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Surface Integrals

  1. Feb 29, 2008 #1
    1) [​IMG]

    For this question, we have to use the surface integral to compute the area, but I just can't picture what is going on geometrically, so I am stuck at the very begining...can someone please help me?

    2) [​IMG]

    I don't quite get the setup for this question as well.
    Firstly, we are given that z=y2-2x, -2<y<2, 0<z<4, but what does it mean geometrically? I can't understand this part...
    Secondly, what does it mean for the normal to point away from the z-axis? Does it even define a unique direction?

    Any general hints or explanation will be greatly appreciated!
    Last edited: Feb 29, 2008
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  3. Feb 29, 2008 #2


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    You might do better to cast it into cylindrical coordinates. The cone is given by [itex]z= \sqrt{x^2+ y^2}[/itex] which, in cylindrical coordinates, is z= r. The lemniscate is given by [itex](x^2+ y^2)^2= x^2- y^2[/itex] which, in cylindrical coordinates is [itex]r^4= r^2 cos^2(\theta)- r^2 sin^2(\theta)= r^2 cos(2\theta)[/itex]. Dividing by r2, the lemniscate is given by [itex]r^2= cos(2\theta)[/itex]. A lemniscate is a two lobed "propeller" looking graph. The area you want is of the surface of the cone above that so use the cone to determine the "differential of surface area" and the lemniscate to determine the limits of integration.

    z= y2- 2x is a like a "chute" with bottom running along the line z= -2x, y= 0. The sides of the chute are parabolic.

    Yes, it certainly does. A surface has 2 unit normals at each point- pointing in opposite directions. Knowing that the normal points away from the z axis (or that it points up, or to the right) determines which of the two you use. Deciding which of the two normals to use at each point (the orientation) is as important in integrating a vector over a surface as deciding which in direction to integrate a path integral-reversing the direction reverses the sign. That plays an important part in calculating the "vector differential of surface area". Do you know how to find the "differential of surface area" and the "vector differential of surface area"?

  4. Mar 1, 2008 #3
    Thanks for your helpful comments, but I still have some further questions:

    1) For r2=cos(2theta), how can I sketch this by hand and figure out the shape without using technology? Is it possible?

    Now I can visualize both surfaces, but what area is the question referring to? I still can't visualize this part...is it the intersection of the two surfaces? But wouldn't it be a curve?

    2) Is it possible to finish this problem without knowing the exact shape of z= y2- 2x?

    Also, there are restrictions -2<y<2, 0<z<4, so we have restrictions on y and z, but how does this change the surface? Again, I can't visualize what is going on...

    What I am thinking is that there are infinitely many directions pointing away from the z-axis, how is it possible to choose a unique one? Shouldn't the question read "...n is the unit normal chosen to be pointing away from the y-z plane..." or something similar to this?

    P.S. and yes, I know the definitions of surface integrals

    Thanks for your great help!
    Last edited: Mar 1, 2008
  5. Mar 1, 2008 #4


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    I always wonder about "using technology"! Isn't a lead pencil and eraser "technology"? Is looking up the value of cosine in a table "technology"? It's "donkey work" but not impossible to graph the function. Those of us who learned math in years "B.C." (before computers) had to do that sort of thing all the time. I might recommend that you look up the work of Gaston Julia. He worked out graphs of things (the "Julia" sets) that became the basis for the Mandlebrot set in 1900- long before computers were invented.

    Yes, there are an infinite number of directions pointing away from the z-axis. But only one of the two normal vectors at a given point on the surface.
  6. Mar 2, 2008 #5
    2) [​IMG]
    Am I right so far? If so, I am stuck now, I don't know how to write this as an iterated integral...restrictions are on y and z, but I have dxdy. This is weird...what is going on?

    Another thing, did I get the correct normal that points away from the z-axis? How can I confirm this?

  7. Mar 2, 2008 #6


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    You are missing the limits on the x-coordinate which you need since you are integrating, as you say, with "dxdy". The equation is z= y2- x and z can range from 0 to 4. If z= 0, then 0= y2- x so x= y2. If z= 4, then 4= y2- x so x= y2- 4. You integral should be
    [tex]\int_{y= -2}^2\int_{x= y^2- 4}^y^2 dx dy[/itex]

    A quick sketch of z= y2, on a yz coordinate system, for various values of x, shows that the surface is a "descending" parabolic cylinder. You should be able to see, from the parabolas, which open upward, that the normals away from the z-axis point downward. No, what you have is not the "normal pointing away from the z-axis". Since the z component is positive, it is pointing toward the z-axis. Multiply it by -1 to get the normal vector pointing away from the z- axis.
  8. Mar 3, 2008 #7
    2) (i) If z=0, x= y2. If z= 4, x= y2- 4. But in general is it possible that at e.g. z=1.8, that x is NOT between y2- 4 and y2, i.e. it goes OUTSIDE of it?

