Find the volume of the region D using spherical coordinates

[s3a]

Homework Statement

The problem and its solution are attached as TheProblemAndSolution.jpg.

Homework Equations

V(D) = ∫∫∫_D ρ^2 sinθ dρ dϕ dθ

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))?

 

[haruspex]

How exactly does the solution get cos α = 1/√(3)?

Good question. I get that it's 1/2.

when the solution goes from the step with two integrals to the step with one integral, is a minus sign forgotten by the author

What's the integral of sin(x)?

shouldn't the solution say "in spherical coordinates (ρ, ϕ, θ)"

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.

 

[s3a]

Good question. I get that it's 1/2.

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.

What's the integral of sin(x)?

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

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.

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?

 

[haruspex]

what you did was the following (or something similar),

Yes, similar: the distance from the origin to a point on the cone is 2z. cos(alpha) = adjacent/hypotenuse = z/2z..

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

That would be true even if (ρ, ϕ, θ) were absolutely standard for polar co-ordinates.

 

[s3a]

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

 

[haruspex]

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

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.

 

[s3a]

Thanks for that and everything above too. :)