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

In summary, Schey skips the part where he finds the limits on x and y. This causes the final integral to be incorrect.
  • #1
Hixy
8
0
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?
 
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  • #2
Welcome to PF!

Hi Hixy! Welcome to PF! :smile:
Hixy said:
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.

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)
 
  • #3
[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].
 
  • #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!
 
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1. What is a surface integral problem from H.M. Schey's book?

A surface integral problem from H.M. Schey's book is a mathematical problem that involves calculating the double integral of a function over a two-dimensional surface in three-dimensional space. This type of problem is often encountered in the study of vector calculus and is used in many areas of science and engineering.

2. How is a surface integral problem solved?

A surface integral problem is solved by first choosing a parameterization for the surface, such as using Cartesian or polar coordinates. Then, the surface is divided into small regions, and the function is evaluated at each point within these regions. The double integral is then calculated over the entire surface by summing up the contributions from each region.

3. What are the applications of surface integrals?

Surface integrals have many applications in physics, engineering, and other areas of science. They are used to calculate the flux of a vector field through a surface, the mass of a thin sheet, the work done by a force field on a surface, and the surface area of a three-dimensional object, among others.

4. What is the significance of H.M. Schey's book in the study of surface integrals?

H.M. Schey's book, "Div, Grad, Curl, and All That: An Informal Text on Vector Calculus", is a widely used text for teaching vector calculus and surface integrals. It provides clear explanations and examples that help students understand the concepts and techniques involved in solving surface integral problems.

5. Are there any tips for solving surface integral problems from H.M. Schey's book?

Some tips for solving surface integral problems from H.M. Schey's book include carefully choosing the parameterization of the surface, breaking the surface into smaller regions for easier calculation, and checking for symmetry in the problem to simplify the solution. It is also important to have a solid understanding of vector calculus concepts, such as gradients and line integrals, to successfully solve surface integral problems.

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