    (ii) z = y2 - 2x and restrictions are on y and z. If I solve for x in terms of y and z at the very beginning, then I would have all constant limits of integration, am I right? Will this method work as well?

    (iii) In general, for this type of question, do I even need a UNIT normal? or is it the case that ANY normal pointing in the right direction would work? I am very confused about this idea...
  9. Mar 3, 2008 #8


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    I thought we'd dealt with that before.

    In many books they give the formula as [itex]\int \vec{f}\cdot\vec{n}dS[/itex] where [itex]\vec{n}[/itex] is the unit vector to the surface. Of course, calcuating that requires calculating a normal vector to the surface. The simplest way to do that is to write the "position vector" of a given point on the surface as [itex]\vec{r}(u,v)= x(u,v)\vec{i}+ y(u,v)\vec{j}+ z(u,v)\vec{k}[/itex] in terms of the parameters u and v (if the surface is given as z= f(x,y), we can use x and y as parameters: [itex]\vec{r}(x,y)= x\vec{i}+ y\vec{j}+ f(x,y)\vec{k}[/itex]). (That's often called the "fundamental vector product for the surface".) Then the derivative vectors, [itex]\vec{r}_u[/itex] and [itex]\vec{r}_v[/itex] line in the tangent plane to the surface and so their cross product is normal to the surface: call that [itex]\vec{N}[/itex]. The unit normal vector is given by [itex]\vec{N}/||\vec{N}||[/itex]. But it is also true that the differential of surface area is given by ||\vec{N}||dudv. Of course, it makes no sense to calculate the length of that normal vector because the two lengths will cancel.

    I prefer to talk about the "vector differential of surface area", [itex]d\vec{S}= \vec{N}dudv[/itex]. Then you can write [itex]\int \vec{f}\cdot d\vec{S}[/itex] and only have to calculate [itex]d\vec{S}= \vec{N}dudv[/itex]. However, you can't use just any normal vector- it has to be specifically the "fundamental vector product", the crossproduct [itex]\vec{r}_u\times\vec{r}_v[/itex].
    Last edited: Mar 13, 2008
  10. Mar 12, 2008 #9
    Hi HallsofIvy,

    I can't see the comment at the bottom of your last post, would you mind editing it? Thanks!

    There is still something that I don't understand...

    e.g. Say, a surface is given by F(x,y,z)=0
    Then grad F is a normal N to the surface.

    e.g. Say, a surface is given by x+2y-4z=0 (plane)
    Then (1,2,-4) is a normal N to the surface

    So to get the unit normal n appearing in
    [itex]\int \vec{f}\cdot\vec{n}dS[/itex]
    I can just put n=N/||N||

    But what does dS become in these 2 cases?

    My textbook always says that dS=||Gu x Gv||dudv where G is some parametrization of the surface, then is it also true that
    dS=||grad F||dxdy
    and dS=||N||dxdy where N is any normal (not necessarily equal to Gu x Gv)

    Or does it always have to be the case that dS=||Gu x Gv||dudv? If so, then the norms/lengths cannot cancel in the two cases I mentioned above...

    Please help as I am totally confused on these concepts and I am not able to find answers to these in my textbook...Your help is greatly appreciated!
    Last edited: Mar 12, 2008
  11. Mar 13, 2008 #10


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    I've done that.
    What is F here? If z= F(x,y) then ||grad F||= [itex]\sqrt{F_x^2+F_y^2}[/itex] but dS= [itex]\sqrt{1+ F_x^2+ F_y^2}[/itex]. Or is the surface given by F(x,y,z)= constant? And why would [itex]\nabla F[/itex] work specifically for a projection onto the xy plane? What is you projected onto the yz plane instead?

