How do you find the amplitude and period of #y= 4/3 sin (2/3)x#?
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"How do I change #int_0^1int_0^sqrt(1-x^2)int_sqrt(x^2+y^2)^sqrt(2-x^2-y^2)xydzdydx# to cylindrical or spherical coordinates?"
This question has no sense to me, the x should be INSIDE the sinusfor being a periodic function. If the x is really out the parentesis then is a tricky question. It has no period nor amplitude..
If you really wrote it correctly then #y=4/3sin(2/3)x=0.824x# ot's a linnear equation and it has no period nor amplitude, it's a linnear function.
If your equation is #y=4/3sin(2/3x)# then you need to know that the sin oscillates between -1 and 1 so if it's multipliyed by 4/3 your equaion is oscillating between 4/3 and -4/3 so you have you amplitude, it's 4/3.
for the period it is #3pi#, for x=0 the equation's value is 4/3, for #y=4/3sin(2pi)# you again have y=4/3 betwwen this two point there's no other point with x=4/3 so between these two has passed a period.
The difference between the x's will be the period, then isolating x form #2pi=2/3x# you get #x=3pi# and #3pi-0=3pi# then #T=3pi#
