Volume inside Sphere, outside Cylinder

In summary, someone forgot to add the two areas to their equation and then used the disk method to find the volume.
  • #1
mattmns
1,128
6
Just wondering if I did this correct.



Find the volume of the region that lies inside the sphere [tex]x^2 + y^2 + z^2 = 2 [/tex] and outside the cylinder [tex]x^2 + y^2 = 1[/tex]
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Using cylindrical coordinates, and symmetry, I got:

I went up the z-axis, hitting z = 0 first, then exiting at [tex]z = \sqrt{2-r^2}[/tex]

So, the projection is two circles, one with r=1 and the other [tex]r=\sqrt{2}[/tex]

[tex]2 \int_{0}^{2\pi}\int_{1}^{\sqrt{2}}\int_{0}^{\sqrt{2-r^2}} r dz dr d\theta[/tex]

Which is then

[tex] 4\pi \int_{1}^{\sqrt{2}} r \sqrt{2-r^2} dr [/tex]

Let[tex] u = 2-r^2
=> du = -2rdr
=> \frac{-du}{2} = rdr [/tex]

Then I got:

[tex] -2\pi \int_{r=1}^{r=\sqrt{2}} \sqrt{u}du = \left[ -2\pi \frac{2}{3} u^\frac{3}{2} \right]_{r=1}^{r=\sqrt{2}}[/tex]

[tex]\left[ \frac{-4\pi}{3}(2-r^2)^\frac{3}{2} \right]_{1}^{\sqrt{2}} [/tex]

Which equals [tex]\frac{4\pi}{3}[/tex]

Look ok? Thanks.
 
Last edited:
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  • #2
Looks good. Could also have been done by the methods of "concentric cylindrical shells" or "washers" ususually used to introduce volumes of revolution, but multiple integration gets you to the same place.
 
  • #3
It's not really necessary to use calculus at all! The cylinder is given by x2+ y2= 1. That will intersect the sphere when x2+ y2+ z2= 1+ z2= 2: that is, when z= -1 and 1 so the cylinder has height 2. The volume of a sphere with radius [tex]\sqrt{2}[/tex] is [tex]\frac{4}{3}\pi (\sqrt{2})^3= \frac{16\sqrt{2}}{3}\pi[/tex]. The volume of a cylinder with radius 1 and height 2 is [tex]\pi (1)^2(2)= 2\pi[/tex]. The volume of the region between them is [tex]\frac{16\sqrt{2}}{3}\pi-2\pi= \frac{16\sqrt{2}-6}{3}\pi[/tex].
 
  • #4
Well it may not be necessary to use calculus, but this is the way it is expected to be done for the class I am taking.

Also, I think you forgot the two areas (the top and bottom). Shown here:

http://img307.imageshack.us/img307/9403/areaforgot9ew.gif [Broken]
 
Last edited by a moderator:
  • #5
Of course, the formula I gave was for a cylinder with flat tops. I suspect the problem means the part of the infinite cylinder contained within the sphere- that is, yes, you have to subtract off the two top and bottom volumes. You will need to use calculus to find those. The disk method ought to work.
 

1. What is the formula for calculating the volume inside a sphere and outside a cylinder?

The formula for calculating the volume inside a sphere and outside a cylinder is V = π(2r^3 - h^3)/3, where r is the radius of the sphere and h is the height of the cylinder.

2. How do you determine the volume inside a sphere and outside a cylinder when the radius and height are given?

To determine the volume inside a sphere and outside a cylinder, plug in the given values for the radius and height into the formula V = π(2r^3 - h^3)/3 and solve for V.

3. Can the volume inside a sphere and outside a cylinder be negative?

No, the volume inside a sphere and outside a cylinder cannot be negative. Volume is a measure of space and cannot have a negative value.

4. What is the difference between the volume inside a sphere and outside a cylinder?

The volume inside a sphere is the space contained within the sphere, while the volume outside a cylinder is the space between the cylinder and the sphere. In other words, the volume inside a sphere is the volume of the sphere itself, while the volume outside a cylinder is the space between the sphere and the cylinder.

5. How is the volume inside a sphere and outside a cylinder used in real life?

The volume inside a sphere and outside a cylinder is used in many real-life applications, such as calculating the volume of a water tank or determining the capacity of a storage container. It is also used in engineering and construction to design structures and determine the amount of material needed for a project.

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