    What you can do is take [itex]\nabla F[/itex] and "normalize" to the xy plane by dividing by the z-component. For example, if [itex]z^2= x^2+ y^2[/itex] (the cone), Then we can write F(x,y,z)= x^2+ y^2- z^2= 0. Now [itex]\nabla F= 2x\vec{i}+ 2y\vec{j}-2z\vec{k}[/itex]. Dividing that vector by the [itex]\vec{k}[/itex] component gives [itex]-(x/z)\vec{i}- {y/z}\vec{j}+ \vec{k}[/itex] and its length is [itex]\sqrt{(x^2/z^2)+ (y^2/z^2)+ 1}= \sqrt{(x^2+ y^2+ z^2)/z^2}[/itex]
    [itex]= \sqrt{2(x^2+ y^2)/(x^2+ y^2+ z^2}=\sqrt{2}[/itex] and dS= [itex]\sqrt{2}dxdy[/itex] as we should have.

    It should be obvious that this can't be true. If N is a normal, 2N is also but ||2N||dx= 2||N||dxdy. You must have misunderstood your textbook.

    There may be different ways to calculate it, but the "differential of surace area" is a specific number times dxdy, not just "any number"dx dy which it would be if you could use a normal vector with any length!

    Last edited: Mar 13, 2008
  12. Mar 13, 2008 #11
    Say, if z=f(x,y) and F=z-f(x,y), then would dS be equal to ||grad F||dxdy? My textbook seems to be doing this, but grad F is not the so called "fundamental vector product", right? The "fundamental vector product" is ru x rv

    This paragraph is very helpful and eduactive, but what is the maning of "normalizing" to the xy plane by dividing by the z-component? Why do we have to do this? I think it is important for me to understand this...

    Yes, you are right, I get your point now!

    You're very nice, thanks for your help!
    Last edited: Mar 13, 2008
  13. Mar 13, 2008 #12
    Also, how about the normal vectors that are obtained in different ways: (intuively, geometric properties, etc.)

    e.g. surface given by x+2y-4z=0 (plane)
    Then N=(1,2,-4) is a normal to the surface (immediately clear from the equation of a plane without even having to parametrize it)

    e.g. surface given by x2+y2+z2=4 (sphere)
    Then N=(x,y,z) is a normal to the surface (which should be clear from the geometry even without parametrizing it since it's a sphere centred at origin)

    What would dS be in these cases? Would the norms/lengths cancel nicely in these cases?
  14. Mar 13, 2008 #13


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    A normal vector doesn't do you any good! It does not necessarily have anything to do with the differential of area of the surface which depends entirely on the length of the vector.
    For the plane, you could, as you have suggested, think of the surface as a "level surface" of the function f(x,y,z)= x+ 2y- 4z. Its gradient is [itex]\nabla f= \vec{i}+ 2\vec{j}- 4\vec{k}[/itex]. Now you have to decide in which plane to do the integration! If you choose the usual xy-plane, you have to "normalize" by dividing that gradient vector by the [itex]\vec{k}[/itex] component, -4. That gives us
    [tex]\frac{-1}{4}\vec{i}+ \frac{-1}{2}\vec{j}+ \vec{k}[/tex]
    It's length is
    [tex]\sqrt{\frac{1}{16}+ \frac{1}{4}+ 1}= \frac{\sqrt{21}}{4}[/tex]
    so the differential of area is
    [tex]\frac{\sqrt{21}}{4}}dx dy[/tex]
    Certainly, you can get a unit normal vector by dividing any normal vector by its length. In particular, your original vector, [itex]1\vec{i}+ 2\vec{j}- 4\vec{k}[/itex] has length [itex]\sqrt{21}[/itex] so a unit normal vector is
    [tex]\frac{1}{\sqrt{21}}\vec{i}+ \frac{2}{\sqrt{21}}\vec{j}-\frac{4}{\sqrt{21}}\vec{k}[/tex]
    Then [itex]\vec{n}dS[/itex] is the product of those:
    [tex]\left(\frac{1}{4}\vec{i}+ \frac{1}{2}\vec{j}- \vec{k}\right)dxdy[/tex]
    which, which, except for the factor of -1, the "orientation", is just the "normalized" vector we originally got- we didn't really need to do both of those length calculations.

    But you do need to do that division to "normalize" the vector. If we had wanted to integrate in the yz-plane, since the [itex]\vec{i}[/itex] component is already one, we would use [itex]\left(\vec{i}+ 2\vec{j}- 4\vec{k}\right)dydz[/itex] as [itex]\vec{n}dS[/itex] (which I prefer to call "[itex]d\vec{S}[/itex] rather than [itex]\vec{n}dS[/itex]) and its length, [itex]\sqrt{1+ 4+ 16}dydz= \sqrt{21}dydz[/itex] as the differential of surface area. Notice that that is different from what we got with "dxdy"!

