draw a picture
x^2 + y^2 = 16
is a circle of radius 4
it extends up as a round cylinder from
z = 0
to
z = 4
and we must also integrate over the top and bottom
Now as Mathmate told you it is easier to do this in polar coordinates because it is a cylinder with its centerline up the z axis
in polar coordinates on the top and bottom where z is constant
dA = r dr dT where r is distance from origin in x y plane and T is angle theta counterclockwise from x axis
now at the bottom of the cylinder where z = 0:
we integrate from r = 0 to r = 1 and from theta = 0 o theta = 2 pi
(x^2+y^2 )dS
but x^2 + y^2 = r^2
so we have
r^2 dS but dS = r dr dT
so we have
integral r^3 dR dT from 0 to 4 and 0 to 2pi
2 pi (4)^4/4 = 128 pi on the bottom
SAVE that
on the top, x^2+y^2 = r^2 = 4 and z^2 = 16
so you have
(r^2+16)r dr dT from 0 to 4 and from 0 to 2 pi
so
2 pi [ (4^3)/4 + 16 (4^2)/2 ]
you do that and SAVE it
NOW the sides of the cylinder
x^2+y^2 = 16 all the time but z goes from 1 to 4
ds = 2 pi r dz where r = 4
ds = 8 pi dz
so we have
integral of
(1+z^2)(8 pi dz ) from z = 0 to z = 4
8 pi [ 4 + 4^3/3 ]
you calc and SAVE that
NOW add up the contributions from bottom, sides, and top :)
I have asked a couple questions about surface integrals but I was not able to fully understand what the person who answered was saying and/or their answer was not correct.
If someone could please help me with this problem that would be great!
evaluate the surface integral
double integral
x^2 + y^2 + z^2 dS
S is the part of the cylinder
x^2 + y^2 = 16
that lies between the planes
z = 0 and z = 4,
together with its top and bottom disks
3 answers
I get 618.6666 pi but it is wrong
integral r^3 dR dT from 0 to 4 and 0 to 2pi
2 pi (4)^4/4 = 128 pi on the bottom
SAVE that
on the top, x^2+y^2 = r^2 = 4 and z^2 = 16
so you have
(r^2+16)r dr dT from 0 to 4 and from 0 to 2 pi
so
2 pi [ (4^3)/3 + 16 (4^2)/2 ] note over 3
2 pi (64/3 + 256/2)
= 298.67 pi
you do that and SAVE it
NOW the sides of the cylinder
x^2+y^2 = 16 all the time but z goes from 1 to 4
ds = 2 pi r dz where r = 4
ds = 8 pi dz
so we have
integral of
(1+z^2)(8 pi dz ) from z = 0 to z = 4
8 pi [ 4 + 4^3/3 ]
= 202.67 pi
you calc and SAVE that
NOW add up the contributions from bottom, sides, and top :)
I get 629.33 pi
2 pi (4)^4/4 = 128 pi on the bottom
SAVE that
on the top, x^2+y^2 = r^2 = 4 and z^2 = 16
so you have
(r^2+16)r dr dT from 0 to 4 and from 0 to 2 pi
so
2 pi [ (4^3)/3 + 16 (4^2)/2 ] note over 3
2 pi (64/3 + 256/2)
= 298.67 pi
you do that and SAVE it
NOW the sides of the cylinder
x^2+y^2 = 16 all the time but z goes from 1 to 4
ds = 2 pi r dz where r = 4
ds = 8 pi dz
so we have
integral of
(1+z^2)(8 pi dz ) from z = 0 to z = 4
8 pi [ 4 + 4^3/3 ]
= 202.67 pi
you calc and SAVE that
NOW add up the contributions from bottom, sides, and top :)
I get 629.33 pi