Surface Integral 1 / ( x + y + z )dS. S: x + 43y + 5z = 6, in intersection with axis' coordinates.

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Dumirubbish

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A surface integral 1 / ( x + y + z )dS. S: x + 43y + 5z = 6, in intersection with axis' coordinates.

Please help. I tried to do the substitution z = ( 6 - x - 43y ) / 5. The Surface element sqrt( 1 + p^2 = q^2 ), but didn't get anywhere near a solution.
 
I don't see any problem of using that substitution.

\(\displaystyle \int_{0}^{6} \int_{0}^{\frac{6-x}{43}} \frac{1}{x + y + \left(\frac{6 - x - 43y}{5}\right)} \sqrt{1^2 + 43^2 + 5^2} \ \ dy \ dx \)
 
Dumirubbish said:
Yes

, it was supposed to be “+”. p and q in this case are g’_x and g’_y
Click to expand...
Parametrize the surface as a function of x and y i.e.
[math]r(x,y) =\left(x,y, \dfrac{6 - x - 43y}{ 5}\right)[/math]
Compute the gradient with respect to x and y:
[math]\dfrac{\partial r }{\partial x} = (1, 0, -1/5) \\ \\ \dfrac{\partial r }{\partial y} = (0, 1, -43/5) \\ [/math]
Compute the cross-product of the vectors to determine the element of the surface, dS:
[math]dS = \left|\dfrac{\partial r }{\partial x} \times \dfrac{\partial r }{\partial y} \right|dA =? [/math]
 
nasi112 said:
I don't see any problem of using that substitution.

\(\displaystyle \int_{0}^{6} \int_{0}^{\frac{6-x}{43}} \frac{1}{x + y + \left(\frac{6 - x - 43y}{5}\right)} \sqrt{1^2 + 43^2 + 5^2} \ \ dy \ dx \)
Click to expand...

One thing I forgot in my integral.

nasi112 said:
\(\displaystyle \int_{0}^{6} \int_{0}^{\frac{6-x}{43}} \frac{1}{x + y + \left(\frac{6 - x - 43y}{5}\right)} \frac{\sqrt{1^2 + 43^2 + 5^2}}{5} \ \ dy \ dx \)
Click to expand...
 
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