Are you interesting about average value if so trying here.

In summary, the conversation is about a problem posted by an individual asking for help with finding the average value of a function over a sphere using integrals and polar coordinates. The individual is advised to use proper notation and provide more information in their post to receive accurate help.
  • #1
memomath
13
0
Hello

here you have the same question in the enclosed

What is the average value?

average value for (exp [tex]\alpha[/tex]Z ) = [tex]\int[/tex][tex]\int[/tex] exp[tex]\alpha[/tex]Z ds / [tex]\int[/tex][tex]\int[/tex] ds



Over the sphere S: X^2+Y^2+Z^2=a^2

Also by use the parameterization

X= a sinϴ cos ɸ
Y= a sinϴ sin ɸ
Z= a cos ϴ

And the usual substitution t = cos ϴ in the integral:

[tex]0\int[/tex][tex]\Pi[/tex] f (cos ϴ) sinϴ dϴ =[tex]-1\int[/tex]1 f(t) dt

Thanks
 
Last edited:
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  • #2
No comments or solution until now!

memomath said:
hello

please open the enclosed to find a nice problem and to solve this interesting question

thanks
 
  • #3
Again, Word files are notorious for harboring viruses. I will not open one from someone I do not know.
 
  • #4
Hello here the same question in the enclosed

What is the average value?

average value for (exp [tex]\alpha[/tex]Z ) = [tex]\int[/tex][tex]\int[/tex] exp[tex]\alpha[/tex]Z ds / [tex]\int[/tex][tex]\int[/tex] ds



Over the sphere S: X^2+Y^2+Z^2=a^2

Also by use the parameterization

X= a sinϴ cos ɸ
Y= a sinϴ sin ɸ
Z= a cos ϴ

And the usual substitution t = cos ϴ in the integral:

[tex]0\int[/tex][tex]\Pi[/tex] f (cos ϴ) sinϴ dϴ =[tex]-1\int[/tex]1 f(t) dt

Thanks
 
  • #5
I can kind of guess what you mean, but your notation is pretty bad. I do not know what exp[tex]\:^\alpha Z[/tex] is supposed to be. Your substitution seems wrong, your notation is unorthodox and you write factors like -1 instead of just a - . This is probably the reason why people don't answer.

You might have better chances if you wrote something like:

"I am trying to get the average [tex]\left< f(z) \right>_{S_a}[/tex] of a function [tex]f(z)= \alpha ^ z[/tex] over a sphere [tex]S_a[/tex] of radius [tex]r=\sqrt{a}[/tex]. I swear this is not for homework.

What I have so far is this:

[tex]\left< f(z) \right>_{S_a} = \frac{\int_{S_a} \alpha^z\,\mathrm{d}\Omega}{\int_{S_a} \,\mathrm{d}\Omega}[/tex]

I tried to express the integral in polar coordinates:
X= a sinϴ cos ɸ
Y= a sinϴ sin ɸ
Z= a cos ϴ
[tex]\theta \in \left[0,\pi \right][/tex]
[tex]\phi \in \left[0,2\pi \right][/tex]

and found:
[tex] \int_{S_a} f(z) \,\mathrm{d}\Omega = \int_0^{\pi} f(\cos \theta) \sin \theta \, \mathrm{d}\theta [/tex]

Is this correct? How do I proceed?"

Then people would have told you what you did wrong.
 
  • #6
Thanks for reply but it is not correct hint and I think you don’t proceed with the right way for the real problem you probably fix some thing else many things missing from the real posted thank you
0xDEADBEEF said:
I can kind of guess what you mean, but your notation is pretty bad. I do not know what exp[tex]\:^\alpha Z[/tex] is supposed to be. Your substitution seems wrong, your notation is unorthodox and you write factors like -1 instead of just a - . This is probably the reason why people don't answer.

You might have better chances if you wrote something like:

"I am trying to get the average [tex]\left< f(z) \right>_{S_a}[/tex] of a function [tex]f(z)= \alpha ^ z[/tex] over a sphere [tex]S_a[/tex] of radius [tex]r=\sqrt{a}[/tex]. I swear this is not for homework.

What I have so far is this:

[tex]\left< f(z) \right>_{S_a} = \frac{\int_{S_a} \alpha^z\,\mathrm{d}\Omega}{\int_{S_a} \,\mathrm{d}\Omega}[/tex]

I tried to express the integral in polar coordinates:
X= a sinϴ cos ɸ
Y= a sinϴ sin ɸ
Z= a cos ϴ
[tex]\theta \in \left[0,\pi \right][/tex]
[tex]\phi \in \left[0,2\pi \right][/tex]

and found:
[tex] \int_{S_a} f(z) \,\mathrm{d}\Omega = \int_0^{\pi} f(\cos \theta) \sin \theta \, \mathrm{d}\theta [/tex]

Is this correct? How do I proceed?"

Then people would have told you what you did wrong.
 
Last edited by a moderator:
  • #7


YES memomath I believe there are some mistakes and the good hint to the correct answer for this problem would be in ENGINEERING MATHEMATICS JOHN BIRD

memomath said:
Thanks for reply but it is not correct hint and I think you don’t proceed with the right way for the real problem you probably fix some thing else many things missing from the real posted thank you
 

FAQ: Are you interesting about average value if so trying here.

What is an "average value" in scientific terms?

An average value is a measure of the central tendency of a set of data. It is calculated by adding together all the values in the data set and dividing by the number of values. This provides a single value that represents the overall trend of the data.

How is the average value used in scientific research?

The average value is often used to summarize large amounts of data and make comparisons between different groups or conditions. It can also be used to identify outliers or anomalies in the data.

What are some limitations of using the average value?

The average value can be affected by extreme values or outliers in the data, which can skew the overall result. It also does not provide information about the distribution or variability of the data, which may be important in some cases.

How is the average value different from the median and mode?

The median is the middle value in a data set, while the mode is the most frequently occurring value. These measures of central tendency can be used in conjunction with the average value to gain a better understanding of the data.

Can the average value be used for all types of data?

No, the average value is most commonly used for numerical data. It may not be appropriate for categorical data or data that is not normally distributed. In these cases, other measures of central tendency may be more suitable.

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