Posted by andy on Thursday, July 18, 2013 at 3:45pm.
One easy way to do this is to consider the figure to be a solid of revolution formed by revolving the parabola z = 1-x^2 around the z-axis.
v = ∫[0,1] πr^2 dz
where r = x
v = π∫[0,1] (1-z) dz
= π/2
The other, more direct way, of course, is to do a volume integral:
v = 4∫[0,1]∫[0,√(1-x^2)]∫[0,1-x^2-y^2] dz dy dx
= 4∫[0,1]∫[0,√(1-x^2)] 1-x^2-y^2 dy dx
= 4∫[0,1] (1-x^2)y - 1/3 y^3 [0,√(1-x^2)] dx
= 4∫[0,1] (1-x^2)^(3/2) - 1/3 (1-x^2)^(3/2) dx
= 8/3∫[0,1] (1-x^2)^(3/2) dx
= 8/3 * 1/8 (x√(1-x^2) (5-2x^2) + 3arcsin(x)) [0,1]
= π/2
thank you steve!
Another easy way is to use cylindrical coordinates, like polar coordinates with a z-axis.
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 = 1-r^2
v = ∫[0,1]∫[0,2π]∫[0,1-r^2] r dz dr dθ
= ∫[0,1]∫[0,2π] r(1-r^2) dr dθ
= ∫[0,1] 2πr(1-r^2) dr
= 2π(1/2 r^2 - 1/4 r^4)
= π/2