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# Interesting calculus-geometry problem

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The base of a solid is the area between the parabolas $$x=y^2$$ and $$2y^2 = 3 - x$$. Calculate hte volume of the solid given that the cross sections perpendicular to the $$x$$-axis are equilateral triangles!

Thank you.

Dec 24, 2021

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The base of a solid is the area between the parabolas $$x=y^2$$ and $$2y^2 = 3 - x$$ . Calculate hte volume of the solid given that the cross sections perpendicular to the -axis are equilateral triangles!

This is how I see it.

If  the y value at  a given point is k then the side length of the equilateral triangle will be 2k and the vertical height will be (sqrt3)k

The area of the triangle (the cross section) would be 0.5*2k*(sqrt3)k = (sqrt3)k^2

$$\displaystyle \int_0^1 \sqrt3*(\sqrt x)^2\;dx\;+\;\;\int_1 ^3 \sqrt3 *\left[\sqrt{\frac{3-x}{2}}\right]^2\;\;dx\\ =\displaystyle \int_0^1 \sqrt3*x\;dx\;+\;\;\int_1 ^3 \sqrt3 *\;\frac{3-x}{2}\;\;dx\\ =\frac{\sqrt 3}{2} \left[\displaystyle \int_0^1 2x\;dx\;+\;\;\int_1 ^3 \;(3-x)\;\;dx\right]\\ =\frac{\sqrt 3}{2} \left[\displaystyle \left[x^2 \right ]_0^1+\;\;\left[\;(3x-\frac{x^2}{2})\right]_1^3\;\right]\\ =\frac{\sqrt 3}{2} \left[\displaystyle1+\;\;\;(9-\frac{9}{2})-(3-\frac{1}{2})\;\right]\\ =\frac{\sqrt 3}{2} \left[\displaystyle1+\;\;\;(\frac{9}{2})-(\frac{5}{2})\;\right]\\ =\frac{\sqrt 3}{2} \left[\displaystyle3 \right]\\~\\ =\frac{3\sqrt3}{2}\;\;\;u^3$$

Dec 26, 2021