Find   *double integral of* X dS, where S is the part of the parabolic
cylinder z = x^2/2 that lies in the first octant of the cylinder
x^2+y^2=1

---

Request for Question Clarification by mathtalk-ga on 02 Feb 2005 18:15 PST

Hi, chuvak2k-ga:

Is the double integral intended to be the integral of (coordinate) x
over that portion of the surface z = x^2/2 lying inside:

  x >= 0
  y >= 0
  z >= 0
  x^2 + y^2 <= 1

That is, by double integral do you mean the surface integral?

regards, mathtalk-ga

---

Clarification of Question by chuvak2k-ga on 02 Feb 2005 18:31 PST

Yes that is correct, S is part of the parabolic cylinder which resides
in the first octan of x^2+y^2 =1 , and yes it is a surface integral.
ALso yes, it is the double integral of "x dS" the variable x, as in
integral of x = x^2/2, that x =)

INT.[x]ds

where,
>>>dS being a differential of surface area S of the z=(x^2)/2 parabolic cylinder.
>>>x being an x-coordinate of the same parabolic cylinder, or, being
the distance from the yz plane a point in the parabolic cylinder

dS = dy*dz
Then, the integrand x*dS is x*dy*dz. It is cubic in units. The double
integral then calls for a volume or solid---not of a surface area.

The solid in question is in the positive regions of the xyz axes. It is bounded by:
a)yz plane
b)xz plane
c)z = (1/2)x^2 parabolic cylinder
d)another parabolic cylinder, z = (1/2)(1 -y^2)

--------
The other parabolic cylinder is defined by the intersection of the 
circular cylinder x^2 +l^2 = 1 
and the parabolic cylinder z = (1/2)x^2:

x^2 +y^2 = 1  ----(i)
z = (1/2)x^2  ----(ii)
At their intersection, they have common coordinates. The x-coordinate
of one is the same as the x-coordinate of the other. The x^2 of one is
the same as the x^2 of the other. From (ii), x^2 = 2z. Substitute that
in (i),
2z +y^2 = 1
z = (1/2)(1-y^2)  ----(iii), the other parabolic cylinder.

---------------------
We project the solid on the yz plane.

dS = (dy)(dz)

One way is to start first with the dz....

Boundaries of dz are from zero to the parabola y = (1/2)(1 -y^2).
Boundaries of dy then are from zero to one.

dV = x*dS = x*dy*dz

Since we are starting with dz, we need to express the x into z-terms.
z = (1/2)x^2  ----(ii)
2Z = x^2
x = sqrt(2z) = sqr(2)*(z)^(1/2)

Hence, the solid is:
V = INT.INT.[x]dy*dz
= [sqrt(2)]INT(0-->1){INT.(0-->(1/2)(1-y^2))[z^(1/2)]dz}dy
= [sqrt(2)]INT(0-->1){(2/3)[z^(3/2)](0-->(1/2)(1-y^2)}dy
= [(2/3)sqrt(2)]INT(0-->1){((1/2)(1-y^2))^(3/2) -0}dy
= [(2sqrt(2))/3]INT(0-->1){(1/(2sqrt(2))*(1-y^2)^(3/2)}dy
= [1/3]INT(0-->1){(1-y^2)^(3/2)}dy

From The Wolfram Integrator,
INT.[(1 -x^2)^(3/2)]dx 
= (1/8)[x(5 -2x^2)sqrt(1 -x^2) +3arcsin(x)]
So, using that above,

= [1/3]*[(1/8){y(5 -2y^2)sqrt(1 -y^2) +3arcsin(y)}](0-->1)
= [(1/3)(1/8)]*[1(5 -2)sqrt(1 -1) +3arcsin(1) -0]
= [1/24]*[3sqrt(0) +3arcsin(1)]
Since arcsin(1) = 90degrees or pi/2, then,
= (3/24)(pi/2)
= pi/16  cu.units  ----the volume of the solid.

Hi, ticbol-ga:

I have doubts about saying dS = dy*dz.  I agree the parabolic cylinder
z = x^2/2 has a factor of dy in its "infinitesimal element" of surface
area (since it is projected parallel to the y-axis), but the other
factor would be arclength along the parabola z = x^2/2, ie.

