Finding Jacobian in Differential calculus

Question

If $x=r\sin\theta\cos\phi$, $y=r\sin\theta\sin\phi$, $z=r\cos\theta$, show that $$\frac{\delta(x, y, z)} {\delta(r, \theta, \phi)} =r^2 \sin\theta \text.$$

While solving, I am getting $r^2 \sin\theta\cos\phi$.

Answer 1

As we know:

$\dfrac{\delta(x, y, z)} {\delta(r, \theta, \phi)} = \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial \phi}\\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta}& \frac{\partial y}{\partial \phi}\\ \frac{\partial z}{\partial r} & \frac{\partial z}{\partial \theta} & \frac{\partial z}{\partial \phi}\\ \end{vmatrix}$ $=r^2 \sin\theta \begin{vmatrix} sin\theta cos\phi &cos\theta cos\phi & -sin\phi\\ sin\theta sin\phi &cos\theta sin\phi & cos\phi\\ cos\theta &-sin\theta & 0\\ \end{vmatrix}$ $=r^2 \sin\theta\left[cos\theta(cos\theta cos^2\phi+cos\theta sin^2\phi)+sin\theta(sin\theta cos^2\phi+sin\theta sin^2\phi)\right]$ $=r^2 \sin\theta(cos^2\theta+sin^2\theta )$ $=r^2 \sin\theta$,

as you desired.