Calculate Volume of Parabolic Region with Equilateral Triangles

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The base of S is an parabolic region {(x,y)|x^2<y<1}. Cross-sections perpendicular to the y-axis are equilateral trinagles
pls help
no clue
 
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Draw it out. The cross-section in the yz-plane is half of a parabola: the heights of the triangles.The easiest way to approach the volume of this shape is to use your knowledge of the area of each triangle.
 
This is a standard homework problem so I am moving this to the homework section.

It will help you to know that if an equilateral triangle has side length s, then it has altitude \sqrt{3}s and so area (\sqrt{3}/2)s^2.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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