Volume of the space region inside sphere outside coni

In summary, the conversation discusses a problem involving finding the volume of a space region inside a sphere and outside two cones. The solution involves plotting the graph and finding the points of intersection, solving equations in the plane y = 0, and using the triple integral formula in terms of spherical coordinates.
  • #1
melihaltintas
6
0

Homework Statement


Hi everybody i have a problem please help me (sorry for my bad english)

Homework Equations


volume of the space region inside sphere outside coni
z^2=x^2+y^2 coni
x^2+y^2+z^2 =1 sphere

The Attempt at a Solution


I am new in this forum , I search question like this and i found but i didn't solve this question with method that is performed other questions
 
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  • #2
Hi melihaltintas

First, always plot the graph so you can understand how to find the limits.

I have attached the graph of the xz-trace coordinates (in the plane y = 0). The region of volume that you need to find is shaded in blue.

Start by finding the points of intersection of the sphere and cones (in the plane y = 0).

Your equations for the cones and sphere in the plane y = 0, become:
[tex]z^2=x^2
\\x^2+z^2 =1[/tex]
Solve these two equations to find the x and z. From there, you'll be able to find the two required angles for ##\phi## which define the limits.

Finally, use the triple integral formula in terms of spherical coordinates:
[tex]\int \int \int \rho^2 \sin \phi d\rho d\phi d\theta[/tex]
 

Attachments

  • xz.gif
    xz.gif
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  • #3
thanks a lot :)
 

1. What is the formula for calculating the volume of the space region inside a sphere outside a cone?

The formula for calculating the volume of the space region inside a sphere outside a cone is (4/3)πr3 - (1/3)πh2(3r-h), where r is the radius of the sphere and h is the height of the cone.

2. How is this formula derived?

This formula is derived by first finding the volume of the sphere (4/3)πr3 and subtracting the volume of the cone (1/3)πh2(3r-h) from it. This gives us the remaining volume inside the sphere but outside the cone.

3. Can this formula be used for any size sphere and cone?

Yes, this formula can be used for any size sphere and cone as long as the cone fits inside the sphere. The radius and height of the sphere and cone can be any value, as long as they are consistent in the formula.

4. How is this formula useful in real-world applications?

This formula is useful in real-world applications such as architecture and engineering, where precise calculations of volumes are necessary for designing structures. It can also be used in physics and astronomy to calculate the volumes of objects in space.

5. Are there any limitations to this formula?

One limitation of this formula is that it assumes the cone is completely inside the sphere. If the cone is only partially inside the sphere, the volume of the overlapping region must be subtracted from the total volume to get an accurate result.

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