Double Integrals: Solve Volume Problem

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Double Integrals - Please Help!

Hi, I have a question and it asks me to evaluate the volume generated by the area bounded by y = x^3 and y = x^(1/3) and the function z = x^2.y

I'm just having a few problems with setting up the ranges of my variables x and y. I drew a sketch of the area in the x-y plane but I'm not sure what my ranges should be. My guess is: 0 <= x <= 1 and 0 <= y <= [x^(1/3) - x^3].

Help appreciated.
 
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"Hi, I have a question and it asks me to evaluate the volume generated by the area bounded by y = x^3 and y = x^(1/3) and the function z = x^2.y"

To do dbl int's always choose one of the variables to put definite boundaries on (put them on the outer-most int), then put limits on the second int in terms of the first. You know that the graphs intersect at (0,0) and (1,1), so I'd suggest using 0 and 1 as the bounds of x, then choose y = x^3 and y = x^1/3 for the bounds on y. Then, integrate!
 
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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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