# Simple surface integration blues {again}

1. Jan 3, 2005

### brendan_foo

Hi there,

I'm still reading on about surface integrals and stuff, and I've arrived at a formula that goes like this {based on plane projection methods} :

$$\iint_S G(x,y,z) dS = \iint_R G(x,y,z) \sqrt{1 + \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y}}$$

So this is the question at hand :

Evaluate the surface integral : $\iint_S G(x,y,z) dS$

where S is the portion of the plane $x + y + z = 1$ in the first octant, and $G(x,y,z) = z$

So, finding the partial derivatives etc.. i arrive at :

$$\sqrt{3}\int_0^1 \int_0^1 z dx dy$$ which is equivalent to : $$\sqrt{3}\int_0^1 \int_0^1 (1 - x - y) dx dy$$...I chose the limits for dx and dy as so.....

x + y + z = 1.... in the x plane... y+z = 0 thus x = 1, y = 0, and the same reasoning for y...

Computing this i get :

$$\sqrt{3}\int_0^1 (1 - \frac{1}{2} - y) dy = \sqrt{3}\left[ y - \frac{y}{2} - \frac{y^2}{2} \right]_{0}^{1} = 0$$ if im not mistaken..

The official solution however is $$\frac{\sqrt{3}}{6}$$

Help :( :(

2. Jan 3, 2005

### HallsofIvy

Well, actually, that should be
$$\iint_S G(x,y,z) dS = \iint_R G(x,y,z) \sqrt{1 + \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y}}dx dy$$

Here's how I like to do problems like this: the surface is given by x+ y+ z= 1 and we can think of that as a "level surface" of f(x,y,z)= x+ y+ z. The gradient of f: i+ j+ k is normal to that and its length is the differential of area. I like to think of it as the "vector differential of area" . In this particular case, the differential of area, projected onto the xy-plane, is
$\sqrt{3}dxdy$, just as you have.

HOWEVER, you have calculated the limits of integration wrong. The plane x+ y+ z= 1, in the first quadrant, is a triangle with vertices at (1,0,0), (0,1,0), and (0,0,1). Projecting down to the xy-plane, we get a triangle with vertices at (0,0), (1,0), and (0,1). Of course, the sides of that triangle are the x-axis, the y-axis, and the line x+ y= 1.

If we choose to order the integrals with the x-integral outside, then x ranges from 0 to 1, just as you say. BUT, FOR EACH X, y ranges from 0 up to y= 1-x, the slant line. Your integral should be
$$\sqrt{3}\int_{x=0}^{1}\int_{y= 0}^{1-x}(1-x-y)dydx$$

Integrate that and see what you get.

Last edited by a moderator: Jan 3, 2005
3. Jan 3, 2005

### brendan_foo

Saviour...I suppose I should polish up on the theory of iterated integrals...The book I currently have contains alot of mathematical language, such of that a mere engineering student finds a bit much on first glance.. Could someone just brief me up on the theory of these iterated integrals in which a limit is a function.. I know its analogous to partial derivatives, but thats as far as i go :D

Cheers guys, another triumph!! :D:D

Edit : Just to get the ball rolling, the next question is of the same form; but :

$$G(x,y,z) = \frac{1}{1 + 4(x^2 + y^2)}$$ and $$z = x^2 + y^2$$
from Z = 0 to Z = 1; If someone could just push me in the right direction, regarding limits and whether or not i should convert to polar coordinates.

Thanks again

Last edited: Jan 3, 2005
4. Jan 3, 2005

### dextercioby

Yes,the surface [itex] z=x^{2}+y^{2} [/tex] is the of a rotation paraboloid round the Oz axis,with the lowest point:the origin of the cartesian coordinate system.It's projection on the Oxy plane is a disk of radius 1.
Use polar coordinates in the Oxy plane.The limits of integration will be:
$$0\leq \rho\leq 1;0\leq \phi\leq 2\pi$$
The integrals won't look very pretty,though...

Daniel.

5. Jan 3, 2005

### brendan_foo

I wish there was a real time chat facility

6. Jan 4, 2005

### brendan_foo

:( this integral is pretty tricky, any takers?! I've got the numerical solution but my answers just wont agree...sigh

7. Jan 4, 2005

### dextercioby

It's this "baby":
$$\int_{0}^{1} d\rho\int_{0}^{+\pi} d\phi \frac{\sqrt{1+2\rho(\sin\phi+\cos\phi)}}{1+4\rho^{2}}$$
,which i guess it cannot be expressed through elementary functions.At least the integral of the angle seems to be put in connection with a complete elliptic integral of the second kind of Legendre.

Daniel.

8. Jan 4, 2005

### brendan_foo

Man that's mighty unfair...I arrived at the same integral as you've stated but evaluating it was somewhat nasty and led no where really.. Ah well, onto the next I suppose.. Thanks to everyone though, the aid is much appreciated.

All the best!!