Recent content by dimitrix

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...