Identifying Error in Spherical Coordinates - Homework Problem

[Schwarzschild90]

Homework Statement

Homework Equations

And the volume element dV expressed in spherical coordinates.

The Attempt at a Solution

Can you spot where I've made an error?

 

[SammyS]

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

What's you question?

Oh! I see.

Usually the question is posted in the text of the thread.

 

[Schwarzschild90]

What's you question?

Where is the error? * (Where have I made an error)

-Ok! Edited*

 

[SammyS]

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?

 

[Ray Vickson]

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?

Why do you think you have made an error?

 

[haruspex]

There is a slightly easier way. By symmetry you can change the x to z in the integrand.

 

[Schwarzschild90]

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.

 

[haruspex]

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.

 

[Schwarzschild90]

I turned in the assigment. Giving it another shot tonight.

 

[Ray Vickson]

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.