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Surface integral problem from H.M. Schey's book

  1. Mar 9, 2012 #1
    I've been fooling around by myself with the book "div, grad, curl and all that" by H.M. Schey to learn some vector calculus. However, in the second chapter, when he performs the integrals, he skips the part where he finds the limits on x and y. Here's an example:

    Compute the surface integral [tex]\int \int_{S} (x+y)dS[/tex]
    where S is the portion of the plane x+y+z=1 in the first octant.

    yada yada yada some rewriting etc.

    The final integral is [tex]\sqrt{3}\int \int_{R} (1-y)dxdy[/tex], R being the area of S projected onto the xy-plane. He then says "this is a simple double integral with value 1/√3, as you should be able to verify.

    What I take from this:
    "In the first octant" means the first quadrant in the xy-plane, octant being used because of 3-dimensional space.
    Finding the limits on x and y, I set z=y=0 to find x, and z=x=0 to find y, both limits being 0 to 1 from the x+y+z=1 equation. But I don't get the same value for the integral as the solution says. Where does it go wrong?
  2. jcsd
  3. Mar 9, 2012 #2


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    Welcome to PF!

    Hi Hixy! Welcome to PF! :smile:
    No, you can't do that.

    If your first limit is x (which is from 0 to 1), then your y limits will depend on x

    for example, 0 to x or 0 to 1-x or … ? :wink:

    (or you can do it t'other way round, with y going from 0 to 1, and then the x limits depending on y)
  4. Mar 9, 2012 #3


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    [itex]\int_{x=0}^1\int_{y=0}^1 dydx[/itex] would be an integral over the rectangle with boundaries the lines x=0, x= 1, y= 0, y= 1. To do an integral like this, you must first decide the order in which you will be integrating. If you decide to integrate with respect to y first, then x, since you want a number, rather than a function of x, as your answer, you must integrate from x= 0 to x= 1. But then, for each x, y varies for 0 up to the line x+y= 1.

    Also, because you are integrating over a surface other than the xy-plane, you need to use "dS" for that surface, not just "dydx". There are a number of ways to do that but my favorite is this: if the surface is given by z= f(x,y), we can write any point on the surface as [itex]\vec{r}= x\vec{i}+ y\vec{j}+ z\vec{k}= x\vec{i}+ y\vec{j}+ f(x,y)\vec{k}[/itex]. The derivative of that vector with respect to y, [itex]\vec{r}_y= \vec{j}+ f_y\vec{k}[/itex], is a vector tangent to the surface and whose length gives the rate of change of distances on the surface as y changes. The derivative of that vector with respect to x, [itex]\vec{r}_x= \vec{i}+ f_x\vec{k}[/itex], is a vector tangent to the surface and whose length gives the rate of change of distances on the surface as x changes.

    The cross product of those two vectors, the "fundamental vector product" for the surface, [itex]\vec{r}_y\times\vec{r}_x= f_x\vec{i}+ f_y\vec{j}- \vec{k}[/itex] is perpendicular to the surface and its length gives the rate of change of area as both x and y change:[itex]\sqrt{f_x^2+ f_y^2+ 1}[/itex]. The "differential of surface area" is [itex]\sqrt{f_x^2+ f_y^2+ 1}dydx[/itex]. Here, [itex]z= f(x,y)= 1- x- y[/itex] so that [itex]\vec{r}_y\times\vec{r}_x= -\vec{i}- \vec{j}- \vec{k}[/itex] and so [itex]dS= \sqrt{3}dydx[/itex].
  5. Mar 9, 2012 #4
    Got it .. ;) Seems so trivial now that I understand. Of course, for y, x has to be a variable to cover the whole area. Then, since y has been taken care of, x can just go from 0 to 1. I get the correct answer.

    Thanks for taking the time to write such an elaborate answer, HallsofIvy. That was really helpful. Good point with that way to rewrite dS in terms of dy and dx.

    Thanks to both of you!
    Last edited: Mar 9, 2012
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