Help with a simple cos substitution

[ck99]

Homework Statement

Hi folks, I am sure this is very simple but there are not enough steps given in this calculation for my simple brain to get from the beginning to the end!

σ = ∫ (dσ/dΩ) = ∫ r²sin²θ (no integral limits given)

σ = 2∏r² ∫ (1 - u²) du (integral from -1 to 1)

σ = 8∏r² / 3

Homework Equations

u = cos θ

The Attempt at a Solution

I used Ω = sin θ dθ d∅ and first integrated ∅ from 0 to 2∏ to get

σ = ∫ dσ = r² ∫ sin²θ dΩ

σ = 2∏ r² ∫ sin²θ sin θ dθ

Use sin²θ = 1 - cos²θ to get

σ = 2∏ r² ∫(1 - cos²θ) sin θ dθ

Let u = cos θ so du/dθ = - sin θ and dθ = -arcsin θ to get

σ = 2∏ r² ∫(1 - u²) sin θ -arcsin θ du

I think sin and arcsin cancel to give

σ = 2∏ r² -∫(1 - u²) du

σ = 2∏ r² 2u

From the answer that was given I have the integral limits running from -1 to 1 so the final term becomes [2 - (-2)] = 4 which gives

σ = 8∏ r²

Where am I going wrong please?

 

[ck99]

Oh. I differentiated u instead of integrating. Some days there is not enough coffee in the world . . .

 

[I like Serena]

Hi ck99!

Homework Statement

Hi folks, I am sure this is very simple but there are not enough steps given in this calculation for my simple brain to get from the beginning to the end!

σ = ∫ (dσ/dΩ) = ∫ r²sin²θ (no integral limits given)

You seem to mix up differentials a bit here and elsewhere.

I'm going to assume you actually intended:

σ = ∫ dσ = ∫ r²sin²θ dΩ

σ = 2∏r² ∫ (1 - u²) du (integral from -1 to 1)

σ = 8∏r² / 3

Homework Equations

u = cos θ

The Attempt at a Solution

I used Ω = sin θ dθ d∅ and first integrated ∅ from 0 to 2∏ to get

σ = ∫ dσ = r² ∫ sin²θ dΩ

σ = 2∏ r² ∫ sin²θ sin θ dθ

Use sin²θ = 1 - cos²θ to get

σ = 2∏ r² ∫(1 - cos²θ) sin θ dθ

Let u = cos θ so du/dθ = - sin θ and dθ = -arcsin θ to get

Here's another mix-up.
Differentials always need to balance.

This is not the case in dθ = -arcsin θ, but then, you don't need it.

Let's stick to: u = cos θ so du = - sin θ dθ.

σ = 2∏ r² ∫(1 - u²) sin θ -arcsin θ du

That should be:

σ = 2∏ r² ∫(1 - u²)(-du)

I think sin and arcsin cancel to give

σ = 2∏ r² -∫(1 - u²) du

σ = 2∏ r² 2u

From the answer that was given I have the integral limits running from -1 to 1 so the final term becomes [2 - (-2)] = 4 which gives

σ = 8∏ r²

Where am I going wrong please?