Question

Find the area of the region bounded by the curves y =arcsin(x/2), y = 0, and x = 2 obtained by integrating with respect to y.

Answer 1

Graph it.
Looks like it is bounded from x=0 to x=2
\[\int\limits_{0}^{2}\sin^{-1} (x/2)\]

Answer 2

yes but what do i do about integrating with respect to y, i thought that meant you can only have y expressions in the integrant

Answer 3

respect to y means leaving the y alone
You are referring to respect to x

Answer 4

um but it sort of says integrate with respect to y.... that's the part that im confused about

Answer 5

read what I wrote above

Answer 6

lol yea i read it.

Answer 7

I know this will give me the area
\int\limits_{0}^{2}\sin^{-1} (x/2) dx

im just not sure how to find that same area but with respect to y

Answer 8

If it is respect to y then you do not change the function.
y=arcsin(x/2) does not need to be change do the x= form
The integration boundaries are the range of the x coordinates

Answer 9

It means the same thing. If it is respect to x. It would have about 2 x=f(y) functions but you have to find the inverse of it.

Answer 10

Since it respect to y, the f(x) does not have to change..

Answer 11

the y=f(x) function

Answer 12

okay so then would it be \[\int\limits\limits_{-1}^{0}\ \frac{ \sin(y) }{ 2 } dy\]

Answer 13

am i completely wrong ?

Answer 14

Yes

Answer 15

Graph the function and tell me the area bounded at the x axis. That is respect to y
Respect to x is finding the area bounded at the y axis and converting y=arcsin(x/2) into x= ? which is the inverse function.

Answer 16

the boundaries are 0 and 2 right