## Monday, June 4, 2007

### More Integrals... *whine*. Topic: Calculus.

Definition: (Jacobian) If $T$ is the transformation from the $uv$-plane to the $xy$-plane defined by the equations $x = x(u, v)$ and $y = y(u, v)$, then the Jacobian of $T$ is denoted by $J(u, v)$ or by $\partial(x, y)/\partial(u, v)$ and is defined by

$J(u, v) = \frac{\partial(x, y)}{\partial(u, v)} = \frac{\partial x}{\partial u} \cdot \frac{\partial y}{\partial v} - \frac{\partial y}{\partial u} \cdot \frac{\partial x}{\partial v}$,

i.e. the determinant of the matrix of the partial derivatives (also known as the Jacobian matrix). Naturally, this can be generalized to more variables.

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Theorem: If the transformation $x = x(u, v)$, $y = y(u, v)$ maps the region $S$ in the $uv$-plane into the region $R$ in the $xy$-plane, and if the Jacobian $\partial(x, y)/\partial(u, v)$ is nonzero and does not change sign on $S$, then (with appropriate restrictions on the transformation and the regions) it follows that

$\displaystyle \int_R \int f(x, y) dA_{xy} = \int_S \int f(x(u, v), y(u, v)) \left|\frac{\partial(x, y)}{\partial(u, v)} \right| dA_{uv}$.

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Problem: Evaluate $\displaystyle \int_R \int e^{(y-x)/(y+x)} dA$, where $R$ is the region in the first quadrant enclosed by the trapezoid with vertices $(0, 1); (1, 0); (0, 4); (4, 0)$.

Solution: The bounding lines can be written as $x = 0$, $y = 0$, $y = -x+1$, and $y = -x+4$. Now consider the transformation $u = y+x$ and $v = y-x$. In the $uv$-plane, the bounding lines of the new region $S$ can now be written as $u = 1$, $u = 4$, $v = u$, and $v = -u$.

We can write $x$ and $y$ as functions of $u$ and $v$: simply $x = \frac{u-v}{2}$ and $y = \frac{u+v}{2}$. So the Jacobian $\displaystyle \frac{\partial(x, y)}{\partial(u, v)} = \frac{\partial x}{\partial u} \cdot \frac{\partial y}{\partial v} - \frac{\partial y}{\partial u} \cdot \frac{\partial x}{\partial v} = \frac{1}{2} \cdot \frac{1}{2} - \frac{1}{2} \cdot \left(-\frac{1}{2} \right) = \frac{1}{2}$.

Then our original integral becomes $\displaystyle \int_R \int e^{(y-x)/(y+x)} dA = \frac{1}{2} \int_S \int e^{v/u} dA$. And this is equivalent to

$\displaystyle \frac{1}{2} \int_S \int e^{v/u} dA = \frac{1}{2} \int_1^4 \int_{-u}^u e^{v/u} dv du = \frac{1}{2} \int_1^4 \big[ u e^{v/u} \big]_{v=-u}^u du = \frac{1}{2} \int_1^4 u\left(e-\frac{1}{e}\right) du = \frac{15}{4}\left(e-\frac{1}{e}\right)$.

QED.

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Comment: Note that the above theorem is probably very important in multivariable calculus, as it is the equivalent to $u$-substitution in one variable, which we all know is the ultimate integration technique. It functions in the same way, giving you a lot more flexibility on the function you are integrating and the region you are integrating on.

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Practice Problem: Evaluate $\displaystyle \int_R \int (x^2-y^2) dA$, where $R$ is the rectangular region enclosed by the lines $y = -x$, $y = 1-x$, $y = x$, $y = x+2$.