How Do You Calculate the Volume Between a Cone and a Sphere?

  • Thread starter Thread starter Rijad Hadzic
  • Start date Start date
  • Tags Tags
    Solid Volume
Rijad Hadzic
Messages
321
Reaction score
20

Homework Statement



Find volume of the solid that lies above the cone Φ = π/3 and below the sphere ρ = 4cosΦ

Homework Equations

The Attempt at a Solution



Obviously this is a triple integral. My book tells me that 0 ≤ρ≤ 4cosΦ

but this makes no sense to me.

From the problem, it lies ABOVE the cone Φ = π/3 and below the sphere ρ = 4cosΦ, so wouldn't that imply that ρ is not starting at 0?

What I did was solved ρ = 4cosΦ

arccos(ρ/4) = Φ and set it = to pi/3

arccos(ρ/4) = π/3

ρ/4 = cosπ/3

ρ = 4 * (1/2) = 2

so wouldn't 2≤ρ≤4cosΦ be the bounds? I don't understand how the lower bound can start at 0 when its asking for what's above the cone and below the sphere..
 
Physics news on Phys.org
Rijad Hadzic said:
wouldnt that imply that ρ is not starting at 0?
No. Also, please define your variables.

I suggest you draw an image of what things look like.
 
Orodruin said:
No. Also, please define your variables.

I suggest you draw an image of what things look like.

ρ is the distance to the point from the origin
Φ is the angle from the z axis to the point. 0≤Φ≤pi
θ is the angle to the point from the projection in the xy plane.

I did draw an image, it looks like an icecream cone basically. The part I want is the icecream on the top of the cone. I still don't understand why ρ is 0..
 
Hmm I think I may be graphing it wrong. Maybe I have no clue how to graph ρ = 4cosΦ..
 
Rijad Hadzic said:
ρ is the distance to the point from the origin
Φ is the angle from the z axis to the point. 0≤Φ≤pi
θ is the angle to the point from the projection in the xy plane.
Different textbooks treat spherical coordinates in different ways, particularly the ##\theta## and ##\phi## coordinates. According to this wikipedia article (https://en.wikipedia.org/wiki/Spherical_coordinate_system), physics books consider ##\phi## to be the angle in the "x-y" plane, while math textbooks consider ##\phi## to be measured from the positive z-axis.
 
Rijad Hadzic said:
The part I want is the icecream on the top of the cone.
No, this is not correct. The "cone" that the problem talks about is the mantle surface of the cone. If it was just the top that was intended, the problem would talk about a plane, not about a cone.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top