Find the volume of the region bounded by y=sqrt(z-x^2) and x^2+y^2+2z=12?

A graphic contribution.

this is symmetric about z axis so we use polar co-ords

first paraboloid

# y=sqrt(z-x^2)#
# y^2=z-x^2#
#z_1 = y^2 + x^2 = r^2#

second

#x^2+y^2+2z=12#
#z_2=(12 - (x^2+y^2))/2 = 6 - r^2/2#

#z_1 = z_2 implies 6 - r^2/2 = r^2#
# implies r = 2#
(and # z_1 = z_2 = 4 #)

here is a 2-D plot:

We need to find the volume that is made by revolving the area between the red and blue curves around the z-axis. That is the area under the purple curve, which is entered as #z_3 = z_2 - z_1# in the box on the left.

Ie, we need to find volume

#V = int_A \ (z_2 - z_1) dA#

where volume element in polar is #dA = r \ dr \ d theta #

# implies int_0^(2 pi) int_0^2 (6 - 3/2 r^2) r \ dr \ d theta#

# = int_0^(2 pi) \ d theta * \ int_0^2 6r - 3/2 r^3 \ dr#

# = 2 pi [ 3r^2 - 3/8 r^4 ]_0^2#

# = 2 pi [ 3r^2( 1 - 1/8 r^2) ]_0^2#

# = 2 pi ( 12( 1 - 1/2) ) #

# =12 pi#

As an aside, @abubakar wanted this expressed as a triple integral . I originally skipped that step because it just adds extra notation but if we start with the general volume element and switching straight to cylindrical co-ordinates, we can say that

#int_V dV = int_(theta_0)^(theta_1) \ int_(r_0)^(r_1) \ int_(z_0)^(z_1) \ r dr d theta dz#

now neither r or z depend upon #theta# so we can immediately, using Fubini's theorem, lift #theta# outside the integration as it is independent so we have

#int_V dV =( int_(0)^(2 pi) d theta )*( \ int_(r_0)^(r_1) \ int_(z_0)^(z_1) \ r dr dz )#

#=2 pi \ int_(r_0)^(r_1) \ int_(z_0)^(z_1) \ r dr dz #

I did that to simplify the next bit, because we have #r = r(z)# and (z = z(r)#. z and r are inter-dependent. clearly!

In general terms, to do this we need either do

#=2 pi \ int_(z_0)^(z_1) \ int_(r = r_0(z))^(r_1(z)) \ r \ dr \ dz #

OR

#=2 pi \ int_(r_0)^(r_1) \ int_(z = z_0(r))^(z_1(r)) \ r \ dz \ dr #

The explanation and drawings here are helpful

The key to this is that these are iterated integrals. The #dA = dx \ dy = r \ dr \ d theta# element notation is very useful but perhaps misleading.