Hey guys! Bit stumped here. Hopefully, there is someone who can answer this on this website (ik its for less advanced math but I'm gonna ask anyways). Here it is:
Let S be the solid whose base is the region in the x, y-plane bounded by y=x^2 and y=1 whose cross-sections parallel to the y-axis are equilateral triangles. Find the volume of S. Thanks!
- BasicMaths (lol what a paradox)
Where it says cross-sections parallel to the y-axis, can you confirm that this really means cross-sections parallel to the yz plane ?
For everyone who is helping me solve this problem, yes, the cross-sections are parallel to the yz-plane.
The bounds on integration are from x -1 to x = 1 (these are the x intersections of the graphs)
The base of each triangle = ( 1 - x^2) = the side of each triangle
The height of each triangle = side * sqrt (3) / 2
Thus.......the area for each triangle = (1/2) base * ( height) =
(1/2) ( 1 - x^2) (1 - x^2) *sqrt (3) /2 =
(sqrt (3) / 4) ( 1 - x^2) ( 1 - x^2) = (sqrt (3)/ 4) ( 1 - x^2)^2 = (sqrt (3) / 4) (x^4 - 2x^2 + 1)
And summing the areas of all the triangles from x = -1 to x = 1 will give us the volume of S
Since the volume is symmetric on each side of the y axis, we can actually just integrate this
1
(2 * sqrt (3) / 4) ∫ x^4 - 2x^2 + 1 dx =
0
1
(sqrt (3) / 2 ) [ x^5 / 5 - (2/3)x^3 + x ] =
0
(sqrt (3) / 2 ) [ 1/5 - 2/3 + 1 ] =
(sqrt (3) / 2) [ 6/5 - 2/3 ] =
(sqrt (3) / 2 ) [ 18 - 10] /15 =
(sqrt (3) / 2 ) [ 8 / 15] =
(4/15)sqrt (3) ≈ .4619 units ^3 = volume of S
0