1. Section 17.7 Surface Integrals

2. Suppose f is a function of three variables whose domain includes the surface S. We divide S into patches S ij with area ΔS ij. We evaluate f at a point in each patch, multiply it by the area ΔS ij, and form the sum Then we take the limit as the patch size approaches 0 and define the surface integral of f over the surface S as THE SURFACE INTEGRAL

3. Suppose the surface S is a graph of a function of two variables g(x, y), where (x, y) are in some region D. Divide D into smaller rectangles R ij of equal size. The patch S ij lies directly above the rectangle R ij and the point in S ij is of the form. We use the approximation If f is continuous on S and g has continuous derivatives, then the definition on the previous slide becomes.

4. SUMMARY NOTE: Similar formulas apply when it is more convenient to project S onto the yz-plane or xz- plane. If z = g(x, y), then

5. 1.Evaluate the surface integral where S is the first octant portion of the plane 2x + y + 2z = 6. 2.Evaluate the surface integral where S is the first octant portion of the cylinder y 2 + z 2 = 9 between x = 0 and x = 4. EXAMPLES

6. Suppose a thin sheet has the shape of the surface S and that that ρ(x, y, z) is the density function for the sheet. Then the mass of the sheet is Its center of mass is MASS AND CENTER OF MASS

7. A cone-shaped surface lamina S is given by At each point on S, the density is proportional to the distance between the point and the z-axis. Find the mass m of the lamina. EXAMPLE

8. ORIENTED SURFACES We only want to consider two-sided, orientable surfaces, We start with a surface that has tangent planes at every point (x, y, z) on S. There are two unit normal vectors n 1 and n 2 = −n 1. If it is possible to choose a unit normal vector n at every point (x, y, z) so that n varies continuously over S, then S is called an oriented surface and the given choice of n provides S with an orientation. Thus, there are two possible orientations for any orientable surface which correspond to the choice for normal vectors.

9. For a surface z = g(x, y) given as the graph of g, we see that the induced orientation is given by the unit normal vector Since the k-component is positive, this gives the upward orientation of the surface.

10. CLOSED SURFACES AND ORIENTATION A closed surface is a surface the is the boundary of a (closed) solid region E. The positive orientation of a closed surface is the one for which the normal vectors point outward from the region E.

11. FLUX INTEGRAL Suppose that S is an oriented surface with unit normal vector n, and imagine a fluid with density ρ(x, y, z) and velocity field v(x, y, z) flowing through S. Then the rate of flow (mass per unit time) per unit area is ρv. If we divide S into small patches S ij, then S ij is nearly planar and so we can approximate the mass of fluid crossing S ij in the direction of the normal n per unit time by the quantity (ρv · n)A(S ij ) where ρ, v, and n are evaluated at some point on S ij. By summing these quantities and taking the limit we get the surface integral of the function ρv · n over S: and this is interpreted physically as the rate of flow through S.

12. If we write F = ρv, then F is also a vector field on and the integral on the previous slide becomes A surface integral of this form occurs frequently in physics, even when F is not ρv, and is called the surface integral (or flux integral) of F over S. FLUX INTEGRAL (CONTINUED)

13. SURFACE INTEGRAL OF A VECTOR FIELD If F is a continuous vector field defined on an oriented surface S with unit normal vector n, then the surface integral of F over S is This integral is also called the flux of F across S. In words, this definition says that the surface integral of a vector field over S is equal to the surface integral of its normal component over S.

14. EXAMPLE Let S be the portion of paraboloid z = g (x, y) = 4 − x 2 − y 2 lying above the xy-plane. Find the upward flux of F(x, y, z) = xi + yj + zk across the surface S.

15. In the case of a surface S given by a graph z = g(x, y), we can find n by noting that S is also the level surface f (x, y, z) = z − g(x, y) = 0. We know that the gradient is normal to this surface at (x, y, z) and so a unit normal vector is Using the formula for the surface integral from Slide 4 to evaluate the flux integral with F(x, y, z) = P(x, y, z)i + Q(x, y, z)j + R(x, y, z)k we get the following:

17. EXAMPLE Evaluate the flux for the vector field F(x, y, z) = xi + yj + zk across S which is the first octant portion of the sphere x 2 + y 2 + z 2 = 16, taking n to be the upward normal.