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I understood that when using substitution for multiple variables, the Jacobian determinant should be added.

\begin{vmatrix} \frac{\partial f_1}{\partial v_1} & \frac{\partial f_1}{\partial v_2} \\ \frac{\partial f_2}{\partial v_1} & \frac{\partial f_1}{\partial v_2} \\ \end{vmatrix}

I do not understand which function should be donated as $f$ and which the variables.

Taking the polar-cartesian transformation for example:

$x=r\cos \theta$

$y=r\sin \theta$

Which one is $f$ and which one are the variables? if I have the Jacobian determinant of transformation from cartesian to polar, Jacobian determinant of of the polar to cartesian while be its inverse?

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    $\begingroup$ $f_1 = x$, $f_2 = y$, $v_1 = r$, $v_2 = \theta$. So for example, $\partial f_1/ \partial v_1 = \partial x/\partial r = \cos\theta$ $\endgroup$ – John Martin Apr 6 '16 at 15:21
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When you use the change of variables formula in multiple integration, you are relating the familiar $x,y$ coordinates to the unfamiliar $u,v$ coordinates in some way. It is best to think about it by writing $x$ and $y$ as functions of $u,v$, that is write $x=x(u,v)$ and $y=y(u,v)$. The Jacobian Matrix of the Transformation is: $$\begin{bmatrix} \frac{\partial x}{\partial u} &&\frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u}&&\frac{\partial y}{\partial v} \end{bmatrix}.$$

Now suppose you do a double integral like this: $$\int_{-1}^{1}\int_{0}^{\sqrt{1-x^2}}1\ dy \ dx,$$ and you don't want to evaluate it directly. You would want to use a change of variables, and the best way is using polar coordinates. Let: $$x(r,\theta)=r\cos(\theta),y(r,\theta)=r\sin(\theta).$$ Now, when you do apply change of variables formula, you are transforming the double integral originally in the $x,y$ coordinates into a double integral in the $r,\theta$ coordinates. But how do we do that ? The first thing to ask is what does it mean to integrate over the top half of the unit disk in the $r,\theta$ plane ? The visual representation of polar coordinates tells us we must integrate over the rectangle with length $1$ in the positive $r$ direction and width $\pi$ in the positive $\theta$ direction. Now how does $dxdy$ transform into $drd\theta$ ? The change of variables formula says: $$\iint_{R}f(x,y)dxdy=\iint_{S}f(x(r,\theta),y(r,\theta))\left|\frac{\partial (x,y)}{\partial (r,\theta)}\right| drd\theta$$ where $R$ is the original region of integration and $S$ is the transformed region (after doing polar coordinates). When we compute $\left|\frac{\partial (x,y)}{\partial (r,\theta)}\right|$, the determinant of the Jacobian Matrix, we compute: $$\begin{vmatrix} \frac{\partial x}{\partial r} &&\frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r}&&\frac{\partial y}{\partial \theta} \end{vmatrix},$$ which turns out to be $r$ (Verify this). So putting everything together, our double integral becomes: $$\int_{0}^{\pi}\int_{0}^{1} r\ dr \ d\theta=\frac{\pi}{2}.$$

To reiterate, what we did was we transformed a double integral in the $x,y$ plane into a double integral in the $r,\theta$ plane by relating $x,y$ to $r,\theta$ explicitly.

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