Identifying Error in Spherical Coordinates - Homework Problem

  • Thread starter Schwarzschild90
  • Start date
  • Tags
    Error
In summary: Do it the easy way: ##V = \int_0^1 x A(x) \, dx##, where ##A(x)## is the area of the slice at ##x## parallel to the ##yz## plane. The slice is a quarter circle of radius ##\sqrt{1-x^2}##, so ##A(x) = \frac{1}{4} \pi (1-x^2)##. This gives ##V = \pi/16##.
  • #1
Schwarzschild90
113
1

Homework Statement


Problem statement.PNG


Homework Equations


Spherical Coordinates.PNG

And the volume element dV expressed in spherical coordinates.

The Attempt at a Solution


1.PNG


Can you spot where I've made an error?
 
Last edited:
Physics news on Phys.org
  • #2
  • #3
SammyS said:
What's you question?
Where is the error? * (Where have I made an error)

-Ok! Edited*
 
  • #4
Schwarzschild90 said:

Homework Statement


View attachment 81041

Homework Equations


View attachment 81042
And the volume element dV expressed in spherical coordinates.

The Attempt at a Solution


View attachment 81043

Can you spot where I've made an error?
I don't see the error after going through what you did twice.

What is the correct answer?
 
  • #5
  • #6
There is a slightly easier way. By symmetry you can change the x to z in the integrand.
 
  • #7
Ray Vickson said:
Why do you think you have made an error?
My study friends have gotten a different result than I. (Pi/8). And my intuition tells me that by integrating over radius=1, I should get something that looks like 1/8 the volume of a sphere of radius 1 (Pi/6). Would someone like to clear this up? I'd like to know if I'm doing something wrong.
 
Last edited:
  • #8
Schwarzschild90 said:
My study friends have gotten a different result than I. (Pi/8). And my intuition tells me that by integrating over radius=1, I should get something that looks like 1/8 the volume of a sphere of radius 1 (Pi/6). Would someone like to clear this up? I'd like to know if I'm doing something wrong.
Sure but the integrand, x, is on average less than 1/2.
You can check your answer by looking up the mass centre of a hemisphere.
 
  • #9
I turned in the assigment. Giving it another shot tonight.
 
  • #10
Schwarzschild90 said:
My study friends have gotten a different result than I. (Pi/8). And my intuition tells me that by integrating over radius=1, I should get something that looks like 1/8 the volume of a sphere of radius 1 (Pi/6). Would someone like to clear this up? I'd like to know if I'm doing something wrong.

Do it the easy way: ##V = \int_0^1 x A(x) \, dx##, where ##A(x)## is the area of the slice at ##x## parallel to the ##yz## plane. The slice is a quarter circle of radius ##\sqrt{1-x^2}##, so ##A(x) = \frac{1}{4} \pi (1-x^2)##. This gives ##V = \pi/16##, as you obtained.
 

Related to Identifying Error in Spherical Coordinates - Homework Problem

1. What is the most common cause of errors in scientific experiments?

The most common cause of errors in scientific experiments is human error. This includes mistakes in measurement, calculations, or recording data.

2. How can I identify the source of an error in my experiment?

To identify the source of an error, it is important to carefully examine all steps of the experiment and review any data collected. This can help pinpoint where mistakes may have occurred.

3. Can equipment malfunctions cause errors in scientific experiments?

Yes, equipment malfunctions can certainly cause errors in scientific experiments. It is important to regularly calibrate and maintain equipment to minimize the risk of errors.

4. Are there any common errors that can be avoided in scientific experiments?

Yes, some common errors that can be avoided include using expired or contaminated materials, not following proper procedures, and not properly controlling variables in the experiment.

5. How can I prevent errors from occurring in my scientific experiments?

To prevent errors, it is important to carefully plan and design experiments, follow proper procedures, maintain equipment, and carefully record and analyze data. It can also be helpful to have a colleague review your work and provide feedback.

Back
Top