# Find the volume of the solid that lies inside both of the spheres? x^2 + y^2 + z^2 + 4x − 2y + 4z + 5 = 0 & x^2 + y^2 + z^2 = 4

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Feb 11, 2018

$V = \frac{11}{12} \pi$

#### Explanation:

Given the sphere:

x^2 + y^2 + z^2 + 4x − 2y + 4z + 5 = 0

To obtain the form ${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} + {\left(z - c\right)}^{2} = {r}^{2}$, we complete the squares:

x^2 + 4x + a^2 + y^2 − 2y + b^2+ z^2 + 4z + c^2 = -5 +a^2+b^2+c^2

$- 2 a x = 4 x$

$a = - 2$

$- 2 b y = - 2 y$

$b = 1$

$- 2 c z = 4 z$

$c = - 2$

The standard equation of the first sphere is:

${\left(x - \left(- 2\right)\right)}^{2} + {\left(y - 1\right)}^{2} + {\left(z - \left(- 2\right)\right)}^{2} = {2}^{2}$

The center is $\left(- 2 , 1 , - 2\right)$ and the radius, $r = 2$

The standard equation of second sphere is:

${\left(x - 0\right)}^{2} + {\left(y - 0\right)}^{2} + {\left(z - 0\right)}^{2} = {2}^{2}$

The center is $\left(0 , 0 , 0\right)$ and the radius, $R = 2$

The distance between the two centers is:

$d = \sqrt{{\left(- 2 - 0\right)}^{2} + {\left(1 - 0\right)}^{2} + {\left(- 2 - 0\right)}^{2}}$

$d = 3$

This the same volume as between the two spheres:

${x}^{2} + {y}^{2} + {z}^{2} = {2}^{2}$
${\left(x - 3\right)}^{2} + {y}^{2} + {z}^{2} = {2}^{2}$

I found equations for this special case where $R = r$ and d lies along the x axis at the following Wolfram Mathworld reference source; equation 17:

$V = \frac{1}{12} \pi \left(4 R + d\right) {\left(2 R - d\right)}^{2}$

$V = \frac{1}{12} \pi \left(4 \left(2\right) + 3\right) {\left(2 \left(2\right) - 3\right)}^{2}$

$V = \frac{11}{12} \pi$

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