Surface integral of dome

[vik121]

Hello everyone, so this is an example from a physics calculus exam that I'm working through and I would really need some help setting up the entire thing, since I'm very confused as to how am I even going to start this example. I am even confused as to what some of the specific parameters are - like n̂. Any kind of help would be useful!

The example is in german so I tried my best to translate it as clearly as possible (including the notation for surface):

"You are working on a project in which a building will have a glass-dome roof built on top. The building itself is cyllindrical with a radius of r=50 m and a height of h. The dome represents a circle segment of a sphere which has a radius R = 100 m. ....edited (see picture)

a) Calculate the height h
b) You are asked for the requiered amount of glass needed. Therefore calculate the surface of the dome.
Hint: Calculate in spherical coordinates. In general it applies:

[MATH]A = \int_{0}^{\vartheta }\int_{0}^{\varphi }\hat{n}\cdot \mathrm{d}\vec{\sigma }[/MATH]
where n̂ is the unit normal vector to the given surface. The vectorial surface element dσ is given as:

[MATH]\mathrm{d}\vec{\sigma }=\hat{n}\mathrm{d}\vec{\sigma }=\left ( \frac{\partial \vec{r}}{\partial \vartheta }\mathrm{d}\vartheta \right )\times \left ( \frac{\partial \vec{r}}{\partial \varphi }\mathrm{d}\varphi \right )[/MATH]"

I am certain you can calculate the height very simply (that is worth only 1 point and the integral is 5 points) even though I currently can't even think of how to do that. Thank you in advance.

 

[Deleted member 4993]

Hello everyone, so this is an example from a physics calculus exam that I'm working through and I would really need some help setting up the entire thing, since I'm very confused as to how am I even going to start this example. I am even confused as to what some of the specific parameters are - like n̂. Any kind of help would be useful!

The example is in german so I tried my best to translate it as clearly as possible (including the notation for surface):

"You are working on a project in which a building will have a glass-dome roof built on top. The building itself is cyllindrical with a radius of r=50 m and a height of h. The dome represents a circle segment of a sphere which has a radius R = 50 m. (see picture)
View attachment 20465
a) Calculate the height h
b) You are asked for the requiered amount of glass needed. Therefore calculate the surface of the dome.
Hint: Calculate in spherical coordinates. In general it applies:

[MATH]A = \int_{0}^{\vartheta }\int_{0}^{\varphi }\hat{n}\cdot \mathrm{d}\vec{\sigma }[/MATH]
where n̂ is the unit normal vector to the given surface. The vectorial surface element dσ is given as:

[MATH]\mathrm{d}\vec{\sigma }=\hat{n}\mathrm{d}\vec{\sigma }=\left ( \frac{\partial \vec{r}}{\partial \vartheta }\mathrm{d}\vartheta \right )\times \left ( \frac{\partial \vec{r}}{\partial \varphi }\mathrm{d}\varphi \right )[/MATH]"

I am certain you can calculate the height very simply (that is worth only 1 point and the integral is 5 points) even though I currently can't even think of how to do that. Thank you in advance.

Please review your problem statement and make sure that it is GIVEN that r = 50 m and R = 50 m [edited]

You can calculate 'h' using Pythagorus's Theorem.

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520​

Please share your work/thoughts about this problem.

 

[yoscar04]

Hi. Why don't you show us how to compute h?
Have you taken any course of multiple variables?

 

[vik121]

Sorry about the h question, I see how stupid it was now - for some reason I didn't notice that I have both the radii and can just put it into the pythagorian theorem.
And yes, that was a typing error, R is = 100m and r = 50m. I can't edit the post anymore.

I have done quite some work with multiple variables, but I haven't done a lot of work with spherical coordinates and that's the biggest problem here. I can't even set up the example. Which parameters should I use to describe the r vector when doing the partial derivatives? Which paramaters should I use for n̂?
I am familiar with the classic x=rsinϑcosφ, y=rsinϑsinφ and z=rcosϑ transformation to spherical coordinates, and I have done some integrals with them where I was given a function, but here I have no clue how to set up the n and r.

 

[yoscar04]

I recommend you to read the section on surface integrals of any multi variable calculus book, i.e. Tom Apostol, Vol 2. Then try to write your integral and see if you can complete the computation by yourself.

 

[Deleted member 4993]

Sorry about the h question, I see how stupid it was now - for some reason I didn't notice that I have both the radii and can just put it into the pythagorian theorem.
And yes, that was a typing error, R is = 100m and r = 50m. I can't edit the post anymore.

