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?

calculus (point me in the right direction please?) 
Steve,
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 
calculus (point me in the right direction please?) 
andy,
thank you steve!

calculus (point me in the right direction please?) 
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