## anonymous 4 years ago How to put y=4cos(x) in X= form?

1. anonymous

$x=\cos^{-1} (y/4)$

2. anonymous

It is only allowed in said domain in order to make it the inverse function (It needs to pass the horizontal line test)

3. anonymous

Inverse y = 4cos(x) x/4 = 4cos(x)/4 y/4 = cos(x) take inverse of each side arccos(y/4) = arccos(cos(x)) arccos(y/4) = 1x arccos(y/4) = x Remember that arccos is the inverse function of cos(x) and is limited to the domain [0, pi]

4. anonymous

ignore the mistake i made in the variables

5. anonymous

Here is the problem I'm working at:

6. anonymous

Find the volume of the solid generated by revolving the described region about the given axis: The region in the first quadrant bounded above by the line y=4 and by the curve y=4sin(x) for the interval 0≤x≤π2 about the line y=4

7. anonymous

I think I'm using the cross-section method, but am not too sure on the radius.

8. anonymous

Which I think would be (4-4sin(x)

9. anonymous

Wait do you want 4sin(x) to equal 4?

10. anonymous

I'm honestly not too sure on where to start with this problem.

11. anonymous

I'm confused by your question but if you want it to equal four you can use the unit circle and think where is cos(x) = 1, the answer being pi

12. anonymous

Are you familiar with solids of rotation?

13. anonymous

tbh no I know trig functions though but meh you should ask in chat for help

14. anonymous

haha alright, thanks for the help

15. anonymous
16. anonymous

17. anonymous

@Cinar, do you know which method I would use here?

18. anonymous

little bit (: I am trying to find it

19. anonymous

Sweet, thanks

20. anonymous

what is the rotation axis?

21. anonymous

it is y=4

22. anonymous

so I'm thinking the radius is (4-4sin(x))

23. anonymous

Any luck?

24. anonymous

nope

25. anonymous

$V=\pi \int\limits_{0}^{2\pi}(4-4\sin x)^2dx$

26. anonymous