  SQRT( (dx)^2 + (dz)^2 ) = SQRT( 1 + (dz/dx)^2 ) dx

                          = SQRT( 1 + x^2 ) dx

From there it helps that we are integrating x, so the indefinite
integral could be stated:

INTEGRAL x SQRT(1 + x^2) dx  = (1/3) (1 + x^2)^(3/2)

Of course the definite version of the double integral is a bit tricky
to evaluate, due to the bounds of integration.  Since you projected
onto the yz-plane and my approach needs a projection onto the
xy-plane, our limits of integration will probably be different.

After doing the inner integration, I get:
            ____
   1       ?1-y²     ____                   1
INTEGRAL INTEGRAL x ?1+x² dx dy  = (1/3) INTEGRAL[(2 - y²)^(3/2) - 1]dy
   0        0                               0


regards, mathtalk-ga

Hello, mathtalk-ga.

Why dS = dy*dz.

Let us review the question as posted by chukva-ga. 
"Find   *double integral of* X dS, where S is the part of the parabolic
cylinder z = x^2/2 that lies in the first octant of the cylinder
x^2+y^2=1."

So S is a surface area, the portion of the z=(x^2)/2 parabolic
cylinder that is included by the circular cylinder x^2 +y^2 = 1 in the
1st octant. Hence, dS is also an area, an infinitesimal area portion
of the said S.
So the integrand "X ds" is about the product of ds and of this ds's
distance from the yz plane. Area times distance will have cubic units.
So, INT.INT.[x]ds will produce a volume of a solid---not an area of a
surface area.

Now ds can be:
1) If S is projected on the xy plane, ds = dx*dy
2) If s is projected on the yz plane, dS = dy*dz. 
3) If S is projected on the xz plane, ...cannot be. Here the
projection of S is only that of a line, a parabola z = (x^2)/2.

dS = dx*dy cannot be used in the integrand X*dS if we are looking for
a volume. We'd get maybe moment of inertia of S about the y-axis.

Only dS = dy*dz will give a volume for x*dS.

--------------------------------

I see that you saw the question as an "integration of X". 
I cannot follow that. Granted there is an integration of a coordinate,
would that not lead to a length only? Not even an area. Much less, a
volume.
And the integrand should have been in the form [a]dx, where a is a
constant, not a variable.

If you interpreted the question as an intergation for an area, then
the integrand should be in the form x*dy. Not even x*dx would give an
area. Surely, not x*dS.

----------------
If you integrate x*(differential of the arclength of parabola z =
(x^2)/2} then you'd get infinitely many areas parallel to the xz
plane, and each of these areas are bounded by the xy plane, yz plane
and the z = (x^2)/2 parabolic cylinder.

--------------
Also, I cannot follow your
"  SQRT( (dx)^2 + (dz)^2 ) = SQRT( 1 + (dz/dx)^2 ) dx

                          = SQRT( 1 + x^2 ) dx "

How did (dz/dx)^2 become x^2?

regards, 
ticbol

Sorry. Some corrections.

In my first comment,
"Boundaries of dz are from zero to the parabola y = (1/2)(1 -y^2)."

should have been ".....parabola z = (1/2)(1 -y^2)."


In my 2nd comment,
"If you integrate x*(differential of the arclength of parabola z =
(x^2)/2} then you'd get infinitely many areas parallel to the xz
plane, and each of these areas are bounded by the xy plane, yz plane
and the z = (x^2)/2 parabolic cylinder."

should have been
"....and each of these areas are bounded are bounded by the yz plane,
the parabolic cylinder z = (x^2)/2, and the parabolic cylinder z =
(1-y^2)/2."

Hi, ticbol-ga:

Okay, first the easy question.  How did (dz/dx)^2 become x^2?

We are on the "surface" of the parabolic cylinder z = x^2/2, so:

    dz
    -- = x
    dx 

and therefore the square of this derivative is x^2.

*  *  *  *  *  *  *  *  *  *  *  *  *  *  *  *  *  *

Next, let's consider the idea of projecting a patch of surface area dS
lying on the parabolic cylinder onto one of the planes.  The effect of
this is to reduce the area, as the orthogonal projection of dS onto,
say the yz-plane, will be less than the original surface area.  [The
extreme case would be projecting onto the xz-plane, where this surface
collapses to the curve z = x^2/2.]

The integral of x dS has units of volume, but this doesn't imply dS = dy*dz.

