# 21. Multiple Integrals in Curvilinear Coordinates

## d. Integrating in 2D Curvilinear Coordinates

## 5. Diamond Shaped Examples

Below is an example of using curvilinear coordinates to compute an integral over a “diamond shaped” region.

Compute the integral \(\displaystyle \iint xy\,dA\) over the
diamond shaped region, \(R\), in the first quadrant bounded by

\(y=1+x^2\) and \(y=2+x^2\) in red
and
\(y=3-x^2\) and \(y=5-x^2\) in blue.

We look at the boundary curves to determine a useful coordinate system (\(u\) and \(v\)) for this problem. The first two curves, \(y=1+x^2\) and \(y=2+x^2\), can be written as: \[ y-x^2=1 \quad \text{and} \quad y-x^2=2 \] If we let \(u=y-x^2\), then the bounds for \(u\) become \[ u=1 \quad \text{and} \quad u=2 \] Similarly, the other two curves, \(y=3-x^2\) and \(y=5-x^2\), can be written as: \[ y+x^2=3 \quad \text{and} \quad y+x^2=5 \] If we let \(v=y+x^2\), then the bounds for \(v\) become \[ v=3 \quad \text{and} \quad v=5 \] Summarizing, \[ u=y-x^2 \qquad \text{and} \qquad v=y+x^2 \qquad \text{(*)} \] Next we need to find the Jacobian \(J=\left|\dfrac{\partial(x,y)}{\partial(u,v)}\right|\). This requires us to know \(x\) and \(y\) as functions of \(u\) and \(v\), but unfortunately, we know \(u\) and \(v\) as functions of \(x\) and \(y\). So we need to solve (*) for \(x\) and \(y\). There are two way to do this:

First, we can solve one equation for one variable and plug into the other. For example, if we solve the first equation for \(y=u+x^2\) then the second equation becomes \(v=u+2x^2\) or \(x^2=\dfrac{v-u}{2}\) or \(x=\sqrt{\dfrac{v-u}{2}}\). We plug this back into \(y=u+x^2\) to get \(y=u+\dfrac{v-u}{2}=\dfrac{v+u}{2}\).

Alternatively, we can add and subtract the equations. If we add, we get \(u+v=2y\) or \(y=\dfrac{v+u}{2}\). If we subtract, we get \(u-v=-2x^2\) or \(x=\sqrt{\dfrac{v-u}{2}}\).

By either method we get: (It's beneficial to split off the \(\sqrt{2}\) in the denominator since we are going to differentiate.) \[ x=\sqrt{\dfrac{v-u}{2}}=\dfrac{\sqrt{v-u}}{\sqrt{2}} \qquad \text{and} \qquad y=\dfrac{v+u}{2} \qquad \text{(**)} \]

Now we can compute the Jacobian: \[\begin{aligned} J&=\left|\begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial y}{\partial u} \\[8pt] \dfrac{\partial x}{\partial v} & \dfrac{\partial y}{\partial v} \end{vmatrix}\right| =\left|\begin{vmatrix} \dfrac{-1}{2\sqrt{2}\sqrt{v-u}} & \dfrac{1}{2} \\[8pt] \dfrac{1}{2\sqrt{2}\sqrt{v-u}} & \dfrac{1}{2} \end{vmatrix}\right| \\ &=\left|\dfrac{-1}{4\sqrt{2}\sqrt{v-u}}-\,\dfrac{1}{4\sqrt{2}\sqrt{v-u}}\right| =\left|\dfrac{-1}{2\sqrt{2}\sqrt{v-u}}\right|=\dfrac{1}{2\sqrt{2}\sqrt{v-u}} \end{aligned}\] We dropped the minus sign because a square root is always posiitve. The last thing we need is the integrand: \[ xy=\dfrac{\sqrt{v-u}(v+u)}{2\sqrt{2}} \] We can now compute the integral: \[\begin{aligned} I&=\iint xy\,dA =\iint xy \,J\,du\,dv \\ &=\int_3^5\int_1^2 \dfrac{\sqrt{v-u}(v+u)}{2\sqrt{2}}\dfrac{1}{2\sqrt{2}\sqrt{v-u}}\,du\,dv \\ &=\dfrac{1}{8}\int_3^5\int_1^2 (v+u)\,du\,dv \\ &=\dfrac{1}{8}\left[\dfrac{v^2}{2}\dfrac{}{}\right]_3^5\left[u\dfrac{}{}\right]_1^2 +\dfrac{1}{8}\left[v\dfrac{}{}\right]_3^5\left[\dfrac{u^2}{2}\right]_1^2 \\ &=\dfrac{1}{8}\left(\dfrac{25-9}{2}(1)+(2)\dfrac{4-1}{2}\right) =\dfrac{1}{8}\left(8+3\right)=\dfrac{11}{8} \end{aligned}\]

Notice how easy this integral was once we had the coordinate system and the Jacobian. Imagine doing it in rectangular coordinates. We would have to find all the intersection points, break the integral into \(3\) regions and integrate with bounds which were quadratic.

The next page on applications of 2D curvilinear coordinates has lots of additional examples and exercises using diamond shaped regions.

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