Question

How do you find the volume of the solid with base region bounded by the curve `9x^2+4y^2=36` if cross sections perpendicular to the `x`-axis are isosceles right triangles with hypotenuse on the base?

Answer 1

The volume of the solid can be found by
`V=1/4int_{-2}^2(36-9x^2)dx=24`.

Let us look at some details.
If we rewrite `9x^2+4y^2=36` as (by dividing by 36)
`x^2/2^2+y^2/3^2=1`,
we realize that it is en ellipse that spans from `x=-2` to `x=2`.

By solving for `y`, we have
`y=pm1/2sqrt{36-9x^2}`
Since the area `A(x)` of the cross-sections can be expressed as

`A(x)=1/2`(base)(height)
`=1/2 cdot sqrt{36-9x^2}cdot 1/2sqrt{36-9x^2}`
`=1/4(36-9x^2)`

So, the volume of the solid can be found by
`V=1/4int_{-2}^2(36-9x^2)dx`
by symmetry about the y-axis,
`=1/2int_0^2(36-9x^2)dx`
`=1/2[36x-3x^3]_0^2=1/2(72-24)=24`