One can express an infinitesimal part dS of surface area in terms of
dy and dz, though I thought it would be easier to do in terms of dy
and dx.  Here's how to do it in terms of dy and dz.

The parabolic cylinder's surface z = x^2/2 runs "parallel" to the
y-axis, so we can "coordinatize" the surface in terms of a product dy
times the infinitesimal arclength ds of the parabola z = x^2/2:

  ds = SQRT( (dx)^2 + (dz)^2 )

     = SQRT( (dx/dz)^2 + 1 ) dz

[I'm using lowercase s for arclength to distinguish it from surface area S.]

Now x = SQRT(2z) and dx/dz = 1/SQRT(2z), so:

  ds = SQRT( 1/(2z)  +  1 ) dz

and combining this with the factor dy gives the infinitesimal surface area:

  dS = SQRT( 1/(2z)  +  1 ) dz dy

Now you can integrate:

  x dS = SQRT(2z) SQRT( 1/(2z) + 1 ) dz dy

       = SQRT( 1 + 2z ) dz dy

using the limits of integration that you set up before.

regards, mathtalk-ga

Hello, mathtalk-ga.

Your dz then is the derivative of z of the parabola z = (x^2)/2. But
that is only the slope of the tangent line to any point on the
parabola. It is not an increment/differential of the surface of the
cylinder z = (x^2)/2.
As you know, the dz we use in integration is an infinitesimal portion
of a line that is in z-term. Not the slope/rate of change/derivative
of that line.

Others would call your dz as D(z). Those same others will still call
the increment as dz.

------------------
When we project the image of a line/surface/solid onto a plane we want
to see the boundaries/limits of the differentials and we want to find
which sequence to use to get an integrand that is easier to integrate.
Say, for dA = dx*dy, sometimes it is easier if we start with the dx
first, or vice versa.

Sure, the orthogonal projection is lesser in area than the actual
area, if the actual area is not flat and parallel to the projected
one. We are not trying to find the area of the projected one. We use
it only for the boundaries of the increments and for the choice of the
sequence of integration, to repeat.

See again my second comment as to why the volume cannot be derived
from using dx*dy. Only through dy*dz can the volume be possible.

I have shown how to find the volume through the dy*dz, or the area of
S through dy*dz.

Your way, the idea, is okay. But here, in this S, it is not
applicable. Your way is correct if the length of the arc is constant
throughout the boundaries of dy. Here the arclength varies from
longest at the xz plane to zero at the y=1 plane.
That is why I said that if you use the dx*dy you'd get those
infinitely many areas---not in one integration only. If you use the
arclength when y=1/2, you get only the area at the y=1/2 plane.

regards,
ticbol

Again, let me correct something I said. It was early this morning when
I did my last (4th} comment above. Meaning, I was not fully awake. So,
some mistakes.
It was on my way to work later, on the road, when it came to me that I
was wrong on my understanding mathtalk-ga's dz. Now, back from work,
let me take back that portion of my 4th comment.

Of course, mathtalk's dz is the same as anybody's dz, mine included.
His dz is a z-increment of the parabola z = (x^2)/2. It is not D(z).

D(z) is dy/dx or z'.It is the derivative of z.
I misunderstood his (dz/dx) as his dz. 

His sqrt[1 +(dz/dx)^2]dx then is really an increment of the length of
the parabola. If L is that length, then dL = sqrt[1 +(dz/dx)^2]dx.

He called this dL as his ds. 

--------------------

The rest of my 4th comment stays.

Hi, ticbol-ga:

Perhaps a way to reconcile our difference is to consider a "simpler"
problem, namely how to integrate 1 dS rather than x dS.

As the original poster clarified, his/her problem was that of
integrating x over the surface of the parabolic cylinder, up to
boundaries defined by intersections with the circular cylinder x^2 +
y^2 = 1 and the first octant.

Replacing x in the integrand by 1 should instead result in the surface
area of this same region of the parabolic cylinder.

To find surface area requires integrating the infinitesimal "patch"
dS.  Operationally we must reformulate dS in terms of coordinates and
match the limits of the resulting "double integral" to the boundaries
of the region.

I hope that in this context it is clear one cannot simply substitute:

  dS = dy*dz

as the "patch" dS on the parabolic cylinder is sloping, and thus
strictly larger in area than the corresponding projection dy*dz on the
yz-plane.

regards, mathtalk-ga

Hello, mathtalk-ga.

chuvak2k-ga's question was for the double integral od Xds so that is for a solid.

