Simple surface integration blues {again}

brendan_foo
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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 I am not mistaken..

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

Help :( :(
 
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brendan_foo said:
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}}

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


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 I am not mistaken..

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

Help :( :(

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.
 
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Saviour...I suppose I should polish up on the theory of iterated integrals...The book I currently have contains a lot 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 that's 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 :biggrin: :biggrin:
 
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Yes,the surface 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&#039;s projection on the Oxy plane is a disk of radius 1.<br /> Use polar coordinates in the Oxy plane.The limits of integration will be:<br /> 0\leq \rho\leq 1;0\leq \phi\leq 2\pi<br /> The integrals won&#039;t look very pretty,though...<br /> <br /> Daniel.
 
I wish there was a real time chat facility :biggrin:
 
:( this integral is pretty tricky, any takers?! I've got the numerical solution but my answers just won't agree...sigh :frown:
 
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.
 
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!
:biggrin:
 
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