figure 1

Figure 1 shows a surface S lying above its "shadow" region R in the plane beneath it. The surface is defined by the equation f (x, y, z) = c. If the surface is smooth (Ñ f is continuous and never vanishes on S), then we can define and calculate its area as a double integral over R. First partition the region R into small rectangles D A _k. (See figure 2) Directly above each D A _k lies a patch of the surface D s _k that we may approximate with a portion of D p _k of the tangent plane. Suppose D p _k is a portion of the plane that is tangent to the surface at the point T_k (x _k, y _k, z _k) directly

above the back corner c _k of D A _k. If the tangent plane is parallel to R, then D p _k will be congruent to D A _k. If it is not, then it will be a parallelogram whose area is somewhat larger than the area of D A _k.

This view of D s _k and D p _k (See figure 3) showing the gradient vector Ñ f (x _k, y _k, z _k) at T_k and a unit vector p that is normal to R. This figure also shows the angle g _k between Ñ f and p. There are also two other vectors in this picture, and they are u _k and v _k that lie along the edges of the patch D p _k in the

figure 2

tangent plane. Therefore u _k ´ v _k and Ñ f are normal to the tangent plane. Using a fact from

figure 3

advanced vector geometry, | (u _k ´ v _k)· p| equals the area of the projection of the parallelogram determined by u _k and v _k onto any plane whose normal is p. That implies | (u _k ´ v _k)· p| = D A _k and | u _k ´ v _k | is the area of D p _k. Therefore | u _k ´ v _k | | p | | cos (angle between u _k ´ v _k and p)| = D A _k. Since Ñ f and u _k ´ v _k are perpendicular to the tangent plane, then | cos (angle between u _k ´ v _k and p)| = |cos g _k |. D p _k | cos g _k | = D A _k implies

provided cos g _k ¹ 0. We will have cos g _k ¹ 0 as long as Ñ f is not parallel to the ground plane and Ñ f · p ¹ 0. Since the patches D p _k approximate the surface patches D s _k that fit together to make S, then

As the partition becomes finer, this sum will become the double integral

Now how do we handle the integrand of this integral? For any surface f (x, y, z) = c, we will have | Ñ f · p | = | Ñ f || p ||cos g | or

where p is the unit normal vector to R and Ñ f · p ¹ 0. Therefore,

EXAMPLE 1:

Find the area of the surface cut from the paraboloid x ² + y ² - z = 0 by the plane x = 2.

SOLUTION: x ² + y ² - z = 0 ® z = x ² + y ² and its shadow is in the xy-plane, so let p = k. Now find the gradient vector of f = x ² + y ² - z and its length.

Now find Ñ f · p. Ñ f · p = (-1)(1) = -1, so | Ñ f · p | = 1. Therefore the surface area is

The region of integration is the circle x ² + y ² = 2 (z = 2), so I will use polar coordinates.

EXAMPLE 2:

Find the area of the region cut from the plane x + 2y + 2z = 5 by the cylinder whose walls are x = y ² and x = 2 - y ².

SOLUTION: Since the area of the region is cut from the plane by the curves x = y ² and x = 2 - y ², then the shadow is in the xy-plane and p = k. Now find the gradient of f = x + 2y + 2z - 5 and its length.

Now find Ñ f · p. Ñ f · p = 2.

Therefore, the surface area is

Here is the region of integration R. The upper curve is x = 2 - y ² and the lower curve is x = y ².

So the bounds of integration are x goes from y ² to 2 - y ² and y goes from -1 to 1.

EXAMPLE 3:

Find the area of the surface cut from the paraboloid x ² + y + z ² = 2 by the plane y = 0.

SOLUTION: x ² + y + z ² = 2 ® y = 2 - x ² - y ² and its shadow is in the xz-plane, so let p = j. Now find the gradient of f = x ² + y + z ² - 2 and its length.

When y = 0, the region in the xz-plane is the circle x ² + z ² = 2, so I am going to use polar coordinates.

Now we will learn how to integrate a function over a surface. Suppose we have an electrical charge distributed over a surface f (x, y, z) = c. Let g (x, y, z) be the function for the charge density, and what we want to calculate the total charge on S. if we divide the shadow region up like we did before, and determine D A _k, D p _k, and D s _k, we will get

As the partitions become finer, the surface integral becomes

EXAMPLE 4:

Integrate g (x, y, z) = x + y + z over the surface of the cube cut from the first octant by the planes x = a, y = a, z = a.