I have done quite some work with multiple variables, but I haven't done a lot of work with spherical coordinates and that's the biggest problem here. I can't even set up the example. Which parameters should I use to describe the r vector when doing the partial derivatives? Which paramaters should I use for n̂?
I am familiar with the classic x=rsinϑcosφ, y=rsinϑsinφ and z=rcosϑ transformation to spherical coordinates, and I have done some integrals with them where I was given a function, but here I have no clue how to set up the n and r.

So .... h = ?

 

[LCKurtz]

The OP translated the problem from German for us. I suppose the picture was included in the problem. But suppose, just for kicks, that the picture wasn't included and the OP just made the assumption that the center of the spherical glass is at the center of the base of the building. What if that wasn't given? Then you cannot calculate [MATH]h[/MATH] from the given information. But you can still solve the problem because it doesn't depend on the height of the building. Same amount of glass.

 

[Deleted member 4993]

The OP translated the problem from German for us. I suppose the picture was included in the problem. But suppose, just for kicks, that the picture wasn't included and the OP just made the assumption that the center of the spherical glass is at the center of the base of the building. What if that wasn't given? Then you cannot calculate [MATH]h[/MATH] from the given information. But you can still solve the problem because it doesn't depend on the height of the building. Same amount of glass.

The problem specifically asked for 'h'.

I think, that will help to establish the limit of integration.

 

[LCKurtz]

The problem specifically asked for 'h'.

I think, that will help to establish the limit of integration.

Yes, they asked for [MATH]h[/MATH]. But it isn't needed.

 

[Deleted member 4993]

Yes, they asked for [MATH]h[/MATH]. But it isn't needed.

True....

 

[vik121]

So after some consultation with a colleague of mine, I found a way to set up the integral, using the same spherical coordinate form for both the r vector and the n̂ normal unit vector. The problem now is I get a rather interesting situation where all φ terms in the integral cancel out, and at that point, I just didn't integrate at all for φ and just integrated for ϑ. I get a result which I know is wrong (I checked online in a calculator what the surface of the spherical cap should be).
Here is what I did:




Another thing to add, another colleague of mine had a friend of his calculate his problem, and their and my calculation match exactly until the φ-integral, and I can't understand what exactly happened; so I'll post their solution as well (their solution skips a lot of the calculation so it's much shorter):



They pulled out a "2pi" factor in front of the integral, and their end solution is CORRECT - that's the obvious difference between their work and mine, but I don't know how that happened? How do you exactly integrate a function by a variable that the function doesn't even have?

P.S
The notation for area used here is "F" (german) and the colleague's friend used small r to notate the R (sphere radius).

 

[vik121]

I just now noticed, I wrote my result as m^2 even though actually it is m^3 (radius^3). Even the friend's solution is in m^3 even though it is the correct value. We're calculating area, not volume, so there has to be some kind of mistake somewhere, or something has to be re-written?

 

[LCKurtz]

The element of surface area for this problem is [MATH]R^2\sin\phi ~d\phi d\theta[/MATH] where [MATH]\phi[/MATH] is the azimuth angle and [MATH]\theta[/MATH] is the polar coordinate angle. I would agree with your last image with [MATH]r^2[/MATH] instead of [MATH]r^3[/MATH] although, as a math guy, I would switch the [MATH]\phi[/MATH] and [MATH]\theta[/MATH] variables.

 

[vik121]

In the meantime I figured out what happens to the 2π integral - the boundaries are just set in even without the variable there, but my question is still, how do I actually get to having r^2 and not r^3? I mean whichever way I differentiate I get a r^3 constant in front of the integral.

 

[LCKurtz]

In the meantime I figured out what happens to the 2π integral - the boundaries are just set in even without the variable there, but my question is still, how do I actually get to having r^2 and not r^3? I mean whichever way I differentiate I get a r^3 constant in front of the integral.

Here's how I would have set it up, using math conventions for [MATH]\phi[/MATH] and [MATH]\theta[/MATH] and radius [MATH]a[/MATH]. The vector equation for the surface is [MATH]\vec R(\phi,\theta) = \langle a\sin\phi \cos\theta, a\sin\phi\sin\theta,a\cos\phi \rangle[/MATH]. If you calculate [MATH]\vec R_\phi[/MATH] and [MATH]\vec R_\theta[/MATH] and cross them you will get
[MATH]\vec R_\phi \times \vec R_\theta = \langle a^2\sin^2\phi\cos\theta, a^2\sin^2\phi\sin\theta,a^2\cos\phi\sin\phi \rangle[/MATH]. Note the factor of [MATH]a^2[/MATH]. When you calculate its length you will have a factor of [MATH]a^4[/MATH] under the square root sign and you can check that you get [math]dS = a^2\sin\phi~d\phi d\theta[/MATH]. There is no [MATH]a^3[/MATH] to be found.