I have to find the volume of the solid described as follows. The base is an elli

Question

I have to find the volume of the solid described as follows. The base is an elliptical boundary curve:
x^2 + 4y^2 = 4
Cross sections perpendicular to x-axis are isosceles right triangles with hypotenuse in the base.
I will comment back if I have any questions, Thanks! I have to find the volume of the solid described as follows. The base is an elliptical boundary curve:
x^2 + 4y^2 = 4
Cross sections perpendicular to x-axis are isosceles right triangles with hypotenuse in the base.
I will comment back if I have any questions, Thanks!
x^2 + 4y^2 = 4
Cross sections perpendicular to x-axis are isosceles right triangles with hypotenuse in the base.
I will comment back if I have any questions, Thanks!

Explanation / Answer

Solution:

x^2 + 4y^2 = 4

Solving for y:

y = (±1/2)(4 - x^2)

So, the length of a hypotenuse is

=> (1/2)(4 - x^2) - (-1/2)(4 - x^2) = (4 - x^2).

Let a be the length of a leg in this isosceles right triangle.
Then, a^2 + a^2 = [(4 - x^2)]^2
==> a = (1/2) (4 - x^2)

Hence, one of these triangles has area
1/2 * (1/2) (4 - x^2) * (1/2)) (4 - x^2)
= (1/4) (4 - x^2).

Since x is in [-2, 2] (from the equation of the ellipse), the volume equals
(x = -2 to 2) (1/4) (4 - x^2) dx
= 2 (x = 0 to 2) (1/4) (4 - x^2) dx, since the integrand is even
= (1/2) (4x - x^3/3) {for x = 0 to 2}
= (1/2) {(4*2 - 2^3 / 3) - 0}
= (1/2) (16/3) = 8/3

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