Question

The base of a solid is the region in the first quadrant bounded by the ellipse x^2/a^2 + y^2/b^2 = 1. Each cross-section perpendicular to the x-axis is an isosceles right triangle with the hypotenuse as the base. Find the volume of the solid in terms of a and b.

I'm having trouble figuring out the area equation and the height and base that goes in it so I can integrate in terms of a and b to find volume.

Answer 1

when the cross-section has a base of 2y, its height is thus y√2

So, the volume is

v = ∫[-a,a] 1/2 bh dx
= ∫[0,a] (2y)(y√2) dx
= 2√2b^3/a^3 ∫[0,a] (a^2 - x^2)^(3/2) dx
now it's just straightforward integration and evaluation

Answer 2

area of triangle at x with base y:

altitude of triangle is y/2, same as half the base

so area = (y/2)(y/2) = y^2/4

y^2 = b^2 (1 - x^2/a^2)
so
y^2/4 = (b^2/4) (1 - x^2/a^2)

integrate that from x = 0 to x = a

Answer 3

It is all in the first quadrant, base = y

Answer 4

Damon is right about the quadrant. My integral used the whole ellipse as the solid's base, not just the 1st quadrant. So, adjust the values as needed.

However, Damon's triangles are not evaluated with the base as hypotenuse. So, the altitude will be base/√2.

It is unclear, however, whether the entire ellipse is to be used as the base, so each triangle's altitude is 2y/√2, or wheher only the top half of the ellipse is to be used, making each triangle's altitude y/√2.

Answer 5

sure, the base is y, the hypotenuse

the center of the base is at y/2

half the base = the altitude

Answer 6

Dang! I was figuring the size of the legs, not the altitude.

As Emily Latella would say, "Never mind."