    Similarly, we could write the surface as a vector equation, using x and y as parameters.
    [tex]\vec{r}(x,y)= x\vec{i}+ y\vec{j}+ z\vec{k}= x\vec{i}+ y\vec{j}+ \frac{1}{4}(x+ 2y)\vec{k}[/tex]
    [tex]\vec{r}_x= \vec{i}+ \frac{1}{4}\vec{k}[/tex]
    [tex]\vec{r}_y= \vec{j}+ \frac{1}{2}\vec{k}[/itex]
    The cross product of those two vectors is
    [tex]-\frac{1}{4}\vec{i}- \frac{1}{2}\vec{j}+ \vec{k}[/itex]
    which is the same as we got before so we have
    [tex]\left(-\frac{1}{4}\vec{i}- \frac{1}{2}\vec{j}+ \vec{k}\right)[/itex]
    as the "vector differential of area", [itex]d\vec{S}[/itex], and
    [tex]\frac{\sqrt{21}}{4} dxdy[/tex]
    as the differential of area as before.

    You will notice that the two "vector differentials" are not exactly the same. One is -1 times the other. That depends on an arbitrary choice of order for the cross multiplication and reflects the "orientation" of the surface. The first, where the [itex]\vec{k}[/itex] component is positive, is "oriented upward" and the second, where it is negative, is "oriented downward".
  15. Mar 13, 2008 #14


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    (Added when I had more time.)

    Now let's look at the sphere, [itex]x^2+ y^2+ z^2= R^2[/itex]. Treating this as a "level curve" of [itex]F(x,y,z)= x^2+ y^2+ z^2[/itex], the gradient is normal to the surface. [itex]\nabla F= 2x\vec{i}+ 2y\vec{j}+ 2z\vec{k}[/itex]. If you want to integrate in the xy-plane (with respect to x and y), you need to divide by the [itex]\vec{k}[/itex] coefficient, 2z, to "normalize":
    [tex]\frac{x}{z}\vec{i}+ \frac{y}{z}\vec{j}+ \vec{k}[/tex]
    The length of that is
    [tex]\sqrt{\frac{x^2}{z^2}+ \frac{y^2}{z^2}+ 1}= \sqrt{\frac{x^2+ y^2+ z^2}{z^2}}[/tex]
    [tex]= \frac{R}{\sqrt{R^2- x^2- y^2}}[/tex]
    That tells us that the "differential of surface area" is
    [tex]dS= \frac{R}{\sqrt{R^2- x^2- y^2}}dxdy[/itex]
    and the "vector differential of surface area is
    [tex]\vec{n}dS= d\vec{S}= \left(\frac{x}{\sqrt{R^2- x^2- y^2}}\vec{i}+ \frac{y}{\sqrt{R^2- x^2- y^2}}\vec{j}+ \vec{k}\right)dx dy[/itex]
    which is the same as
    [tex]\left(\frac{x^2\vec{i}+ y^2\vec{j}+ (R^2- x^2- y^2)\vec{k}}{z}\right)dxdy[/tex]

    Of we could use x and y as parameters to write
    [tex]\vec{r}(x,y)= x\vec{i}+ y\vec{j}+ z\vec{k}= x\vec{i}+ y\vec{j}+ \sqrt{R^2- x^2- y^2}\vec{k}[/tex]
    [tex]\vec{r}_x= \vec{i}- \frac{x}{\sqrt{R^2- x^2- y^2}}\vec{k}[/tex]
    [tex]\vec{r}_y= \vec{j}- \frac{y}{\sqrt{R^2- x^2- y^2}}\vec{k}[/itex]
    and the cross product of those is
    [tex]\frac{x}{\sqrt{R^2- x^2- y^2}}\vec{i}+ \frac{y}{\sqrt{R^2- x^2- y^2}}\vec{j}+ \vec{k}[/tex]

    [tex]d\vec{S}= \right(\frac{x}{\sqrt{R^2- x^2- y^2}}\vec{i}+ \frac{y}{\sqrt{R^2- x^2- y^2}}\vec{j}+ \vec{k}\left)dxdy[/tex]
    as before.
    dS is the length of that vector
    [tex]\frac{R}{\sqrt{R^2- x^2- y^2}}dxdy[/tex]
    as before.

    The advantage of the second method is that it works in exactly the same way for any parameters for the surface.

    That way we could use spherical coordinates, with [itex]\rho[/itex]= R (which you won't let me do!), and it would be much simpler.
  16. Mar 15, 2008 #15
    This is very helpful! Thank you very much!!!
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