In solving the volume of a body of revolution by definite integral, there is a solid, with the ellipse whose major half axis is a = 10 and the minor half axis is b = 5 as the bottom, and the section perpendicular to the major axis All equilateral triangles, find the volume of solid. Correct answer (1000 √ 3) / 3. Thank you for your help

In solving the volume of a body of revolution by definite integral, there is a solid, with the ellipse whose major half axis is a = 10 and the minor half axis is b = 5 as the bottom, and the section perpendicular to the major axis All equilateral triangles, find the volume of solid. Correct answer (1000 √ 3) / 3. Thank you for your help

(there's something wrong with the title. It shouldn't be a rotator. But for the sake of explanation, it's called a rotator.)
Let the major axis of the ellipse be the X axis and the minor axis be the Y axis. Then take a small section of △ x, and the shape of the solid cut by the plane perpendicular to the X axis should be a triangular prism with an equilateral bottom, a side length of 2Y and a height of △ X
∴△V=(1/2*2y*√3*y)*△x=√3*y^2*△x
Quarter DV = √ 2 * y ^ 2 * DX
It can be seen from the symmetry of the figure that the volume of the body of revolution in the positive half axis of X axis is equal to that in the negative half axis of X axis
The total volume is v = 2 ∫ √ 3 * y ^ 2 * DX (upper limit is 10, lower limit is 0)
The parameter form of elliptic equation is as follows
x=10cosθ
y=5sinθ
Substituting v = 2 ∫ √ 3 * y ^ 2 * DX = ∫ √ 3 * (5sin θ) ^ 2D (10cos θ) (at this time, the upper and lower limits of integral corresponding to θ are 0, π / 2 respectively)
V=500√3∫(1-cosθ^2)d(cosθ)
∫ (1-cos θ ^ 2) d (COS θ) = 2 / 3
So v = (1000 √ 3) / 3
If not very clear, we can discuss it again

What is the name of the three-dimensional figure obtained by rotating an ellipse around its axis of symmetry? How to calculate its volume?

Ellipsoid v = 4 π ABC / 3
Two of ABC in your question are equal, that is, the major axis or minor axis of the ellipse

The definite integral is used to find the area where the volume of the body of revolution y = x ^ 2 intersects y = x ^ 0.5, and the volume is obtained by rotating around the X axis

y=x^2 y=x^0.5
x^2=x^0.5
X = 0 or x = 1 [0,1] x ^ 2