Math Problem Statement

Find the volume of the solid that lies inside both of the spheres, x^2 + y^2 + z^2 +4x - 2y + 4z + 5 = 0 and x^2 + y^2 + z^2 =4

Solution

To find the volume of the solid inside both spheres, we need to analyze the equations of the two spheres and their intersection. The equations of the spheres are:

  1. First sphere: x2+y2+z2+4x2y+4z+5=0x^2 + y^2 + z^2 + 4x - 2y + 4z + 5 = 0

  2. Second sphere: x2+y2+z2=4x^2 + y^2 + z^2 = 4

Step 1: Completing the square for the first sphere's equation

To simplify the first equation into the standard form of a sphere, we complete the square for the xx, yy, and zz terms:

  • For x2+4xx^2 + 4x, complete the square:
    x2+4x=(x+2)24x^2 + 4x = (x+2)^2 - 4

  • For y22yy^2 - 2y, complete the square:
    y22y=(y1)21y^2 - 2y = (y-1)^2 - 1

  • For z2+4zz^2 + 4z, complete the square:
    z2+4z=(z+2)24z^2 + 4z = (z+2)^2 - 4

Substitute these into the original equation of the first sphere:

(x+2)24+(y1)21+(z+2)24+5=0(x+2)^2 - 4 + (y-1)^2 - 1 + (z+2)^2 - 4 + 5 = 0

Simplify:

(x+2)2+(y1)2+(z+2)2414+5=0(x+2)^2 + (y-1)^2 + (z+2)^2 - 4 - 1 - 4 + 5 = 0

(x+2)2+(y1)2+(z+2)24=0(x+2)^2 + (y-1)^2 + (z+2)^2 - 4 = 0

(x+2)2+(y1)2+(z+2)2=4(x+2)^2 + (y-1)^2 + (z+2)^2 = 4

Thus, the first sphere has center (2,1,2)(-2, 1, -2) and radius 22.

Step 2: Understanding the geometry of the intersection

  • The second sphere is centered at the origin (0,0,0)(0, 0, 0) and has radius 22.
  • The first sphere is centered at (2,1,2)(-2, 1, -2) and also has radius 22.

Since the two spheres have the same radius and their centers are (20)2+(10)2+(20)2=4+1+4=3\sqrt{(-2-0)^2 + (1-0)^2 + (-2-0)^2} = \sqrt{4 + 1 + 4} = 3 units apart, the spheres intersect.

Step 3: Volume of the intersection of two spheres

For two intersecting spheres of the same radius rr and the distance dd between their centers, the volume of their intersection is given by the formula:

V=π(4rd)2(d2+2rd3r2)12dV = \frac{\pi (4r - d)^2 (d^2 + 2rd - 3r^2)}{12d}

Where:

  • r=2r = 2 (radius of the spheres),
  • d=3d = 3 (distance between the centers of the spheres).

Substituting these values:

V=π(4(2)3)2(32+2(2)(3)3(2)2)12(3)V = \frac{\pi (4(2) - 3)^2 (3^2 + 2(2)(3) - 3(2)^2)}{12(3)}

V=π(83)2(9+1212)36V = \frac{\pi (8 - 3)^2 (9 + 12 - 12)}{36}

V=π(5)2(9)36V = \frac{\pi (5)^2 (9)}{36}

V=π(25)(9)36V = \frac{\pi (25)(9)}{36}

V=225π36V = \frac{225\pi}{36}

V=25π4V = \frac{25\pi}{4}

Thus, the volume of the solid that lies inside both spheres is 25π4\frac{25\pi}{4} cubic units.


Would you like further details or have any questions about this solution?

Here are some related questions:

  1. How do you find the volume of intersection for spheres with different radii?
  2. What is the formula for the volume of intersection between two spheres?
  3. How does the distance between two sphere centers affect the intersection volume?
  4. How can you find the center and radius of a sphere from its general equation?
  5. What happens when the distance between two spheres is greater than the sum of their radii?

Tip: When completing the square for sphere equations, remember to group the terms of xx, yy, and zz separately to make the process easier.

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Math Problem Analysis

Mathematical Concepts

Geometry
Spheres
Volume of Intersection of Spheres
Coordinate Geometry
Completing the Square

Formulas

Volume of intersection of two spheres: V = (π (4r - d)^2 (d^2 + 2rd - 3r^2)) / 12d

Theorems

Formula for volume of intersection of two spheres

Suitable Grade Level

Grades 11-12