Question

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

Answer 1

\[r=\sqrt \theta , 0\le \theta \le \frac{ \pi }{ 4 }\]

Answer 2

\[\int\limits_{0}^{\frac{ \pi }{ 4 }}\frac{ 1 }{ 2 } \theta d \theta\]

Answer 3

its simple integration .. give it a try..

Answer 4

\[\frac{ 1 }{ 2 }\frac{( \frac{ \pi }{ 4 } )^3}{ 3 }\]

Answer 5

\[\frac{ \frac{ \pi }{ 64 } }{ 6 }\]

Answer 6

the answer is \[\frac{ \pi^2 }{ 64 }\]

Answer 7

im just not getting that

Answer 8

can u help me i'm begging u i dont want to fail please

Answer 9

please

Answer 10

with what?

Answer 11

this

Answer 12

I am the one that posted this question hun. I need help too

Answer 13

OK me to and i have no idea on how to get the help?

Answer 14

just hang tight, maybe someone will come through for us

Answer 15

i hope so

Answer 16

on the urge for crying

Answer 17

you have made some mistake in performing integration

Answer 18

ok...?

Answer 19

\[\int\limits \theta d \theta = \theta^2 /2\]

Answer 20

i hope this help now try to do it correctly. with your question.

Answer 21

\[\int\limits_{0}^{\frac{ \pi }{ 4 }}\frac{ 1 }{ 2 }(\theta)^2d \theta= \frac{ 1 }{ 2 }\int\limits_{0}^{\frac{ \pi }{ 4 }}\theta d \theta = \frac{ 1 }{ 2 }(\frac{ \theta^2 }{ 2 })\]\[= \frac{ 1 }{ 2 }(\frac{ (\frac{ \pi }{ 4 })^2 }{ 2 })= \frac{ 1 }{ 2 }(\frac{ \pi^2 }{ 32 })= \frac{ \pi^2 }{ 64 }\]

Answer 22

is this correct?

Answer 23

yes

Answer 24

thank you so much!

Answer 25

welcome..

Answer 26

@home_work check out my steps here hun