Step1

\(\displaystyle{x}^{{2}}+{y}^{{2}}+{z}^{{2}}+{4}{x}-{2}{y}+{4}{z}+{5}={0}\)

\(\displaystyle{\left({x}^{{2}}+{4}{x}\right)}+{\left({y}^{{2}}-{2}{y}\right)}+{\left({z}^{{2}}+{4}{z}\right)}+{5}={0}\)

\(\displaystyle{\left({x}^{{2}}+{4}{x}+{4}-{4}\right)}+{\left({y}^{{2}}-{2}{y}+{1}-{1}\right)}+{\left({z}^{{2}}+{4}{z}+{4}-{4}\right)}+{5}={0}\)

\(\displaystyle{\left({x}^{{2}}+{4}{x}+{4}\right)}-{4}+{\left({y}^{{2}}-{2}{y}+{1}\right)}-{1}+{\left({z}^{{2}}+{4}{z}+{4}\right)}-{4}+{5}={0}\)

\(\displaystyle{\left({x}+{2}\right)}^{{2}}+{\left({y}-{1}\right)}^{{2}}+{\left({z}+{2}\right)}^{{2}}={4}\)

This is a sphere with centre at (-2,1,-2) and radius 2

Step 2

I am going to use the following formula :

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

Where d is distance between the centres of two spheres

And R is the radius of the two spheres, that is 2 in this case

Link to the proof of this formula is given in comment section.

Step 3

Distance between the two centres is

\(\displaystyle{d}=\sqrt{{{\left(-{2}\right)}^{{2}}+{1}{\left({1}\right)}^{{2}}+{\left(-{2}\right)}^{{2}}}}=\sqrt{{{4}+{1}+{4}}}=\sqrt{{{9}}}={3}\)

Step 4

Substitute d=3 and R=2, to get the volume

\(\displaystyle{V}={\frac{{\pi}}{{{12}}}}{\left({4}\cdot{2}+{3}\right)}{\left({2}\cdot{2}-{3}\right)}^{{2}}\approx{2.88}\)

Result

\(\displaystyle{V}\approx{2.88}\)

\(\displaystyle{x}^{{2}}+{y}^{{2}}+{z}^{{2}}+{4}{x}-{2}{y}+{4}{z}+{5}={0}\)

\(\displaystyle{\left({x}^{{2}}+{4}{x}\right)}+{\left({y}^{{2}}-{2}{y}\right)}+{\left({z}^{{2}}+{4}{z}\right)}+{5}={0}\)

\(\displaystyle{\left({x}^{{2}}+{4}{x}+{4}-{4}\right)}+{\left({y}^{{2}}-{2}{y}+{1}-{1}\right)}+{\left({z}^{{2}}+{4}{z}+{4}-{4}\right)}+{5}={0}\)

\(\displaystyle{\left({x}^{{2}}+{4}{x}+{4}\right)}-{4}+{\left({y}^{{2}}-{2}{y}+{1}\right)}-{1}+{\left({z}^{{2}}+{4}{z}+{4}\right)}-{4}+{5}={0}\)

\(\displaystyle{\left({x}+{2}\right)}^{{2}}+{\left({y}-{1}\right)}^{{2}}+{\left({z}+{2}\right)}^{{2}}={4}\)

This is a sphere with centre at (-2,1,-2) and radius 2

Step 2

I am going to use the following formula :

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

Where d is distance between the centres of two spheres

And R is the radius of the two spheres, that is 2 in this case

Link to the proof of this formula is given in comment section.

Step 3

Distance between the two centres is

\(\displaystyle{d}=\sqrt{{{\left(-{2}\right)}^{{2}}+{1}{\left({1}\right)}^{{2}}+{\left(-{2}\right)}^{{2}}}}=\sqrt{{{4}+{1}+{4}}}=\sqrt{{{9}}}={3}\)

Step 4

Substitute d=3 and R=2, to get the volume

\(\displaystyle{V}={\frac{{\pi}}{{{12}}}}{\left({4}\cdot{2}+{3}\right)}{\left({2}\cdot{2}-{3}\right)}^{{2}}\approx{2.88}\)

Result

\(\displaystyle{V}\approx{2.88}\)