Question

Calculate the volume of a solid whose base is the ellipse `4x^2 + y^2 = 4` and has vertical cross sections that are square?

Answer 1

I got `64/3` `"u"^3`, interpreting the solid as having an elliptical base, ultimately looking kind of like an armored vest (thanks, Steve!). :D

So, it is similar to this:

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An ellipse is defined as:

`x^2/a^2 + y^2/b^2 = 1`

Rewrite your ellipse equation as:

`1/4 xx (4x^2 + 1y^2 = 4)`

`=> ul(x^2/1^2 + y^2/2^2 = 1)`

Thus, `a = 1`, `b = 2` defines your ellipse, i.e. it has `x` length `1` and `y` length `2`. The projection would look like this:

Now what I would do is:

1. Obtain an equation for one-fourth of this ellipse.
2. Project it along the `x` axis from `0` to `1` using `y xx 2y` rectangles (the height spanning `0 harr (z = 2y)`).
3. Multiply the resultant volume by `4` to get the result by symmetry.

(Projecting in `0->1` is half the volume, and I only considered `(+x,+y)` values.)

Try solving for `y` in terms of `x`:

`y_(+) = 2sqrt(1 - x^2)`

In this case, our `dV`, the differential volume, shall be defined as:

`dV = (y xx 2y)dx = 2y^2dx = 2(2sqrt(1 - x^2))^2dx`

`= 8(1 - x^2)dx`

So, the projection gives `1/4` the total volume, and the total volume is:

`color(blue)(4 xx V/4) = 4 xx int_(0)^(1) 8(1 - x^2)dx`

`= 32|[x - x^3/3]|_(0)^(1)`

`= 32[(1 - 1^3/3) - (0 - 0^3/3)]`

`= 32[1 - 1/3]`

`= color(blue)(64/3 "u"^3)`

Answer 2

`64/3 \ \ \ \ \ "units"^3`

Explanation:

Consider a vertical view of the base of the object.

The grey shaded area represents a top view of the square cross section. In order to find the volume of the solid we seek the volume of a generic cross sectional square "slice" and integrate over the entire base (the ellipse)

The equation of the ellipse is:

`4x^2 + y^2 = 4`

So for some arbitrary `x`-value we have:

`y^2=4-4x^2`
`:. y = +-sqrt(4-4x^2)`

So for that arbitrary `x`-value we have the associated `y`-coordinates `y_1 , y_2` as marked on the image:

`y_1 = +sqrt(4-4x^2)`
`y_2 = -sqrt(4-4x^2)`

Thus, the length of a side of an arbitrary cross sectional square slice is:

`l = y_1 - y_2`
`\ \ = sqrt(4-4x^2) - (-sqrt(4-4x^2) )`
`\ \ = 2sqrt(4-4x^2)`

Thus the Area of an arbitrary cross sectional square slice is:

`A_("slice") = l^2`
`" " = (2sqrt(4-4x^2))^2`
`" " = (4)(4-4x^2)`
`" " = 16-16x^2`

Finally, the volume of the entire solid is the sum of those arbitrary cross sectional slices over the elliptical base:

`V = sum_("ellipse") lim_(delta x rarr 0) A_("slice") delta x`
`\ \ \ = int_(-1)^(1) 16-16x^2 dx`

`\ \ \ = [16x-(16x^3)/3]_(-1)^(1)`

`\ \ \ = (16-16/3) - (-16+16/3 )`

`\ \ \ = 16-16/3 +16-16/3`

`\ \ \ = 64/3`