# If x = r sin θ cos ϕ, y = r sin θ sin ϕ and z = r cos θ, then

Question:

If $x=r \sin \theta \cos \phi, y=r \sin \theta \sin \phi$ and $z=r \cos \theta$, then

(a) $x^{2}+y^{2}+z^{2}=r^{2}$

(b) $x^{2}+y^{2}-z^{2}=r^{2}$

(c) $x^{2}-y^{2}+z^{2}=r^{2}$

(d) $z^{2}+y^{2}-x^{2}=r^{2}$

Solution:

Given:

$x=r \sin \theta \cos \phi$,

$y=r \sin \theta \sin \phi$,

$z=r \cos \theta$

Squaring and adding these equations, we get

$x^{2}+y^{2}+z^{2}=(r \sin \theta \cos \phi)^{2}+(r \sin \theta \sin \phi)^{2}+(r \cos \theta)^{2}$

$\Rightarrow x^{2}+y^{2}+z^{2}=r^{2} \sin ^{2} \theta \cos ^{2} \phi+r^{2} \sin ^{2} \theta \sin ^{2} \phi+r^{2} \cos ^{2} \theta$

$\Rightarrow x^{2}+y^{2}+z^{2}=\left(r^{2} \sin ^{2} \theta \cos ^{2} \phi+r^{2} \sin ^{2} \theta \sin ^{2} \phi\right)+r^{2} \cos ^{2} \theta$

$\Rightarrow x^{2}+y^{2}+z^{2}=r^{2} \sin ^{2} \theta\left(\cos ^{2} \phi+\sin ^{2} \phi\right)+r^{2} \cos ^{2} \theta$

$\Rightarrow x^{2}+y^{2}+z^{2}=r^{2} \sin ^{2} \theta(1)+r^{2} \cos ^{2} \theta$

$\Rightarrow x^{2}+y^{2}+z^{2}=r^{2} \sin ^{2} \theta+r^{2} \cos ^{2} \theta$

$\Rightarrow x^{2}+y^{2}+z^{2}=r^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)$

$\Rightarrow x^{2}+y^{2}+z^{2}=r^{2}(1)$

$\Rightarrow x^{2}+y^{2}+z^{2}=r^{2}$

Hence, the correct option is (a).