A cube has 6 sides, therefore we will have 6 surfaces to integrate over.

SIDE 1: x = a and shadow in yz-plane, therefore p = i. f = x - a, Ñ f = i, Ñ f · p = 1, and | Ñ f | = 1. g = x + y + z = y + z + a

SIDE 2: y = a and shadow in xz-plane, therefore p = j. f = y - a, Ñ f = j, Ñ f · p = 1, and the length of Ñ f is 1. g = x + y + z = x + z + a.

SIDE 3: z = a and shadow in xy-plane, therefore p = k. f = z - a, Ñ f = k, Ñ f · p = 1, and the length of Ñ f is 1. g = x + y + z = x + y + a.

SIDE 4: y = 0 and shadow in xz-plane, therefore p = j. f = y, Ñ f = j, Ñ f · p = 1, and the length of Ñ f is 1. g = x + y + z = x + z.

SIDE 5: z = 0 and shadow in xy-plane, therefore p = k. f = z, Ñ f = k, Ñ f · p = 1, and the length of Ñ f is 1. g = x + y + z = x + y.

SIDE 6: x = 0 and shadow in yz-plane, therefore p = i. f = x, Ñ f = i, Ñ f · p = 1, and the length of Ñ f is 1. g = x + y + z = y + z.

Surface Integral = side 1 + side 2 + side 3 + side 4 + side 5 + side 6 = 2a ³ + 2a ³ + 2a ³ + a ³ + a ³ + a ³ = 9a ³

EXAMPLE 5:

Integrate g (x, y, z) = xy + 1 over the surface S, where S is the part of the paraboloid z = x ² + y ² that lies inside the cylinder x ² + y ² = 4.

figure 6

The surface that we are integrating over is the paraboloid z = x ² + y ² that is inside the cylinder x ² + y ² = 4. It is the same as the lower part of the paraboloid that is cut by the plane z = 4. (See figure 6) So we will only have one surface to work with. The shadow is in the xy-plane, so p = k.

Since the region that we will be integrating over is a circle, I will use polar coordinates.

We will call a smooth surface S orientable (or two-sided) if it is possible to define a field n of unit normal vectors on S that varies continuously with position. If a surface is orientable, then so will any patch or sub-portion of it. In fact, spheres and other smooth closed surfaces are orientable. The key here is that we have to pick n. Once we have done that, then we have oriented the surface. Combining the surface together with its normal field, we will say that we have an oriented surface. An example of a surface that is not orientable is the mobius strip because it only has one side.

Suppose that F is a continuous vector field defined over an oriented surface S and n is the chosen unit normal field on the surface. Then we can define the flux across S in the positive direction as the integral of F · n over S.

DEFINITION:

The flux of a three-dimensional vector field F across an oriented surface S in the direction of n is given by the formula

If F is the velocity field of a three-dimensional fluid flow, then the flux of F across S is the net rate at which fluid is crossing S in the chosen positive direction. If S is the part of a level surface g (x, y, z) = c, then n may be taken to be one of the two fields-

depending on which one gives the preferred direction, therefore

EXAMPLE 6:

Find the outward flux of the vector field F = x i - y j across the closed surface S which is the first octant part of the pyramid bounded by the coordinate planes and the plane 2x + 3y + 4z = 12.

We want to find the outward flux over the first octant part of the pyramid defined by 2x + 3y + 4z = 12, so g (x, y, z) = 2x + 3y + 4z - 12 = 0. The shadow of this region is in the xy-plane, so p = k.

We want the outward point normal, so

We will be integrating over the region in the xy-plane, so z = 0, which implies 2x + 3y = 12 or y = 4 - (2/3) x.

EXAMPLE 7:

Find the flux of the field F = y i - x j + k across the portion of the sphere x ² + y ² + z ² = a ² in the first octant in the direction away from the origin.

SOLUTION: g (x, y, z) = x ² + y ² + z ² - a ² and Ñ g = 2x i + 2y j + 2z k.

Since we are moving away from the origin, then

The shadow of this surface is in the xy-plane, so p = k.

Now find F · n.

The shadow in the xy-plane is the quarter circle defined by x ² + y ² = a ². Now converting this region to polar, r = a and 0 £ q £ p /2.