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.

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- anonymous

- schrodinger

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- anonymous

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

- anonymous

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

- anonymous

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

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- anonymous

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

- anonymous

read what I wrote above

- anonymous

lol yea i read it.

- anonymous

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

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

the y=f(x) function

- anonymous

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

- anonymous

am i completely wrong ?

- anonymous

Yes

- anonymous

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.

- thomas5267

##### 1 Attachment

- anonymous

the boundaries|dw:1438909739801:dw| are 0 and 2 right

- anonymous

well @thomas5267 picture is better but you get the point.

- anonymous

Its actually \[\int\limits_{0}^{1.5}whatever the inverse function of y=\arcsin(x/2) is\]

- anonymous

why 1.5 ?

- anonymous

for respect to y you care about x axis so its bounded from x= 0 to x= 2
for respect to x you care about y axis so its bounded from y=0 to y= around 1.5

- anonymous

approaching 2 y=1.5

- anonymous

coordinates are around (2,1.5)

- anonymous

oh i see, okay that makes sense
how do i find the inverse

- anonymous

y=arcsin(x/2)
sin(y)=x/2
x=sin(y)/2
I think. I'm not sure.
You are not supposed to find the inverse to this problem anyways, since it is a bit challenging.
Unless you can use a calculator.

- anonymous

tell me if this is wrong\[\int\limits\limits\limits_{0}^{1.5}\ \frac{ \sin(y) }{ 2 } dy\]

- anonymous

It is correct according to me.
But i do not know if the find the inverse function of arcsin(x/2) correctly, but I am pretty sure.

- anonymous

okay no worris i can probably find the inverse with some reasearch. Thank's for the help!

- anonymous

just remember that the upper and lower limit of integration changes when the x=f(y) or y=f(x) changes.
No problem!

- anonymous

. I made a mistake. You have to get x by itself [x=f(y)]. This is not called an inverse.

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