calculus (point me in the right direction please?)
posted by andy on .
Let V be the volume of the threedimensional structure bounded by the region 0≤z≤1−x^2−y^2. If V=a/bπ, where a and b are positive coprime integers, what is a+b?

One easy way to do this is to consider the figure to be a solid of revolution formed by revolving the parabola z = 1x^2 around the zaxis.
v = ∫[0,1] πr^2 dz
where r = x
v = π∫[0,1] (1z) dz
= π/2
The other, more direct way, of course, is to do a volume integral:
v = 4∫[0,1]∫[0,√(1x^2)]∫[0,1x^2y^2] dz dy dx
= 4∫[0,1]∫[0,√(1x^2)] 1x^2y^2 dy dx
= 4∫[0,1] (1x^2)y  1/3 y^3 [0,√(1x^2)] dx
= 4∫[0,1] (1x^2)^(3/2)  1/3 (1x^2)^(3/2) dx
= 8/3∫[0,1] (1x^2)^(3/2) dx
= 8/3 * 1/8 (x√(1x^2) (52x^2) + 3arcsin(x)) [0,1]
= π/2 
thank you steve!

Another easy way is to use cylindrical coordinates, like polar coordinates with a zaxis.
Just as dA = r dr dθ is the area element in polar coordinates,
dV = r dz dr dθ in cylindrical coordinates.
So, now we have z = 1r^2
v = ∫[0,1]∫[0,2π]∫[0,1r^2] r dz dr dθ
= ∫[0,1]∫[0,2π] r(1r^2) dr dθ
= ∫[0,1] 2πr(1r^2) dr
= 2π(1/2 r^2  1/4 r^4)
= π/2