Finding the area of a disk divided by a parabola function.

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Homework Statement


The parabola y=1/2 x^2 divides the disk x^2+y^2 <or= to 8 into two equal parts. Find the area of both parts.


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The Attempt at a Solution


I have no idea of how to go about solving this. We haven't done any application problems until now and when I transferred colleges they were several chapters ahead so I'm struggling as it is. If someone could give me a system to go about solving these and point me in the right direction I'd be very grateful.
 
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Do you know how to calculate an area under a curve?
Do you know how to calculate an area between two functions?

Those two parts don't look equal to me.
 
Drawing a picture seems like a good start. In particular, at what points do the two curves intersect?
 
mfb said:
Those two parts don't look equal to me.
Agree, this question seems wrong as stated. One of the two "halves" will contain the entire bottom half of the disk and then some.
 
I suspect a transcription error. Maybe the original says "unequal".
 
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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