Question

How to solve this very difficult volume problem involving integration?

I can't even seem to visualize the problem as I have no idea what the latter half of the question is trying to say.

Answer 1

`[8a^6]/[2sqrt2]` units ^`3`

Explanation:

From what I can deduce from the information given, the solid remaining will be a cube of side lengths of `a/sqrt2` units.

Think of square inscribed in a circle of radius `a`, then by Pythagoras, the length of the hypotenuse of the small square in the first quadrant will equal the radius of the circle, and since it is a square it must have two equal side lengths of `a/sqrt2` and so the length of the large square will equal `2a/[sqrt2` giving a volume of `[2a/sqrt2]^3`=`[8a]^3/[2sqrt2]`.= volume of cube.

I base this answer on the fact that the two great circles will go through the centre of the sphere.

I will ask some one to check this out, I could have completely misinterpreted this question.

Answer 2

`V = (8a^3)/3`

Explanation:

Two great circles lying in planes that are perpendicular to each other are drawn on a wooden sphere of radius `a`. Part of the sphere is then shaved off in such a way that each cross section of the remaining solid that is perpendicular to the common diameter of the two great circles is a square whose vertices lie on these circles.

The resulting shape is a hemispherical "dome" shape (of radius `a`) that would be suitable to cap a square building (eg a temple roof):

Then the volume, `V`, is given by:

`V = int_(-a)^(a) \ A(x) \ dx`

Where `A(x)` is the area of the square at `x` where the side would be `sqrt(2(a^2-x^2))`, leading to the area:

`A(x) = 2(a^2-x^2)`

Hence, the volume is:

`V= int_(-a)^(a) \ 2(a^2-x^2) \ dx`
`\ \ \ = 2 [a^2x-x^3/3]_(-a)^(a)`
`\ \ \ = 2 {(a^3-a^3/3) - (-a^3+a^3/3)}`
`\ \ \ = 2 (a^3-a^3/3 +a^3-a^3/3))`
`\ \ \ = 2 (4a^3/3)`
`\ \ \ = (8a^3)/3`