Question

Find the area of the region that is bounded by the given curve and lies in the specified sector

Answer 1

\[r= 3+2 \sin \theta ,..... -\frac{ \pi }{ 2 }\le \theta \le \frac{ \pi }{ 2 }\]

Answer 2

I will type what i have so far... give me a min plz

Answer 3

\[A= \int\limits_{\frac{ -\pi }{ 2 }}^{\frac{ \pi }{ 2 }}\frac{ 1 }{ 2 }(3+2 \sin \theta)^2 d \theta\]=\[A= \frac{ 1 }{ 2 }\int\limits_{\frac{ -\pi }{ 2 }}^{\frac{ \pi }{ 2 }}(9+ 12 \sin \theta + 4 \sin^2 \theta) d \theta\]=\[A= \frac{ 1 }{ 2 }\int\limits_{\frac{ -\pi }{ 2 }}^{\frac{ \pi }{ 2 }}(9 +4 \sin^2 \theta) d \theta\]=\[A= \frac{ 1 }{ 2 }(2)\int\limits_{0}^{\frac{ \pi }{ 2 }}( 9 +4 \sin^2 \theta) d \theta\]

Answer 4

I know a half angle identity goes in at this point, but this is where i get messed up
I also know that the answer is \[\frac{ 22\pi }{ 4 }\]

Answer 5

the sin^2 theta needs to becom \[\frac{ 1 }{ 2 }(1-\cos 2 \theta )\] and this messes me up

Answer 6

\[\int\limits_{0}^{\frac{ \pi }{ 2 }}( 9 +4 [\frac{ 1 }{ 2 }(1-\cos 2 \theta)] )d \theta\]

Answer 7

Not sure I would do that:

\(\int \left(3+2\sin(\theta)\right)^{2}\;d\theta = \theta\cdot\left(3+2\sin(\theta)\right)^{2} - \int \theta\cdot 2\cdot\left(3+2\sin(\theta)\right)\cdot\cos(\theta)\;d\theta\)

After that, your half angle looks a lot more obvious.

Answer 8

I dont know how to do your way, im sorry

Answer 9

he did integration by parts

Answer 10

oh...

Answer 11

which looks really cute too

Answer 12

that explains a lot... i suck at that the most

Answer 13

though I think he is missing a multiple of 2 in the integral part

Answer 14

btw, hi freckles! its been a min

Answer 15

I would prefer to continue the way i started, its easier for me

Answer 16

and hey back at ya

Answer 17

that's fine too

Answer 18

but then again, I am stuck at where i posted... the algebra part

Answer 19

this is what you posted
\[\int\limits_{0}^{\frac{ \pi }{ 2 }}( 9 +4 [\frac{ 1 }{ 2 }(1-\cos 2 \theta)] )d \theta\]
i think doing a little distribution would make this looks tons prettier

Answer 20

first of all 4/2 =2
so you actually have for that mumbo jumbo part the following
9+2(1-cos(2theta)

Answer 21

I forgot to close that

Answer 22

I thought that too,.. but where does the 4 come from in the answer?

Answer 23

but you can now distribute that two there

Answer 24

oh are you following an example from the book?
this isn't your work?

Answer 25

this is my homework online, it gives me the answer...

Answer 26

well, i chose the correct multiple choice at least... the answer is 22pi/4

Answer 27

like as it shows you the work and you are trying to follow it?

Answer 28

it doesnt show me the work, i just chose the right answer :(

Answer 29

lol sorry
but you came up with this
\[\int\limits_{0}^{\frac{ \pi }{ 2 }}( 9 +4 [\frac{ 1 }{ 2 }(1-\cos 2 \theta)] )d \theta\]
and you are asking me where you got the 4 from?

Answer 30

no, the answer is \[\frac{ 22\pi }{ 4 }\] i was wondering about the 4 in the denom

Answer 31

i know im close here, i just get stuck when integrating here...

Answer 32

\[\int\limits_{0}^{\frac{\pi}{2}}(9+\frac{4}{2}(1-\cos(2 \theta)) d \theta \\ \int\limits_0^\frac{\pi}{2}(9+2 (1-\cos(2 \theta)) d \theta \\ \int\limits_0^\frac{\pi}{2}(9+2 -2 \cos(2 \theta)) d \theta \\ \int\limits_0^\frac{\pi}{2}(11-2 \cos(2 \theta)) d \theta \]
so you got to this part right ?

Answer 33

oooh... nope, i didnt do that far, i got confused there

Answer 34

let me know if you don't understand something I did there with the distribution part

Answer 35

if you have no questions with that part
can I see your antiderivative

Answer 36

I see that now! but i forget how to do the integration, is it \[11 \theta - 2(\frac{ 1 }{ 2 }) \cos 2 \theta \]

Answer 37

almost use the the antiderivative of cos(x) is sin(x)

Answer 38

oh yea...

Answer 39

oh yea...\[A= 11 \theta - \sin 2 \theta \rightarrow \frac{ \pi }{ 2 }\]

Answer 40

right
\[[11 \theta-\cancel{2} (\frac{1}{\cancel{2}} ) \sin(2 \theta)]_0^\frac{\pi}{2}\]

Answer 41

and yeah you only need to plug in the upper because the lower makes it all 0

Answer 42

ok so...

Answer 43

\[11(\frac{ \pi }{ 2 })- \sin (\frac{ \pi }{ 2 })= \frac{ 11\pi }{ 2 }-1\]

Answer 44

well the only problem with that is you had a 2 inside that sin thingy

Answer 45

2*pi/2=pi

Answer 46

sin(pi)=0

Answer 47

so you have 11pi/2

Answer 48

multiply top and bottom by 2 and you have your desired unsimplified answer

Answer 49

ok... where does the 22pi/4 come from then?

Answer 50

oh... i guess i dont see where the multiplying by 2 top and bottom came from

Answer 51

well 2/2=1

Answer 52

yes.. OH , where we simplified above?

Answer 53

\[\frac{11\pi}{2} \cdot \frac{2}{2}=\frac{22 \pi}{4}\]
this answer they have is simplified
11pi/2 is actually the simplified version of the answer

don't ask me why they prefer the other :p

Answer 54

I bet you if this wasnt multiple choice and i plugged in 11pi/22, it would say correct!

Answer 55

this answer they have is un-simplified*

Answer 56

weird

Answer 57

you mean 11pi/2

Answer 58

lol thanks AGAIN

Answer 59

np

Answer 60

I may have another.... well c

Answer 61

ok i might be around a little while longer

Answer 62

Post on another thread.
Get better at Integration by Parts. :-)