surface integral

[fqqs]

In order to calculate a surface integral I have to find a normal vector. In my case, I have to find a normal vector (pointing down) to \(\displaystyle z=\sqrt{x^2+y^2}\)

do I calculate this vector directly from definition or is there any problem with a point (peak of the cone) where normal vector doesn't exist



Generally speaking, I have a surface integral \(\displaystyle \int_{S}^{} (y-z)dydz+(z-x)dzdx+(x-y)dxdy\) , when S is given by \(\displaystyle x^2+y^2=z^2\), whereas \(\displaystyle 0\le z\le 9\) (cone without lid)

My result: 0

Is that correct or possible?

thanks

 

[DrPhil]

My result: 0

Is that correct or possible?

thanks

Without looking at the problem in detail, but just looking at symmetry, I would expect the x and y components to average to zero, BUT the normal vector every point on the surface has the same z-component. Thus that part of the integral should be non-zero.

 

[fqqs]

Without looking at the problem in detail, but just looking at symmetry, I would expect the x and y components to average to zero, BUT the normal vector every point on the surface has the same z-component. Thus that part of the integral should be non-zero.

So at evening I'll provide you with my solution step by step ok?

 

[HallsofIvy]

Since the "difficulty" is only at a single point, that will not affect the integral.

 

[DrPhil]

So at evening I'll provide you with my solution step by step ok?

Sure - looking forward to it.

 

[fqqs]

Sure - looking forward to it.

sorry for the delay


\(\displaystyle \vec{n}=\nabla\left(\sqrt{x^2+y^2}-z\right)=\left[\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}},-1\right]\\\\
\vec{f}(x,y,z)=[y-z,z-x,x-y]\\\\
\int_S(y-z)dydz+(z-x)dzdx+(x-y)dxdy=\int_A \vec{f}\cdot\vec{n}\,dxdy=\\\\=\int_A\frac{x(y-z)}{\sqrt{x^2+y^2}}+\frac{y(z-x)}{\sqrt{x^2+y^2}}+y-x\,dxdy=\int_A\frac{z(y-x)}{\sqrt{x^2+y^2}}+y-x\,dS=\\\\=2\int_Ay-x\,dxdy=0\)