Question

part d only

Let S be the surface of the sphere r-+92 + ~2-R-with positive orientation. Let F : IR: → IR3 be a vector field and Fr-F.r its radial component (where r denotes the unit vector in the direction of r). a) Show that F,sin(o) do de 0 0 in case of R - 1 b) Find a corresponding formula for arbitrary values of R c) Find a corresponding formula for real-valued functions f : R → and arbi- trary values of R d) Find corresponding formulas for integration over the surface of the cylinder

Answer 1

d) We have to find the formulas for the surface integral of a scalar field and a vector field over a cylinder.

We have formulas to evaluate the surface integrals over any surface, provided we can find a parametrisation of the surface with 2 parameters. So we use this general formula and substitute the parametrisation for the cylinder we want, to find the surface integrals.

To do this, we first need to parametrise the cylinder of radius R. A cylinder of radius R can be thought of as a stack of circles of radius R. So, the x and y coordinates are parametrised as for a circle x = R cos t , y = R sin t, and the z coordinate kept the same.

In vector notation, this is   $\vec{r}(z,t) = R \cos(t) \vec{i} + R \sin (t) \vec{j} + z \vec{k}$ , where $0 \leq t \leq 2 \pi, a \leq z \leq b$ .

We find the partial derivatives of r with respect to t and z (the parameters).

$\vec{r_t} = -R \sin (t) \vec{i} + R \cos (t) \vec{j} \\ \vec{r_z} = \vec{k}$

We also need the cross product of the above 2 vectors, and the magnitude of the cross product.

$\vec{r_z} \times \vec{r_t} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 0& 0 & 1 \\ -R \sin (t) & R \cos (t) & 0 \end{vmatrix} = -R \cos (t) \vec{i} - R \sin (t) \vec{j}$

$\left \| \vec{r_z} \times \vec{r_t} \right \| = \sqrt{R^2 \cos^2 t + R^2 \sin^2 t} = R$

With this parametrisation and the above data, it is easy to find the formulas for surface integrals.

1. For a scalar field (real valued function), the general formula is

$\int_S \int f(x,y,z) \mathbf{dS} = \int_D \int f(\vec{r} (z,t)) \left \| \vec{r_z} \times \vec{r_t} \right \| dA$ , where D is the range of z and t. So, if we substitute our parametrisation, we find

$\int_S \int f(x,y,z) \mathbf{dS} = \int_{a}^b \int_0^{2 \pi} f(R\cos t, R \sin t, z) R \quad dt dz$

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2. For a vector field  F , $\int_S \int \mathbf{F} d\mathbf{S} = \int_S \int \mathbf{F} \cdot \vec{n} d\mathbf{S}$ , where n is the unit normal vector to the surface. So if we find F.n, which is a scalar, then the integral can be evaluated as for a scalar field.

Since the surface is parametrised by $\vec{r}(z,t)$ , the normal to this surface is simply $\vec{r_z} \times \vec{r_t}$ . If this vector is not the positive orientation, we just need to multiply it by -1 to get the correct orientation.

We find the vector n.

$\vec{n} = \frac{\vec{r_z} \times \vec{r_t}}{\left \| \vec{r_z} \times \vec{r_t} \right \|} = \frac{-R \cos t \vec{i} - R \sin t \vec{j}}{R} = -\cos( t) \vec{i} - \sin (t)\vec{j}$

If the vector field F is of the form

$\mathbf{F}(x,y,z) = F_1 \vec{i} + F_2 \vec{j} + F_3 \vec{k}, \\ \mathbf{F} \cdot \vec{n} = -F_1 \cos (t) - F_2 \sin (t)$

Now our integral is  

$\int_S \int \mathbf{F} d\mathbf{S} = \int_S \int \mathbf{F} \cdot \vec{n} d\mathbf{S} = \int \int_S (-F_1 \cos (t) - F_2 \sin (t)) d\mathbf{S}$

Here, both F₁ and F₂ are real valued functions of x,y,z. So, we can evaluate this integral as we found above. Letting $f(x,y,z) = \mathbf{F} \cdot \vec{n}$ in 1, we can find the formula for the surface integral as

$\int_S \int \mathbf{F} d\mathbf{S} = \int \int_S (-F_1 \cos (t) - F_2 \sin (t)) d\mathbf{S} \\ = \int_a^b \int_0^{2 \pi} (-F_1 (R \cos t , R \sin t, z) \cos (t) - F_2 (R \cos t, R \sin t, z) \sin (t)) R dt dz$