Finding Area of Region Enclosed by y = \sqrt[3]{x}, Tangent Line, and y-Axis

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Homework Statement


Find the area of the region enclosed by y = \sqrt[3]{x}, the tangent line to this curve at x = 8, and the y-axis.


Homework Equations


Definate integral properties, fundamental theorem of calculus



The Attempt at a Solution


I know and understand what the question is asking for--find the area of the region--i'm just having problems actually finding the region I need to calculate. I can't seem to figure out what they want through the tangent line; do I need to simply take the derivative of \sqrt[3]{x} and use it, plug in x=8 into the derivative and then use it, or do something completely different? Any help you can guys can give to point me in the right direction here would be greatly appreciated, thanks in advance.
 
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It might be easier if you think of the function as x= y3. Then dx/dy= 3y2 so dy/dx= 1/(3y). Now what is y when x= 8?
 
Find the equation for the tangent line, then draw both the tangent line and y on the same axes. It will be obvious from there which region you want to calculate.

It will look like this:

http://img29.imageshack.us/img29/32/region.jpg
 
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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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