# Relationship between two surface integral equations

1. May 24, 2012

### rrroach

1. The problem statement, all variables and given/known data

Calculate the surface integral ∫∫S x2z2 dS, where S is the part of the cone z2=x2+y2 between the planes z=1 and z=3

2. Relevant equations

There are two relative equations for calculating surface integrals by transforming them into double integrals, but my question is about the two forms of the magnitude: |ru x rv| (where r is the vector equation and u, v are parameters) and √((dg/dx)2+(dg/dy)2+1) (where x,y represent cartesian coordinates and are the parameters for g(x,y), which defines the z-coordinate).

3. The attempt at a solution

I thought that the two equations were really just two different ways of expressing the same value, but due to this problem, I'm not so sure anymore.

I parameterized the equation of the cone as x=r cos θ, y=r sin θ, z=r, which I believe is the correct parameterization. In using the first form of the equation, for the part that involves the magnitude of the cross product of the partial derivatives of the vector equation (which is formed by the parameterization), I yield r√2. By using the second form of the equation, I yield only √2.

I've double-checked the cross product and the parameterization at least 10 times, but I still do not understand why the two forms of the same equation are giving me two different answers. Is my parameterization wrong or am I falsely thinking that the equations are really the same? Any help is appreciated!

2. May 24, 2012

### sharks

Hi rroach

You should draw the graph of the projection of the surface S onto the xy-plane. It'll be two concentric circles, center origin with radius 1 and radius 3. Just substitute the values of z into the equation of the cone to get the equations of the points of intersection between the two z planes and the cone.
Find the surface area. Remember that you are concerned only with the section above the z axis, so z must be positive, meaning, $z=+\sqrt{x^2+y^2}$
$$\iint_S\,.dS=\iint_R \sqrt{(z_x)^2+(z_y)^2+1} \,.dxdy$$
Then use polar coordinates and parametrize accordingly.
$$\sqrt 2\int^{2\pi}_0 \int^3_1 r^2\cos^2 \theta(r^2)\,.rdrd\theta$$

Last edited: May 24, 2012
3. May 24, 2012

### HallsofIvy

Staff Emeritus
If z= g(x,y) then $\vec{r}(x,y)= x\vec{i}+ y\vec{j}+ g(x,y)\vec{k}$
So $\vec{r}_x= \vec{i}+ g_x\vec{k}$ and $\vec{r}_y= \vec{j}+ g_y\vec{k}$. There cross product is $\vec{r}_y\times \vec{r}_x= g_x\vec{i}+ g_y\vec{j}- \vec{k}$ and the norm of that is, in fact, $\sqrt{\vec{r}_x^2+ \vec{r}_y^2+ 1}$.

Okay, with that parameterization, $\vec{r}(\theta, r)= rcos(\theta)\vec{i}+ r sin(\theta)\vec{j}+ r\vec{k}$ so that $\vec{v}_\theta= -r sin(\theta)\vec{i}+ r cos(\theta)\vec{j}$ and $\vec{v}_r= cos(\theta)\vec{i}+ sin(\theta)\vec{j}+ \vec{k}$. Their cross product is $rcos(\theta)\vec{i}+ rsin(\theta)\vec{j}- r\vec{k}$ and the norm of that is $\sqrt{r^2cos^2(\theta)+ r^2sin^2(\theta)+ r^2}= r\sqrt{2}$ as you say. The area integral would be
$$\int_{r= 1}^3\int_{\theta= 0}^{2\pi} r\sqrt{2} d\theta dr$$

Using $z^2= x^2+ y^2$, we have $2z z_x= 2x$ so that $z_x= x/z$ and $2z z_y= 2y$ so that $z_y= y/z$.

$\sqrt{z_x^2+ z_y^2+ 1}= \sqrt{x^2/z^2+ y^2/z^2+ 1}= \sqrt{\frac{x^2+ y^2+ z^2}{z^2}}= \frac{R}{z}$ where R is now the straight line distance from the origin to (x, y, z). Since the cone makes a 45 degree angle, $R= \sqrt{2}r$ with r measured in the xy-plane. And, of course, $z= \sqrt{x^2+ y^2}= r$ so that is, in fact, $\sqrt{2}$ alone.

However, you are not taking into account the "differential of area" in the xy-plane. In polar coordinates, that is $rdrd\theta$ so that the area is given by
$$\int_{r= 1}^3\int_{\theta= 0}^{2\pi} \sqrt{2}(r d\theta dr)$$
exactly as before.

You were checking the wrong integral.

Last edited: May 24, 2012
4. May 24, 2012

### sharks

But isn't this integrand missing the parametrization of $x^2z^2$ from the original integral?

5. May 24, 2012

### Vargo

The formula dS = |r_u X r_v|dudv is the master formula that implies all other formulas. Your second formula is just what happens when the rectangular coordinates x,y are used as the parameters.

The difference between your two calculations is just the difference between polar coordinates and rectangular coordinates. rdrdtheta = dxdy remember?

Either calculation is correct as long as you have the appropriate limits of integration for the coordinates that you chose. Polar coordinates are the better choice for simplicity though.

6. May 24, 2012

### rrroach

Thank you so much! I see now that by using the cross product version of the magnitude, the extra r would translate to the r in r*drdθ were I to use the cartesian root version of the magnitude. Adding in an additional r in r dr d for the cross product version is unnecessary, and so, we are ultimately left with 2^.5 r dr dθ for either version of the magnitude once we add them into the entire surface integral.

Correct. However, the part of the problem I was having problems with was the magnitude part of the integral. Substituting and integrating the surface integral is the easy part; setting it up is the hard part.