How to solve this very difficult volume problem involving integration?

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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.

2 Answers
Apr 2, 2018

#[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.

Apr 3, 2018

# 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):

http://mathcentral.uregina.ca/

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 #