Volume of Revolution: Solve Assignment Question

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The question I need to solve for an assignment is as follows:
Find the volume of the solid that is obtained by revolving the region R around the x-axis.

I figured that the volume would just be the integral of pi R^2 dx, so that would just be pi R^2 x, but that is not the answer. I suspect I am misinterpreting the question.
 
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Well, R is probably a function of x. If you write the question, as stated, then we'll be more able to help you!
 
My problem is that that is the problem as stated. It doesn't give a function in terms of x.
 
Ah nevermind, I found out that the question was referring to an area already found in a previous question.
 
Yes, and if it didn't specify the function, it would be \pi\int^b_a R^2 dx anyway, where R is the closed interval (a,b).
 
Even that wouldn't make sense! If R is an interval it is not a number! I think both you and ductape are confusing the interval R with "radius" R.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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