Cameronmx9
How to put y=4cos(x) in X= form?
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cinar
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\[x=\cos^{-1} (y/4)\]
BlingBlong
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It is only allowed in said domain in order to make it the inverse function (It needs to pass the horizontal line test)
BlingBlong
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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]
BlingBlong
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ignore the mistake i made in the variables
Cameronmx9
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Here is the problem I'm working at:
Cameronmx9
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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
Cameronmx9
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I think I'm using the cross-section method, but am not too sure on the radius.
Cameronmx9
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Which I think would be (4-4sin(x)
BlingBlong
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Wait do you want 4sin(x) to equal 4?
Cameronmx9
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I'm honestly not too sure on where to start with this problem.
BlingBlong
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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
Cameronmx9
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Are you familiar with solids of rotation?
BlingBlong
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tbh no I know trig functions though but meh you should ask in chat for help
Cameronmx9
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haha alright, thanks for the help
cinar
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Cameronmx9
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@Cinar, do you know which method I would use here?
cinar
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little bit (:
I am trying to find it
Cameronmx9
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Sweet, thanks
cinar
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what is the rotation axis?
Cameronmx9
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it is y=4
Cameronmx9
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so I'm thinking the radius is (4-4sin(x))
Cameronmx9
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Any luck?
cinar
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nope
cinar
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\[V=\pi \int\limits_{0}^{2\pi}(4-4\sin x)^2dx\]
cinar
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