Volume of Triangular Solid: A Solution Attempt

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The attempt at a solution[/b]Question: Find the volume of the solid whose base is the triangular region of the xy-plane with vertices (0,0),(1,0),(0,1) and whose cross sections perpendicular to the y-axis are equilateral triangles.

I have the problem set up. just don't know how to get the cross sections of the triangles. i know the area is 1/2 bh i thought maybe similar triangles where the height would be sqrt(3)/4x and x=1-y. am i close? help
 
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Sure, you're close. The base is x=(1-y). The height is base*sqrt(3)/2. So what's the area?
 
so the area is sqrt(3)/2 (1-y)^2? and i would just integrate that from 0 to 1 right?
 
Almost right. The area is (1/2)*b*h. What happened to the (1/2)?
 
ah. so its sqrt(3)/4 (1-y)^2
 
Yes it 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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