Whoopsie, you're absolutely right, 1/2 stays!
I solved the problem!
For those interested in the solution, taking the integral of the above expression between x=0 and x=2 gives you the answer 1.3334 which is indeed what I was looking for :)
Thanks for your help, Defennder!
Okay, so I've got it up like on the diagram.
http://www.dimitrix.org/diagram2.jpg
Doing my high school math right, Sin \theta = \frac{\sqrt{4-x^2}}{2}
y = Sin(2\theta) = 2 Sin(\theta) Cos(\theta)
So plugging in all I know, after simplification I get: y = \frac{x\sqrt{4-x^2}}{2}
And then I...
Trigonometry is seriously not my strong point so correct me if I am wrong, but since there's two unknowns, \theta and X I am not sure how I can find an expression for Sin \theta
In any case, furthermore, say I did find an expression for Sin \theta, I'm wouldn't really know how I could plug...
Alright, but this integral doesn't seem any easier to integrate:
V= \int Sin(ArcCos(x/2)) d\theta
I was thinking, as this question is part of bigger exercise where I calculated the area to begin with then perhaps this help me in any way to find the volume?
Homework Statement
Let R be the region in the 1st quadrant in the region enclosed by x=2cos(\theta) and y=sin(2\theta) Suppose R is rotated around the x-axis.
Find the volume of the resulting solid.
Homework Equations
The formula for the solid of revolution is:
V= \pi\int...