# Understanding the argument of the surface area integral

#### JD_PM

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

Find $\iint_S ydS$, where $s$ is the part of the cone $z = \sqrt{2(x^2 + y^2)}$ that lies below the plane $z = 1 + y$

2. Relevant equations

3. The attempt at a solution

I have already posted this question on MSE: https://math.stackexchange.com/questions/3155620/surface-integral-of-an-intersection-cone-plane/3155634?noredirect=1#comment6498746_3155634

My issue is with $\iint_S ydS =\sqrt{3} \int_A ydxdy=\sqrt{3}\, \bar{y}|A|$. Concretely, I do not get why $\bar{y}$ shows up.

My issue is that I still do not understand how to deal with the argument of the surface integral. Let's say we had for instance $\iint_S xydS$ or $\iint_S y^2dS$. I wouldn't know how to proceed. I know it is somehow related to the centroid of the figure (in this case an elliptical cylinder).

Robert Z provided a short explanation but I do not get it...

May you please provide an explanation?

Thanks

Related Calculus and Beyond Homework News on Phys.org

#### LCKurtz

Science Advisor
Homework Helper
Gold Member
I think I get what is confusing you. Let's say you have an ellipse for which you know the centroid (center). Say the center is at $(c,d)$ and the major and minor axes are $a$ and $b$ and you know the formula for the area of such an ellipse is $\pi a b$. Now let's say you have an integral that you want to evaluate something like $\iint_A y~ dA$ over the elliptical area. That is going to be a bit of work, but you can save yourself the work because you know the formula for the $y$ centroid of area of a region is$$\bar y = \frac {\iint_A y~dA}{\iint_A 1~dA}$$So instead of working your integral out just note you can solve this equation for $\iint_A y~ dA$ getting$$\iint_A y~ dA = \bar y \cdot \text{Area of ellipse} = d \pi a b$$since $d$ is the $y$ coordinate of the center of the ellipse.

• JD_PM

#### JD_PM

I think I get what is confusing you. Let's say you have an ellipse for which you know the centroid (center). Say the center is at $(c,d)$ and the major and minor axes are $a$ and $b$ and you know the formula for the area of such an ellipse is $\pi a b$. Now let's say you have an integral that you want to evaluate something like $\iint_A y~ dA$ over the elliptical area. That is going to be a bit of work, but you can save yourself the work because you know the formula for the $y$ centroid of area of a region is$$\bar y = \frac {\iint_A y~dA}{\iint_A 1~dA}$$So instead of working your integral out just note you can solve this equation for $\iint_A y~ dA$ getting$$\iint_A y~ dA = \bar y \cdot \text{Area of ellipse} = d \pi a b$$since $d$ is the $y$ coordinate of the center of the ellipse.
Oh so if we were to have:

$$\iint_A yx~ dA$$

Could I do:

$$\iint_A yx~ dA = \bar y \bar x \cdot \text{Area of ellipse} = de \pi a b$$

?

I have just seen $x$, $y$ and $z$ applied individually but not multiplying (that is why I am asking).

#### LCKurtz

Science Advisor
Homework Helper
Gold Member
Oh so if we were to have:

$$\iint_A yx~ dA$$

Could I do:

$$\iint_A yx~ dA = \bar y \bar x \cdot \text{Area of ellipse} = de \pi a b$$

?

I have just seen $x$, $y$ and $z$ applied individually but not multiplying (that is why I am asking).
You can probably answer that for yourself. You would be using a formula like this:$$\bar x \bar y = \frac {\iint_A xy~dA}{\iint_A 1~dA}$$You know the formulas for $\bar x$ and $\bar y$. Put them in there and see if you think it is true.

• JD_PM

#### JD_PM

You can probably answer that for yourself. You would be using a formula like this:$$\bar x \bar y = \frac {\iint_A xy~dA}{\iint_A 1~dA}$$You know the formulas for $\bar x$ and $\bar y$. Put them in there and see if you think it is true.
OK so I think you may agree that:

$$\iint_A xy~ dA = \bar x \bar y \cdot \text{Area of ellipse} = cd \pi a b$$

#### LCKurtz

Science Advisor
Homework Helper
Gold Member
OK so I think you may agree that:

$$\iint_A xy~ dA = \bar x \bar y \cdot \text{Area of ellipse} = cd \pi a b$$
You are just guessing. Until you show me what happens when you do what I suggested in the last line of post #4 you won't be able to do anything but guess. And you won't learn anything.

• JD_PM

#### JD_PM

You are just guessing. Until you show me what happens when you do what I suggested in the last line of post #4 you won't be able to do anything but guess. And you won't learn anything.
We know that:

$$\bar x = \frac {\iint_A x~dA}{\iint_A 1~dA}$$

$$\bar y = \frac {\iint_A y~dA}{\iint_A 1~dA}$$

Then:

$$(\frac {\iint_A x~dA}{\iint_A 1~dA})(\frac {\iint_A y~dA}{\iint_A 1~dA}) = \bar x \bar y = \frac {\iint_A xy~dA}{\iint_A 1~dA}$$

So this equality doesn't hold. Then the following is incorrect:

$$\iint_A xy~ dA = \bar x \bar y \cdot \text{Area of ellipse} = cd \pi a b$$

Then we have no alternative but work the integral out.

#### LCKurtz

Science Advisor
Homework Helper
Gold Member
We know that:

$$\bar x = \frac {\iint_A x~dA}{\iint_A 1~dA}$$

$$\bar y = \frac {\iint_A y~dA}{\iint_A 1~dA}$$

Then:

$$(\frac {\iint_A x~dA}{\iint_A 1~dA})(\frac {\iint_A y~dA}{\iint_A 1~dA}) = \bar x \bar y = \frac {\iint_A xy~dA}{\iint_A 1~dA}$$

So this equality doesn't hold. Then the following is incorrect:

$$\iint_A xy~ dA = \bar x \bar y \cdot \text{Area of ellipse} = cd \pi a b$$

Then we have no alternative but work the integral out.
Good, that's a step in the right direction. So it looks like the equality doesn't hold. You do understand that "looks like" isn't a mathematical argument though, right? So what you should do now to really settle the matter for yourself is actually prove that it doesn't hold by working out a simple example and showing you get different numbers. Then you will KNOW it doesn't hold.

Last edited:
• JD_PM

### Want to reply to this thread?

"Understanding the argument of the surface area integral"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving