How to put y=4cos(x) in X= form?

- anonymous

How to put y=4cos(x) in X= form?

- jamiebookeater

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- anonymous

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

- anonymous

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

- 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]

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- anonymous

ignore the mistake i made in the variables

- anonymous

Here is the problem I'm working at:

- 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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- 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

- anonymous

Are you familiar with solids of rotation?

- anonymous

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

- anonymous

haha alright, thanks for the help

- anonymous

http://www.wolframalpha.com/input/?i=plot+y%3D4sin%28x%29

- anonymous

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- anonymous

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

- anonymous

little bit (:
I am trying to find it

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- anonymous

Sweet, thanks

- anonymous

what is the rotation axis?

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- anonymous

it is y=4

- anonymous

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

- anonymous

Any luck?

- anonymous

nope

- anonymous

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

- anonymous

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