If we want to go to another question, the ds, then we solve for the
area of that portion of the parabolic cylinder. And you are correct up
to "...we must reformulate dS in terms of coordinates and match the
limits of the resulting "double integral" to the boundaries of the
region."

The rest, you misunderstood again the dS on the orthogonal projection
of S. To not repeat myself, please see again my explanation above re
this dS.

For purposes of integration, dS is always considered flat, not curved.
It is an infinitesimal patch of S. It is always considered as parallel
to the xy-, xz-, or yz-plane.

To find S we can use two projections.

>>>(1)At the xy plane. Here dS is dx*dy. 
If we start with dx, the limits of dx are [0,sqrt(1-y^2)]. Then dy is [0,1].
IF we start with dy, dy is [0, sqrt(1-x^2)]. Then dx is [0,1].

>>>(2)At the yz plane. Here dS is dy*dz. 

-------------
Again, back to Xds, ds=dx*dy is not applicable. Only dS=dy*dz is correct.

regards,
ticbol

For the dS on the yz-plane,
If we start with dy, dy is [0,sqrt(1-2z)]. Then dz is [0,1/2].
If we start with dz, dz is [0,(1-y^2)/2]. Then dy is [0,1].

Hi, ticbol-ga:

By evaluating the integrals you've specified, I hope you can see the
inconsistency of claiming:

  "dS is always... considered as parallel to the xy-, xz-, or yz-plane."

Realizing that projecting on the xz-plane gives a zero area, you
rejected this option.

The two integrals you propose for projection "(1)At the xy plane" both
give an area of pi/4.  Notice that this is the area of the quarter
circle lying in the first quadrant given by x^2 + y^2 = 1, ie. the
projection of the parabolic surface onto the xy-plane.

The two integrals you propose for projection "(2)At the yz plane" both
give an area of 1/3.  This is the area under the parabola z = (1 -
y^2)/2 between z = 0 and z = 1.

These values, pi/4 and 1/3, are both strictly less than the true area
of the parabolic cylinder section bounded by the circular cylinder in
the first octant:
              ____
     1       ?1-x²   ____            1
  INTEGRAL INTEGRAL ?1+x² dy dx = INTEGRAL SQRT(1 - x^4) dx
     0        0                      0

which according to Maxima is:

        1  3
   BETA(-, -)
        4  2
  ------------  =  0.87401918476404...
        4

The reason we cannot substitute for dS a "patch" parallel to a plane
defined by a pair of coordinate axes is the same reason one cannot
approximate arclength by infinitesimal "pieces" parallel to a
coordinate axis.  The parts in either case would add up to a
projection on the coordinates, rather than to the desired "sloping"
surface area or arclength.  It is true that dS is "flat" as an
infinitesimal, but it is critical that it be allowed to sit at an
angle to the coordinate planes, tangent to the surface being
integrated (just as for an arclength problem we require the "flat"
infinitesimal segments to have slope equal to the curve being
integrated).

regards, mathtalk-ga

Hello, mathtalk-ga.

I am just back from work. I read your latest comment above and it
seems you are correct about this 1dS.
I quickly integrated my 4 possibilities of dS and, yes, S = pi/4 on
both possibilities when projected on the xy plane, and S = 1/3 on both
possibilities when projected on the xz plane. I was thinking that all
the 4 possibilities should give the same S.

Let me open my old Calculus book and review this area and volume
integrations. I want to know why my dS did not give 4 same values.
Then I will tell you what I would find out. After doing my
obligations, etc, I will look for that book.

Meanwhile, no, I still don't think that my solution to chuvak2k-ga's
question is wrong.
Maybe integrands XdS and 1dS should be analyzed differently. I would see later.

regards,
ticbol

Hello, mathtalk-ga.

I have just read my old Calculus book (Calculus, One and Several
Variables, by S.L.Salas and Einar Hille, 4th Edition).

My computation re volume of the solid is correct. My way of projecting
the solid on the xz plane is okay. My dS is okay, meaning, only
through the projected dS=dy*dz can the volume of the solid of XdS be
solved.

As for the 1dS, the book says it is the volume of the projected dS
that is 1 unit thick. Meaning, it is the area of the projected dS. It
is not the area of the actual S.

