1. ## Surface integrals

Here's the question: Find (double integral over S) of xdS over part of the parabolic cylinder z = (1/2)x^2 that lies inside the first octant of the cylinder x^2 + y^2 = 1.

No idea how to do this. These integrals are killing me. I've taught myself this whole course so I'm really not very good at this stuff.

2. Originally Posted by Zirtco
Here's the question: Find (double integral over S) of xdS over part of the parabolic cylinder z = (1/2)x^2 that lies inside the first octant of the cylinder x^2 + y^2 = 1.

No idea how to do this. These integrals are killing me. I've taught myself this whole course so I'm really not very good at this stuff.
You can write the surface as $(x, y, (1/2)x^2)$ or, as a direction vector, $\vec{r}(x,y)= x\vec{i}+ y\vec{j}+ (1/2)x^2\vec{k}$. The derivatives, with respect to x and y, $\vec{r}_x= \vec{i}+ x\vec{k}$ and $\vec{r}_y= \vec{j}$ are tangent vectors and lie in the tangent plane to the surface. Their cross product, $-x\vec{i}+ \vec{k}$ is normal to the surface and its norm, $\sqrt{x^2+ 1}$ gives the "differential of surface area", $dS= \sqrt{x^2+ 1}dxdy$. In the "first octant of the cylinder $x^2+ y^2= 1$, x goes from 0 to 1 and, for each x, y goes from 0 to $\sqrt{1- x^2}$.

The integral you want is $\int_{x=0}^1\int_{y= 0}^{\sqrt{1- x^2}}x\sqrt{x^2+ 1}dy dx$.

3. ## RE:

Thanks so much for that. I knew I had to take the cross product somewhere, I just wasn't sure where, as I was not able to come up with the r vector parametrization. Thanks.

4. OK, here's what I still don't understand. When do you just take double integral of dS? When is dS simply the root of the sum of the squares of the partials? And when is dS ||Ru x Rv||? Everytime I try and do a question, I get the method wrong. And I have a final in this course tomorrow....The question was : Find the area of that portion of the surface z^2 = x^2 + y^2 above the xy-plane for which x≤1 and y≤2. I was about to start integrating sqrt(x^2 +y^2) over the given bounds and take dS as ||Rx x Ry||. The solution just integrates dS as the root of the sum of the squares partials over the given bounds.

5. Originally Posted by Zirtco
OK, here's what I still don't understand. When do you just take double integral of dS? When is dS simply the root of the sum of the squares of the partials? And when is dS ||Ru x Rv||? Everytime I try and do a question, I get the method wrong. And I have a final in this course tomorrow....The question was : Find the area of that portion of the surface z^2 = x^2 + y^2 above the xy-plane for which x≤1 and y≤2. I was about to start integrating sqrt(x^2 +y^2) over the given bounds and take dS as ||Rx x Ry||. The solution just integrates dS as the root of the sum of the squares partials over the given bounds.
Hallsofivy gave a great explanation as to where the solution for these types of problems comes from. But for more practical purposes, it might be better to understand how/when to apply the formulas as opposed to how to derive them (of course if you understand where they come from, you will understand how to apply them).

In the case of surface integrals, we are projecting an element of area from a surface in 3D space (typically given by z=something) known as dS onto the XY plain where the element of area is known as dA.

So when the question says, "Find the area of some surface over some XY domain" we know this is the same thing as saying

$\iint dS$ where $dS= \sqrt{ 1+ (\frac {dz}{dx})^2 + ( \frac {dz}{dy})^2 }dxdy$

Your question was, when do you have more then just the dS in the integral? Well, this comes from the question. So if the question gives (like in your first example)

$\iint xdS$ then we include X. Otherwise, we just use dS.

Of course the bounds for the XY domain come from the given statement.

I would not remember this in terms of vectors, that complicates it more then I suspect it need be in your case.

6. Thanks for that explanation. It makes sense. But here is what I'm still confused about: for the second problem that I couldn't solve with the cone in the first quadrant, how was I to know not to use r vectors and not to have dS equal ||Rx x Ry||? If they make you integrate fdS, what exactly are you integrating graphically? If anyone could explain when to use these methods, it'd be fantastic:

1. dS = ||Rx x Ry||
2. dS = sqrt((∂f/∂x)^2 + (∂f/∂y)^2 + 1)
3. ndS = (-Fx, -Fy, 1)

When to use each of these methods? And what do each of them mean in 3D? A million thanks!

7. Originally Posted by Zirtco
Thanks for that explanation. It makes sense. But here is what I'm still confused about: for the second problem that I couldn't solve with the cone in the first quadrant, how was I to know not to use r vectors and not to have dS equal ||Rx x Ry||? If they make you integrate fdS, what exactly are you integrating graphically? If anyone could explain when to use these methods, it'd be fantastic:

1. dS = ||Rx x Ry||
2. dS = sqrt((∂f/∂x)^2 + (∂f/∂y)^2 + 1)
3. ndS = (-Fx, -Fy, 1)

When to use each of these methods? And what do each of them mean in 3D? A million thanks!
You do realise dS is dS is dS right? It's just another way of expressing it.

For the first 2, we can use either or but please note that if you look at hallsofivey's explanation

$dS = ||Rx x Ry|| = \sqrt{ (df/dx)^2 + (dF/dy)^2 + 1}dxdy$

We will always need the form on the very most right hand side of the equation, so why not memorise this and be done with it? You're studying for an exam right? If it were for learning purposes (out of interest) I would suggest otherwise, but for your exam simply know

$dS=\sqrt{ (dF/dx)^2 + (dF/dy)^2 + 1}dxdy$

As for

$ndS = +/- (-Fx-Fy + 1)$

This is typically used when we have a Flux integral, i.e.

$\iint F \cdot ndS$

Where N is a unit vector in the direction of flux and again, dS is the element of area.

In the questions you have been asking, this is not a needed quantity.

### surface integral for parabolic

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