- #1

Master1022

- 611

- 117

- Homework Statement
- Demonstrate the validity of the divergence theorem by calculating the surface integral ## \iint_S \vec F \cdot \hat n dS ## over the closed surface defined by ## z = 3 - x^2 - y^2 ## and the plane ## z = 0 ## where ## \vec F = x \hat i + y \hat j ##

- Relevant Equations
- Gauss' Theorem

Hi,

I just had a quick question about a step in the method of calculating the surface integral and why it is valid. I have already done the divergence step and it yields the correct result.

Let us calculate the normal: ## \nabla (z + x^2 + y^2 - 3) = (2x, 2y, 1) ##. Just to double check, am I correct in thinking that I don't need to normalise the normal surface vectors as the magnitude cancel out after being mapped onto the integration variables?

For the z = 0 plane portion, the surface vector will be

$$

\begin{pmatrix}

0 \\

0 \\

-1

\end{pmatrix} $$ and thus the contribution to the surface integral will be 0

Now for the 'downward-facing parabola'

$$ \vec F \cdot \hat n dS = \begin{pmatrix} x \\ y \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 2x \\ 2y \\ 1 \end{pmatrix} = 2(x^2 + y^2) $$

The solution suggests that we can use the fact that ## x^2 + y^2 = 3 ## at ## z = 0 ## such that

$$ \iint_S \vec F \cdot \hat n dS = \iint_S 2(3) dS = 6 \iint_S dS = 18\pi $$

This is where I have the questions:

1) Why are we allowed to just use ## x^2 + y^2 = 3 ## when x and y actually vary

2) How do I know that this has been mapped from the '3D' surface to the ##xy## plane? When I defined ## dS ## it was on the 3D surface but now we are working in the plane... I know the purpose of these is to map to the xy plane where the geometry is much simpler (and sometimes an easy shape). I suppose this is the downside of not normlising - it hides these steps?

Any clarification is greatly appreciated.

I just had a quick question about a step in the method of calculating the surface integral and why it is valid. I have already done the divergence step and it yields the correct result.

**Method:**Let us calculate the normal: ## \nabla (z + x^2 + y^2 - 3) = (2x, 2y, 1) ##. Just to double check, am I correct in thinking that I don't need to normalise the normal surface vectors as the magnitude cancel out after being mapped onto the integration variables?

For the z = 0 plane portion, the surface vector will be

$$

\begin{pmatrix}

0 \\

0 \\

-1

\end{pmatrix} $$ and thus the contribution to the surface integral will be 0

Now for the 'downward-facing parabola'

$$ \vec F \cdot \hat n dS = \begin{pmatrix} x \\ y \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 2x \\ 2y \\ 1 \end{pmatrix} = 2(x^2 + y^2) $$

The solution suggests that we can use the fact that ## x^2 + y^2 = 3 ## at ## z = 0 ## such that

$$ \iint_S \vec F \cdot \hat n dS = \iint_S 2(3) dS = 6 \iint_S dS = 18\pi $$

This is where I have the questions:

1) Why are we allowed to just use ## x^2 + y^2 = 3 ## when x and y actually vary

**from 0 to the radius**(i.e. the surface exists for ## x^2 + y^2 \leq 3 ##)? I remember making a similar mistake in the past where I just used the equation when the surface had values for lower x and y.2) How do I know that this has been mapped from the '3D' surface to the ##xy## plane? When I defined ## dS ## it was on the 3D surface but now we are working in the plane... I know the purpose of these is to map to the xy plane where the geometry is much simpler (and sometimes an easy shape). I suppose this is the downside of not normlising - it hides these steps?

Any clarification is greatly appreciated.