I have
$$\begin{aligned}x_{1}&=r\sin(\theta_{1}),\\ x_{2}&=r\cos(\theta_{1})\sin(\theta_{2})\\ x_{3}&=r\cos(\theta_{1})\cos(\theta_{2}). \end{aligned} $$
I know how to compute the Jacobian $$\frac{\partial(x_{1},x_{2},x_3)}{\partial(\theta_{1},\theta_{2},r)}$$ directly.

The thing is, there is a way to get this Jacobian that involves a ratio of an upper triangular determinant and a lower triangular determinant. I just cannot figure out how to get this. I'm guessing it's some chain rule thing. Any help will be appreciated. Thanks.

  • 1
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    – joriki
    Sep 28, 2012 at 6:59
  • $\begingroup$ It is the determinant of a 3 by 3 matrix of partial derivatives of variables $(x, y, z)$ with respect to the other variables $(r, \theta_1, \theta_2)$. I get $r^2\sin(\theta_1)\cos(\theta_2)^2$. $\endgroup$ Sep 28, 2012 at 16:32
  • $\begingroup$ That's what I got too - the thing is, the actual problem is for $x_{1}, \cdots, x_{n}$ - that's why that trick I was talking about would be very useful $\endgroup$
    – Ken Dunn
    Sep 28, 2012 at 16:47

1 Answer 1


$$\frac{\partial(x_1,x_2,x_3)}{\partial(\theta_1,\theta_2,r)}=\begin{bmatrix} \frac{\partial x_1}{\partial\theta_1} & \frac{\partial x_1}{\partial\theta_2} & \frac{\partial x_1}{\partial r}\\ \frac{\partial x_2}{\partial\theta_1} & \frac{\partial x_2}{\partial\theta_2} & \frac{\partial x_2}{\partial r}\\ \frac{\partial x_3}{\partial\theta_1} & \frac{\partial x_3}{\partial\theta_2} & \frac{\partial x_3}{\partial r}\\ \end{bmatrix}$$ $$=\begin{bmatrix} r\cos\theta_1 & 0 & \sin\theta_1 \\ -r\sin\theta_1\sin\theta_2 & r\cos\theta_1\cos\theta_2 & \cos\theta_1\sin\theta_2 \\ -r\sin\theta_1\cos\theta_2 & -r\cos\theta_1\sin\theta_2 & \cos\theta_1\cos\theta_2 \\ \end{bmatrix}$$ $$=r^2(\cos^3\theta_1\cos^2\theta_2+\cos^3\theta_1\sin^2\theta_2+\sin^2\theta_1\cos\theta_1\sin^2\theta_2+\sin^2\theta_1\cos\theta_1\cos^2\theta_2)$$ $$=r^2(\sin^2\theta_1+\cos^2\theta_1)(\sin^2\theta_2+\cos^2\theta_2)(4\cos\theta_1)$$ $$=4r^2\cos\theta_1$$


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