As for the area of the actual S, it can be solved in two ways:
>>>by dL*dy. That leads to your solution above.
>>>by y*dL. Which is the same as your solution also.
dL = sqrt(1 +x^2) dx.

All these showed me that XdS and 1dS really are not related. Surface
area integration then requires that dL be considered as the hypotenuse
of the legs dx and dy in this case. But in the integration for volume,
dS need not be the inclined surface S---dS is always flat and parallel
to the plane it is projected onto. The element of the dV that is
bounded by the S and the plane of projection, in this case that
element is X or sqrt(2z), will take care of the slope of the S
corresponding to the projected dS.

--------------
Could you solve chuvak2k-ga's question? Please give a numerical value.
Let us compare our answers.

regards,
ticbol

Hi, ticbol-ga:

pi/8

Remember that I carried out the inner integration above in one
Comment, with respect to x, leaving an integral with respect to y.

The "solution" is a good bit more obvious to work out if the
integration with respect to y is done first and the outer integral is
with respect to x.  Indeed I showed the advantage of that
"interchange" of orders in working out the surface area 1 dS (which
technically is more difficult than the work for x dS).

regards, mathtalk-ga

Hello, mathtalk-ga.

So you got the volume to be pi/8, or 0.3927 cu.units.

That is too big.

>>>Let us see if the solid were a prism that is bounded by:
xz plane
z=1/2 plane
y=1 plane 
xy plane
yz plane 
and x=1 plane.

That is a box whose base is a square 1X1 and whose height of 1/2. Its
volume is 1/2 or 0.5000 cu.units

>>>Then we cut that with the z=(x^2)/2 cylinder. We get the volume of
the solid (prism) that is bounded by:
xz plane
z=1/2 plane
y=1 plane 
z=(x^2)/2 parabolic cylinder
and yz plane. 

The volume of this smaller prism is 0.333 cubic units.
And that is even less than your pi/8 cu.units you got for only the
solid in chuvak2k-ga's question.

>>>Then we cut further this smaller prism above with the cylinder x^2 +y^2 = 1.
We get now a solid that is bounded by:
xZ plane
z=1/2 plane
x^2 +y^2 = 1 circular cylinder
z=(x^2)/2 parabolic cylinder
and yz plane.

The volume of this solid is less than the volume of the smaller prism
above, of course. V < 0.333 cu.units

>>>Finally, we cut this solid above with the parabolic cylinder z = (1 -y^2)/2.
Now we get the solid mentioned in chuvak2k-ga's question.
My answer is v = pi/16 or 0.1963 cu.units.

----------------
Another way of looking at the volume of the solid.

Approximately, the solid is a pyramid whose base is 1/3 sq.units and
whose height is 1.
Approx. volume = (Base)(1/3 height) = (1/3)(1/3 *1) = 1/9 = 0.1111 cu units.


regards,
ticbol-ga

Hi, ticbol-ga:

Please recall my earlier point, that the infinitesimal area dS is
"larger" than its projection onto any pair of coordinate axes. 
Therefore you should at least give me credit for consistency, the
answer I get is larger than the one you got!

While the integral of x dS has "units" of volume, it does not
correspond to the the volume bounded by the surfaces of the parabolic
cylinder, the circular cylinder, and the first orthant.

Possibly this is one source of confusion (or disagreement).  Notice
that I went through an exchange of clarifications with chuvak2k-ga
earlier, and the "double integral" x dS was specified to be a surface
integral, the integral of x over a portion of the surface of the
parabolic cylinder.

If dS is replaced by dx*dy (projection on the xy-plane), then the
integral would correspond to the volume bounded by the surfaces of the
parabolic cylinder, the circular cylinder, and the first orthant.  And
that result would be smaller than the integral chuvak2k-ga inquired
about.

regards, mathtalk-ga

Also, if this was not evident from the simplified case of 1 dS, one
gets different answers from "integral" of x dS depending on whether dS
is replaced by dx*dy (the volume between the curves in this case), by
dy*dz (your formulation), or by dx*dz (which give zero here).  All of
these will be less than the value pi/8.

regards, mathtalk-ga

Hello, mathtalk-ga.

Maybe it's time to end this discussion. We are going round-and-around or nowhere.

regards,
ticbol