Calc 3 Lecture Notes Section 14.6 Page 7 of 7

Section 14.6: Surface Integrals

Big idea: A surface integral is used to add up how a quantity varies over a surface.

Big skill: You should be able to evaluate surface integrals using the evaluation theorem.

Practice:

  1. An arched dome is in the shape of a solid of revolution of an inverted catenary of height 15 feet and radius 30 feet. The shape is obtained by taking the equation and rotating it around the z axis. The constant k is given by: . Compute the weight of the dome if its surface area density decreases linearly with height from 100 lb/ft2 at the base to 40 lb/ft2 at the top.


Definition 6.1: Surface Integral

The surface integral of a function g(x, y, z) over a surface S Ì R3, written as , is defined by

,

provided the limit exists and is the same for all choices of evaluation points .

Notice that we have to do two things to evaluate this integral:

·  Write g(x, y, z) as a function of two variables, since a double integral involves only two variables.

·  Write dS, the differential element of surface area on the surface S, in terms of dA, the differential element of surface area in the x-y plane over which the double integration is performed.

In section 13.4, we found the differential surface element dS for a surface S defined by z = f(x, y) to be . Here is a review of the derivation:

To compute the area of a surface defined by z = f(x, y), subdivide the surface into small areas Ti (DSi, = Ti), each of which lies above some area Ri of area DAi = (Dxi)(Dyi) in the region of integration, as shown in the two pictures below.


To compute the area of the parallelogram Ti, find the vectors ai and bi, which form the edges of the parallelogram Ti:

The area of Ti is the magnitude of the cross-product of the vectors ai and bi:

Thus,

In this section, the book makes the point that is normal to the plane Ti, and so we can define a normal vector to that small area element . The upshot of this approach is the we can define dS as: , which will be useful for thinking about parametrically defined surfaces. This expression simplifies to.

Theorem 6.1: Evaluation Theorem for Surface Integrals

If a surface S is given by z = f(x, y) for (x, y) in the region R Ì R2, and f has continuous first partial derivatives, then


Practice:

  1. Compute the surface integral over the surface defined by that is over the region bounded by the curves x = y2 and y = 2 – x in the x-y plane.

  1. Set up and evaluate a surface integral for the dome problem from the first page using the evaluation theorem. The equation that describes the inner surface in x-y coordinates is , where k is still: . Recall the surface area density decreases linearly with height from 100 lb/ft2 to 40 lb/ft2, which means .


Parametric Representation of Surfaces

For a surface defined parametrically as , define

and .

ru and rv will be in the tangent plane at any point, which means .

Practice:

  1. Repeat the dome example starting with the equation in cylindrical coordinates. è è .


Definition 6.2: Flux

Let be a continuous vector field defined on an oriented surface S with unit normal vector n. The surface integral of F over S (or the flux of S over S) is given by

Practice:

  1. Show that the flux of the unit vector field over the hemisphere of radius 1 above the x-y plane is equal to the flux through a circle of radius 1 in the x-y plane.