Wednesday

April 16, 2014

April 16, 2014

Posted by **andy** on Thursday, July 18, 2013 at 3:45pm.

- calculus (point me in the right direction please?) -
**Steve**, Thursday, July 18, 2013 at 4:04pmOne 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

- calculus (point me in the right direction please?) -
**andy**, Thursday, July 18, 2013 at 4:40pmthank you steve!

- calculus (point me in the right direction please?) -
**Steve**, Thursday, July 18, 2013 at 6:54pmAnother 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

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