# Find the volume of the region D using spherical coordinates

1. Dec 7, 2014

### s3a

1. The problem statement, all variables and given/known data
The problem and its solution are attached as TheProblemAndSolution.jpg.

2. Relevant equations
V(D) = ∫∫∫_D ρ^2 sinθ dρ dϕ dθ

3. The attempt at a solution
How exactly does the solution get cos α = 1/√(3)?

Also, when the solution goes from the step with two integrals to the step with one integral, is a minus sign forgotten by the author of the solution (because it seems to me that it should be cos(π/2) - cos α, where cos(π/2) = 0)?

Also, to be pedantic, given the way the problem is approached in the solution, shouldn't the solution say "in spherical coordinates (ρ, ϕ, θ)" (instead of (ϕ, θ, ρ))?

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2. Dec 7, 2014

### haruspex

Good question. I get that it's 1/2.
What's the integral of sin(x)?
There is no universal standard, though it is usual to put radius first. Certainly phi and theta are used with swapped roles by different authors, quite apart from how they're placed within the co-ordinate triple.

3. Dec 7, 2014

### s3a

Okay, so to get 1/2 (instead of 1/√(3)) what you did was the following (or something similar), right?:
[x^2 + y^2 + z^2 ≤ 1] + [4z^2 ≤ x^2 + y^2 + z^2]

x^2 + y^2 + z^2 + 4z^2 ≤ 1 + x^2 + y^2 + z^2

4z^2 ≤ 1

z^2 ≤ 1/4

z ≤ 1/2 (we ignore z ≥ -1/2, because z ≥ 0 is one of the conditions)

Then, using a Cartesian coordinate system where the positive z axis is upward, the positive y axis is into the screen and the positive x axis is toward the right, we draw a right triangle with height/leg z = 1/2, radius/hypotenuse ρ = 1, where α is the angle separating the leg z from the hypotenuse ρ, from which we get that cos α = 1/2.

The integral of sin(x) is -cos(x), and I now see my mistake. :)

So, in other words, what you're saying is that the order in the notation (ρ, ϕ, θ) doesn't have to be the same order as that of the differential components in the integral, right?

4. Dec 7, 2014

### haruspex

Yes, similar: the distance from the origin to a point on the cone is 2z. cos(alpha) = adjacent/hypotenuse = z/2z..
That would be true even if (ρ, ϕ, θ) were absolutely standard for polar co-ordinates.

5. Dec 8, 2014

### s3a

Alright, so, lastly, the final answer is V(D) = π/3, right?

6. Dec 8, 2014

### haruspex

Yes.
In fact, cylindrical co-ordinates are slightly better here. $\int_0^{\frac 12} \pi((1-z^2) - 3z^2).dz$, $\pi((1-z^2) - 3z^2)$ being the area of the annulus at z.

7. Dec 8, 2014

### s3a

Thanks for that and everything above too. :)