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.
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
Explanation:
From what I can deduce from the information given, the solid remaining will be a cube of side lengths of
Think of square inscribed in a circle of radius
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.
# 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
The resulting shape is a hemispherical "dome" shape (of radius
Then the volume,
# V = int_(-a)^(a) \ A(x) \ dx #
Where
# 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 #