The base of a solid is the region between the parabolas x=y^2 and 2y^2 = 3-x . Find the volume of the solid if the cross-sections perpendicular to the -axis are equilateral triangles.
+129850 See the graph here : https://www.desmos.com/calculator/kblqq5fy8r
x = y^2 ⇒ sqrt (x) = y ⇒ 2y = 2sqrt (x) = side length of one group of equilateral triangles
2y^2 = 3 - x
y^2 = (3-x)/2
y = sqrt [ (3 - x) / 2 ] ⇒ 2y = 2 sqrt [ (3 - x) /2 ] = sqrt ( 6 - 2x) side length of the other equilateral triangles
We have two integrals
The height of the first group of equilateral triangles = 2y * sqrt (3) / 2 = y sqrt (3) = sqrt x * sqrt 3
Part of the volume =
1 1
∫ sqrt (x) * sqrt (x) * sqrt (3) dx = sqrt (3) [ x^2 / 2] = sqrt (3) / 2
0 0
The height of the other group of equilateral triangles = sqrt (6 - 2x) sqrt (3) / 2
The other part of the volume is
3
∫ sqrt ( 6 -2 x) / 2 * sqrt (6 - 2x) * sqrt (3) / 2 dx =
1
3
(sqrt 3) / 4 * ∫ ( 6 - 2x) dx =
1
3
sqrt (3)/4 * [ 6x - x^2 ] =
1
sqrt (3) / 4 * [ 18 - 9 - 6 + 1 ] =
sqrt (3) / 4 * [ 4] = sqrt (3)
So.....the total volume is
sqrt (3) / 2 + sqrt (3)
3sqrt (3) / 2
