## anonymous one year ago 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.

1. anonymous

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

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

3. anonymous

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

4. anonymous

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

5. anonymous

6. anonymous

7. 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

8. 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

9. 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.

10. anonymous

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

11. anonymous

the y=f(x) function

12. anonymous

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

13. anonymous

am i completely wrong ?

14. anonymous

Yes

15. 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.

16. thomas5267

17. anonymous

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

18. anonymous

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

19. anonymous

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

20. anonymous

why 1.5 ?

21. 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

22. anonymous

approaching 2 y=1.5

23. anonymous

coordinates are around (2,1.5)

24. anonymous

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

25. 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.

26. anonymous

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

27. 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.

28. anonymous

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

29. anonymous

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

30. anonymous

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