SKKU-Calculus-Sec-15-7 Surface Integrals and Flux, SGLee+ 이원준 -- Sage

15.7 Surface Integrals and Flux by SGLee, 이원준

(Ref. http://math1.skku.ac.kr/home/pub/1884/ 2013.12.11)

Surface integrals of scalar fields

To find an explicit formula for the surface integral, we need to parameterize the surface of interest, S, by considering a system ofcurvilinear coordinates on S, like the latitude and longitude on a sphere. Let such a parameterization be x(s, t), where (s, t) varies in some region T in the plane. Then, the surface integral is given by

$\iint_{S} f \,dS = \iint_{T} f(\mathbf{x}(s, t)) \left\|{\partial \mathbf{x} \over \partial s}\times {\partial \mathbf{x} \over \partial t}\right\| ds\, dt$

where the expression between bars on the right-hand side is the magnitude of the cross product of the partial derivatives of x(s, t), and is known as the surface element.

Surface integrals of vector fields

A vector field on a surface

Consider a vector field v on S, that is, for each x in S, v(x) is a vector.

The surface integral can be defined component-wise according to the definition of the surface integral of a scalar field; the result is a vector. This applies for example in the expression of the electric field at some fixed point due to an electrically charged surface, or the gravity at some fixed point due to a sheet of material.

Alternatively, if we integrate the normal component of the vector field, the result is a scalar. Imagine that we have a fluid flowing through S, such that v(x) determines the velocity of the fluid at x. The flux is defined as the quantity of fluid flowing through S in unit amount of time.

This illustration implies that if the vector field is tangent to S at each point, then the flux is zero, because the fluid just flows inparallel to S, and neither in nor out. This also implies that if v does not just flow along S, that is, if v has both a tangential and a normal component, then only the normal component contributes to the flux. Based on this reasoning, to find the flux, we need to take the dot product of v with the unit surface normal to S at each point, which will give us a scalar field, and integrate the obtained field as above. We find the formula

$\iint_S {\mathbf v}\cdot \,d{\mathbf {S}} = \int_S {\mathbf v}\cdot {\mathbf n}\,dS=\iint_T {\mathbf v}(\mathbf{x}(s, t))\cdot \left({\partial \mathbf{x} \over \partial s}\times {\partial \mathbf{x} \over \partial t}\right) ds\, dt.$

The cross product on the right-hand side of this expression is a surface normal determined by the parametrization.

(Ref. http://en.wikipedia.org/wiki/Surface_integral 2013.12.11)

1.Evaluate the surface integral , where is the surface defined by the vector function , where , .

The graph of is called sphere.

Substituting , and into the integrand and we have

and

2. Evaluate , where is the surface with parametric equations , and , .

Sol) $X=(2u,u\mathrm{sin}v,u\mathrm{cos}v)$ 라 하고 $u,v$ 에 대해서 각각 미분한다.

So, $\frac{\partial X}{\partial u}=(2,\mathrm{sin}v,\mathrm{cos}v)$ and $\frac{\partial X}{\partial v}=(0,u\mathrm{cos}v,-u\mathrm{sin}v)$ .

=> = $\iint_A (u\mathrm{sin}v)(u\mathrm{cos}v)\left \| \frac{\partial X}{\partial u}\times \frac{\partial X}{\partial v} \right \| dudv$

=> $\iint_A (u\mathrm{sin}v)(u\mathrm{cos}v)\left \| \frac{\partial X}{\partial u}\times \frac{\partial X}{\partial v} \right \| dudv = \frac{\sqrt{5}\pi}{2}$ .