2

my work =\[A= \frac{ 1 }{ 2 }\int\limits_{\frac{ \pi }{ 2 }}^{\frac{ 2\pi }{ 3 }} \frac{ 1 }{ 2 }(1-\cos 2 \theta) d \theta\]

Answer 3

\[A= \frac{ 1 }{ 4 }\int\limits_{\frac{ \pi }{ 2 }}^{\frac{ 2\pi }{ 3 }}1-\cos2 \theta d \theta\]

Answer 4

this is where i get lost... i think

Answer 5

shoot... all of those pi/2 are supposed to be pi/3

Answer 6

is it \[A= \frac{ 1 }{ 2 }[\theta - \frac{ 1 }{ 2 }\sin 2 \theta] \frac{ \pi }{ 3 }\rightarrow \frac{ 2\pi }{ 3 }\]

Answer 7

@freckles hi!

Answer 8

*\[\int\limits_{\frac{\pi}{3}}^{\frac{2\pi}{3}}1d \theta - \int\limits_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \cos(2 \theta)d \theta\]

Answer 9

looks like i may have integrated the sin wrong, according to what you posted

Answer 10

Sorry, that was just a typo.

Answer 11

is 1/2 sin 2 theta right?

Answer 12

jey @Jlockwo3 why did you change your 1/4 outside the integral to a 1/2?

Answer 13

oops, that was a typo as well

Answer 14

let me clear this up...

Answer 15

is it... \[\frac{ 1 }{ 4 }[ \theta- \frac{ 1 }{ 2 }\sin 2 \theta]\] evaluated from\[\frac{ \pi }{ 3 }\rightarrow \frac{ 2\pi }{ 3 }\]

Answer 16

that is awesome! :) no complaints here anymore.

Answer 17

ok... so, i get real confused doing this part... i will type to you my work

Answer 18

\[A= \frac{ 1 }{ 4 }[ (\frac{2 \pi }{ 3 }-\frac{ 1 }{ 2 }\sin 2*\frac{ 2 \pi}{ 3 })-(\frac{ \pi }{ 3 }-\frac{ 1 }{ 2 }\sin2\frac{ \pi }{ 3 })]\]

Answer 19

\[A= \frac{ 1 }{ 4 }[(\frac{ 2\pi }{ 3 }-\frac{ 1 }{ 2 }\sin \frac{ 4\pi }{ 3 })-(\frac{ \pi }{ 3 }-\frac{ 1 }{ 2 }\sin \frac{ 2\pi }{ 3 })]\]

Answer 20

is that right so far?

Answer 21

groovy looking to me
(Yes)

Answer 22

so much to type lol... ok, so is sin 2pi/3 = \[\sqrt{3}/2\]

Answer 23

im not sure what to put for sin 4pi/3

Answer 24

and you will use that sin(4pi/3) is -sqrt(3)/2
and you will also use that sin(2pi/3) is sqrt(3)/2

Answer 25

should i break that 2 back off and do \[2*\frac{ \sqrt{3} }{ 2 }\]

Answer 26

oh! theres a negative sign there... ok

Answer 27

lets see here...\[A= \frac{ 1 }{ 4 }[(\frac{ 2\pi }{ 3 }-\frac{ 1 }{ 2 }(\frac{ -\sqrt{3} }{ 2 }))-(\frac{ \pi }{ 3 }-\frac{ 1 }{ 2 }(\frac{ \sqrt{3} }{ 2 }))]\]

Answer 28

A=\[\frac{ 1 }{ 4 }[(\frac{ 2\pi }{ 3 }+\frac{ \sqrt{3} }{ 2 }-\frac{ \pi }{ 3 }+\frac{ \sqrt{3} }{ 4 })\]?

Answer 29

that first squar root 3/2 should be divided by 4

Answer 30

sorry my openstudy was failing
but i'm here again

Answer 31

or no?

Answer 32

thats ok, i just finished typing anyway

Answer 33

well...

Answer 34

I think you forgot to multiply something
one sec
I will show

Answer 35

k

Answer 36

\[A= \frac{ 1 }{ 4 }[(\frac{ 2\pi }{ 3 }-\frac{ 1 }{ 2 }(\frac{ -\sqrt{3} }{ 2 }))-(\frac{ \pi }{ 3 }-\frac{ 1 }{ 2 }(\frac{ \sqrt{3} }{ 2 }))] \\ A=\frac{1}{4}[(\frac{2 \pi}{3}+\frac{\sqrt{3}}{\color{red}4})-(\frac{\pi}{3}-\frac{\sqrt{3}}{4})] \\ A=\frac{1}{4}[\frac{2 \pi}{3}-\frac{\pi}{3}+\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}]\]
now you can add your terms with pi together (or subtract)
and add you terms with no pi in them together

Answer 37

on you said that

Answer 38

i'm sorry

Answer 39

you fixed it :p

Answer 40

ok...do the square roots combine as like terms? \[\frac{ 1 }{ 4 }(\frac{ \pi }{ 3 }+\frac{ \sqrt{3} }{ 4 })\]

Answer 41

\[\frac{1}{4}+\frac{1}{4}=\frac{2}{4} =\frac{1}{2} \\ \frac{1}{4}+\frac{1}{4}=\frac{1}{2} \\ \text{ now multiply both sides by } \sqrt{3} \\ \sqrt{3}(\frac{1}{4}+\frac{1}{4})=\sqrt{3}\frac{1}{2} \\ \frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{2}\]

Answer 42

ohh... i c....

Answer 43

so you have
\[A=\frac{1}{4}(\frac{\pi}{3}+\frac{\sqrt{3}}{2})\]

Answer 44

\[A= \frac{ \pi }{ 12 }+\frac{ \sqrt{3} }{ 8 }\]

Answer 45

is that the final?

Answer 46

yep

Answer 47

wow... that one is difficult! thanks!

Answer 48

np

Answer 49

you did all the work

Answer 50

I have to run to the bank and grocery store
you have fun with calculus :)

Answer 51

ok!