Question

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

Answer 1

\[r= \sqrt \sin \theta , 0\le \theta \le \frac{ \pi }{ 2 }\]

Answer 2

you know it is in the first quadrant and that you have to integrate

Answer 3

i know that \[A= \int\limits_{0}^{\frac{ \pi }{ 2 }}\frac{ 1 }{ 2 }\sin \theta\] d theta

Answer 4

do i use half angle identity?

Answer 5

your integral is wrong

Answer 6

it should be sin theta ^ (1/2)

Answer 7

which would become integral of \[\frac{ 1 }{ 4 }\cos \theta\]

Answer 8

well if you are treating r as if it where y then it would be the same thing as y = sqrt(sin theta)

Answer 9

then you would convert it to sin theta ^ (1/2) and integrate from there

Answer 10

\[A= \frac{ 1 }{ 2 }r^2 \rightarrow A= \frac{ 1 }{ 2 }(\sqrt{\sin \theta})^2d \theta \rightarrow A= \frac{ 1 }{ 2 }\sin \theta\]

Answer 11

that is what i meant

Answer 12

A= \[\int\limits_{0}^{\frac{ \pi }{ 2 }}\frac{ 1 }{ 2 }\sin \theta d \theta\]

Answer 13

to my knowledge the way you find the area under a curve is by simply integrating, and assuming it is the area between two curves then you would subtract the top curve from the bottom curve when you integrate

Answer 14

ok

Answer 15

so then..... i'm not sure where to begin. I know the answer is 1/2