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On the Jacobian determinant for conversion to cylindrical coordinates
By Sarah Richards
I have an exercise that asks to compute the triple integral over a region $E$ (a paraboloid with a horizontal slice), which I know to be $$\iiint_E x\,dV=\int_{-1}^1\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\int_{6y^2+6z^2}^6x\,dx\,dz\,dy=12\pi$$ I computed the result via a conversion to cylindrical coordinates at first by using the following parameters: $$\begin{cases}x=x\\ y=r\sin t\\ z=r\cos t\end{cases}$$ To compute the determinant of the Jacobian, I have $$|J|=\begin{vmatrix} \frac{\partial x}{\partial x}&\frac{\partial y}{\partial x}&\frac{\partial z}{\partial x}\\ \frac{\partial x}{\partial r}&\frac{\partial y}{\partial r}&\frac{\partial z}{\partial r}\\ \frac{\partial x}{\partial t}&\frac{\partial y}{\partial t}&\frac{\partial z}{\partial t} \end{vmatrix}= \begin{vmatrix}1&0&0\\0&\sin t&\cos t\\0&r\cos t&-r\sin t\end{vmatrix}=-r$$ However, if I were to swap $x$ and $z$ (so that the paraboloid were oriented along the $z$ axis in place of the $x$ axis) and use the usual parameters, $$\begin{cases}x=r\cos t\\y=r\sin t\\z=z\end{cases}$$ I get the usual Jacobian, $|J|=r$. I felt I trusted this result more, so I stuck with this one to get $12\pi$.
Why is there a discrepancy here? Does the order of the partial derivatives in the Jacobian matter?
$\endgroup$ 82 Answers
$\begingroup$Indeed you must take the absolute value of the jacobian. Both the change of variables are correct.
Take a closer look to the different diffeomorphisms of the two changes of variables. They are $$\begin{array}{ccc} \Phi:& \mathbb R^3& \to & \mathbb R^3\\ & \begin{pmatrix} r\\t\\z\end{pmatrix}&\mapsto & \begin{pmatrix} r\cos(t)\\r\sin(t)\\z\end{pmatrix}\end{array} $$ $$\begin{array}{ccc} \Psi:& \mathbb R^3& \to & \mathbb R^3\\ & \begin{pmatrix} r\\t\\x\end{pmatrix}&\mapsto & \begin{pmatrix} x\\r\sin(t)\\r\cos(t)\end{pmatrix}\end{array} $$
As you can see, $$A\Phi((r,t,z)^T)= \begin{pmatrix}0&0&1\\0&1&0\\1&0&0 \end{pmatrix} \begin{pmatrix} r\cos(t) \\r\sin(t) \\z \end{pmatrix} = \begin{pmatrix} z\\r\sin(t)\\r\cos(t)\end{pmatrix} = \Psi((r,t,z)^T) $$
The matrix $A$ has determinant $-1$ and is a simmetry of $\mathbb R^3$ in itself. Since $A$ does not depend on the variables and because of Binet's theorem, $$J_{\Psi}=AJ_{\Phi}\implies \text{det}(J_{\Psi})=-\text{det}(J_{\Phi})$$ Since we usually want volume to be positive we usually take $|\text{det}(J)|$ over $\text{det}(J)$ as you did: this follows intuitively if you believe that symmetrical shapes have the same volume.
Moreover, a simmetry flips the normal vectors of surfaces in certain directions. For instance ,the normal vector $(0,1)$ on top of a circle in $\mathbb R^2$ becomes $(0,-1)$ when you apply the simmetry $(x,y)\mapsto(x,-y)$.
A minus sign in your calculations of $\text {det}(J)$ shows that you have inverted the orientation of your surface while changing coordinates, so you have "flipped" some normal vectors. Nothing to worry about when calculating volumes!
$\endgroup$ $\begingroup$It seems the only detail I had overlooked when I first posted this question was that one takes the absolute value of the Jacobian determinant.
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