Calculating rotation integral

Question

How do I calculate integral: $\int_{F}^{} \! rot a\cdot \, dn $ if it is given vector field: $a=[−2x−z,−y+z,−2x+z]^T$, sphere equation $x^2+y^2+z^2=9$ and ground plane $z=0$

I was thinking to calculate normal vector and express $a$ with some parameters, and then to multply a with normal vector. But I dont know how excatly should I represent $a$ and what to do next?

Edit: Correct solution is $0$.

Answer 1

In English, the term rotor has fallen out of use. Instead we use curl because the short-hand rot has an alternative meaning (to decompose in a smelly way).

I shall make use of the nabla operator. In this way, our expressions can be 'all mathsy'.

$$\mathbf{rot}\left(\vec{a}\right)\equiv\mathbf{curl}\left(\vec{a}\right)\equiv\vec{\nabla}\times \vec{a}$$

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Normal vectors

Assume $\vec{a}$ is expressed with respect to the standard basis. Proceed by calculating the normal vector $\vec{n_1}$ to the sphere. You could eye-ball it, or calculate the gradient of the surface $F$:

$$\vec{n_1}=\mathbf{grad}(x^2+y^2+z^2-9)=\vec{\nabla}(x^2+y^2+z^2-9)=2\begin{bmatrix}x \\y\\z\end{bmatrix}.$$

The exterior normal vector for the plane $z=0$ is $\vec{n_2}$.

$$\vec{n_2}=\begin{bmatrix}0 \\0\\-1\end{bmatrix}$$

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Express surface integral as double integral

Now

$$I\equiv\int\int_F\left(\vec{\nabla}\times \vec{a}\right)\cdot\mathrm{d}\vec{F}=\int\int_{F_1}\left(\vec{\nabla}\times \vec{a}\right)\cdot\hat{n}_1\ \mathrm{d}F_1+\int\int_{F_2}\left(\vec{\nabla}\times \vec{a}\right)\cdot\hat{n}_2\ \mathrm{d}F_2$$

where the 'hat' denotes a unit vector. We will express this surface integral with Cartesian differentials as a double integral,

$$\mathrm{d}F_i=\left|\vec{n_i}\right|\mathrm{d}x\mathrm{d}y$$

from which

$$I=\int\int_{D_1}\left(\vec{\nabla}\times \vec{a}\right)\cdot\vec{n_1}\ \mathrm{d}x\ \mathrm{d}y+\int\int_{D_2}\left(\vec{\nabla}\times \vec{a}\right)\cdot\vec{n_2}\ \mathrm{d}x\ \mathrm{d}y.$$

The projections $D_1,D_2$ on the $xy$-plane are equal, denote them both with $D$. Let us find the curl.

$$\vec{\nabla}\times \vec{a}=\begin{bmatrix}-1 \\1\\0\end{bmatrix}$$

Immediately spot that the second dot product is zero, hence also the second double integral.

$$\left(\vec{\nabla}\times \vec{a}\right)\cdot\vec{n_2}=\begin{bmatrix}-1 \\1\\0\end{bmatrix}\cdot \begin{bmatrix}0 \\0\\-1\end{bmatrix}=0\implies\int\int_{D}\left(\vec{\nabla}\times \vec{a}\right)\cdot\vec{n_2}\ \mathrm{d}x\ \mathrm{d}y=0$$

So we need to evaluate

$$I=\int\int_{D}\left(\vec{\nabla}\times \vec{a}\right)\cdot\vec{n_1}\ \mathrm{d}x\ \mathrm{d}y.$$

Similarly

$$\left(\vec{\nabla}\times \vec{a}\right)\cdot\vec{n_1}=\begin{bmatrix}-1 \\1\\0\end{bmatrix}\cdot 2\begin{bmatrix}x \\y\\z\end{bmatrix}=2(y-x).$$

We are left to calculate the standard double integral:

$$I=2\int\int_D(y-x)\ \mathrm{d}x\ \mathrm{d}y.$$

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Compute double integral

In the spirit of your other question here, try this first on your own.

The projection of the sphere in the $xy$-plane (where $z=0$) is a disc. Therefore,

$$D=\{(x,y)\ |\ \text{all points on and inside the circle}\ x^2+y^2=9\}.$$

In polar coordinates,

$$D=\{(r,\theta)\ |\ 0\leq r\leq3,\ 0\leq \theta\leq2\pi\}.$$

In conclusion,

$$I=2\int_0^3\int_0^{2\pi}r(\sin\theta-\cos\theta)\ r\ \mathrm{d}\theta\ \mathrm{d}r={0}.$$

Alternatively, you could see directly via a symmetry argument

$$I=2\int\int_D(y-x)\ \mathrm{d}x\ \mathrm{d}y=0.$$