Tricky Calculus III Problem

[themadhatter1]

Homework Statement

Find the volume of the solid that lies in between both of the spheres:

x²+y²+z²+4x+2y+4z+5=0
and
x²+y²+z²=4

Homework Equations

This is the first chapter of the calculus III material so no double or triple integrals are needed to solve this problem.

The Attempt at a Solution

I completed the square on the first equation and obtained:

(x+2)²+(y-2)²+(z+2)²=4

So I just have 2 intersecting spheres. Finding the volume of a "slice" of a sphere can be done using a solid of revolution, however, I would need know how far one sphere is intersecting into the other in order to calculate the volume.

My idea is to minimize the distance between (x+2)²+(y-2)²+(z+2)²=4 and the point (0,0,0). Then I will be able to figure out how "thick" the 2 "slices" of the spheres are.

However, I'm not quite sure how to do this. I know how to find the distance between points in space but solving for each variable I will wind up with 2 solutions because of the radical. Is there an easier way to do this that I am missing?

Thanks.

 

[micromass]

Well, the easiest way to solve this is by calculating a double integral with spherical coordinates, but you said that this is not needed.

Your method is also nice, but a little more work. Let's see if we can do it right.

We need to find the shortest distance between (0,0,0) and the equation $(x+2)^2+(y-2)^2+(z+2)^2=4$.
We can do that by calculus by minimizing a certain function, but that might get tedious. Here's another method: the line which has the shortest distance connect (0,0,0) with (-2,2,-2) (the center of the circle). So you only need to set up the equation of the line and see where it intersects the sphere.

 

[themadhatter1]

Genius! I love it. Can't believe I didn't think of that.

so for the line I I created the equation x=-2t,y=t,z=-2t

Next I plugged the equation into the sphere with it's center at (-2,1,-2) obtaining...

$$(-2t+2)^2+(t-1)^2+(-2t+2)^2=4$$$$2(4t^2-8t+4)+(t^2-2t+1)=4$$$$9t^2-18t+5=0$$$$t=\frac{1}{3},\frac{5}{3}$$

The first t value is the one that will help me so I plugged it back into my equation for a line, obtaining$$(\frac{-2}{3},\frac{1}{3},\frac{-2}{3})$$

The distance from (0,0,0) to this point turns out conveniently be 1. So one half of the solid section has a distance of 1 from the edge of the sphere.

Now I integrate to find the volume of the whole solid.

$$2\pi\int^{2}_{1}4-y^2 dy = 2\pi[4y-\frac{y^3}{3}]^{2}_{1}=\frac{10\pi}{3}$$

Does this look correct?