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
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 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
To evaluate the given surface integral, we need to calculate the double integral of the function x^2 + y^2 + z^2 over the given surface S.
In this case, our surface S consists of three parts: the curved surface of the cylinder, and the two top and bottom disks. To evaluate the surface integral over each part, we can use parametrization.
Let's start by parametrizing the curved surface of the cylinder. Since x^2 + y^2 = 16, we can use cylindrical coordinates to parametrize this surface. Let's represent the point on the surface as P(r, θ, z), where 0 ≤ r ≤ 4, 0 ≤ θ ≤ 2π, and 0 ≤ z ≤ 4.
In cylindrical coordinates, the position vector is given by:
r = <rcos(θ), rsin(θ), z>
Now, we can calculate the unit normal vector, which is perpendicular to the surface at each point P. The unit normal vector is obtained by taking the cross product of the partial derivatives of r with respect to θ and r:
N = (∂r/∂θ) × (∂r/∂r)
Calculating the partial derivatives and taking their cross product, we get:
N = <-rcos(θ), -rsin(θ), 0>
Since our function is x^2 + y^2 + z^2, we can substitute the parametric equations for x, y, and z into the function to obtain x^2 + y^2 + z^2 = (rcos(θ))^2 + (rsin(θ))^2 + z^2 = r^2 + z^2.
The surface integral over the curved surface can be written as the double integral over the parameter domain D:
∬(r^2 + z^2) ||N|| dA,
where ||N|| is the magnitude of the normal vector N and dA is the surface area element in cylindrical coordinates.
To calculate the surface integral over the top and bottom disks, we can parametrize them using polar coordinates to establish an appropriate parameter domain. The area element in polar coordinates can be written as r dr dθ.
Therefore, the surface integral over each disk can be written as ∬(r^2 + z^2) r dr dθ, where the limits of integration for r and θ are determined by the size of the disk.
To calculate the total surface integral over S, we need to evaluate the three separate integrals:
1) The integral over the curved surface of the cylinder, using the cylindrical parameter domain.
2) The integral over the top disk, using the appropriate polar parameter domain.
3) The integral over the bottom disk, also using the corresponding polar parameter domain.
Finally, we can add the results of these three integrals to get the total value of the surface integral.