What Is the Volume of the Object Described by These Triple Integral Limits?

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Homework Statement



http://img12.imageshack.us/img12/7181/integral.th.jpg

Homework Equations


The Attempt at a Solution



Well my first attempt is to convert this to a cylindrical coordinate first, which I believed to be:

\int_0^1 \int_0^{2\pi} \int_0^1 1 \, dr \,d\theta \,dz

is this correct?
 
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Yep, that's the cylindrical coordinate version of the stated problem.
 
and when I evaluate this I got 2*pi*r, but the answer doesn't seem to be correct...
 
Keep in mind that after getting the antiderivative of the function you need to evaluate it at the upper and lower limits of integration and subtract each time. It looks like you did this for \theta and z, but what about r? Since r is not a limit of integration for either \theta or z, you're right in thinking that it shouldn't appear in the answer.

In addition, we can figure out what the answer will be without actually doing the integral if we think about it as the volume of a solid object. If you can figure out what sort of object is described by these limits of integration, then you'll be able to check your solution by comparing it to the formula for the volume of whatever shape this is.
 
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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