 # Find the volume of the solid that lies inside both of the spheres x^2+y^2 Aneeka Hunt 2021-10-28 Answered
Find the volume of the solid that lies inside both of the spheres
$$\displaystyle{x}^{{2}}+{y}^{{2}}+{z}^{{2}}+{4}{x}-{2}{y}+{4}{z}+{5}={0}$$
and
$$\displaystyle{x}^{{2}}+{y}^{{2}}+{z}^{{2}}={4}$$

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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
$$\displaystyle{V}\approx{2.88}$$