Given a vector field \(\bar{F}\) with unit normal vector \(\hat{n}\), then the surface integral of \(\bar{F}\) over the surface \(S\) is given by:
\[ \int\limits_{S} \bar{F} \cdot d\bar{S} = \iint\limits_{S} \bar{F} \cdot \hat{n} \, dS \]where the right-hand side represents a surface integral, also called the flux of \(\bar{F}\) across \(S\).
Cartesian Form
Let \(\bar{F} = F_1\hat{i} + F_2\hat{j} + F_3\hat{k}\) where \(F_1, F_2,\) and \(F_3\) are continuous and differentiable functions of \(x, y, z\), then:
\[ \iint\limits_{S} \bar{F} \cdot \hat{n} \, dS = \iint\limits_{S} F_1 dy dz + F_2 dz dx + F_3 dx dy \]Projection Method
Let \(R_1\) be the projection of \(S\) on the \(xy\)-plane, then:
\[ \iint\limits_{S} \bar{F} \cdot \hat{n} \, dS = \iint\limits_{R_1} \frac{\bar{F} \cdot \hat{n}}{|\hat{n} \cdot \hat{k}|} \, dx dy \]Similarly for other coordinate planes:
\[ \iint\limits_{S} \bar{F} \cdot \hat{n} \, dS = \iint\limits_{R_2} \frac{\bar{F} \cdot \hat{n}}{|\hat{n} \cdot \hat{i}|} \, dy dz \] \[ \iint\limits_{S} \bar{F} \cdot \hat{n} \, dS = \iint\limits_{R_3} \frac{\bar{F} \cdot \hat{n}}{|\hat{n} \cdot \hat{j}|} \, dz dx \]where \(R_2\) and \(R_3\) are the projections of \(S\) on the \(yz\)- and \(zx\)-planes respectively.
Surface Area of a Curved Surface
Let \(S\) be a surface represented by the equation \(f(x,y,z) = 0\), then the unit normal to the surface \(S\) is given by:
\[ \hat{n} = \frac{\nabla f}{\|\nabla f\|} = \frac{f_x \hat{i} + f_y \hat{j} + f_z \hat{k}}{\sqrt{f_x^2 + f_y^2 + f_z^2}} \]where \(f_x, f_y, f_z\) are the partial derivatives of \(f\) with respect to \(x, y, z\) respectively.
Important Notes:
- The surface integral represents the flux of the vector field through the surface
- The choice of projection plane (\(xy\), \(yz\), or \(zx\)) depends on which gives the simplest integral
- The unit normal vector \(\hat{n}\) is crucial for correct flux calculation
- For closed surfaces, the convention is to take the outward normal
Example 1: Evaluate \(\iint\limits_{S} \bar{F} \cdot \hat{n} \, dS\) where: \[ \bar{F} = z\hat{i} + x\hat{j} - 3y^{2}z\hat{k} \] and \(S\) is the surface of the cylinder \(x^{2} + y^{2} = 16\) in the first octant between \(z = 0\) and \(z = 5\).
Solution:
Given the cylinder surface \(S: x^{2} + y^{2} = 16\), we define:
\[ f(x,y,z) = x^{2} + y^{2} - 16 = 0 \]The unit normal vector is:
\[ \hat{n} = \frac{\nabla f}{\|\nabla f\|} = \frac{2x\hat{i} + 2y\hat{j}}{\sqrt{4x^{2} + 4y^{2}}} = \frac{x\hat{i} + y\hat{j}}{4} \](since \(x^{2} + y^{2} = 16\) on \(S\))
We project \(S\) onto the \(yz\)-plane (rectangle OBDE in first octant):
\[ \iint\limits_{S} \bar{F} \cdot \hat{n} \, dS = \iint\limits_{R} \frac{\bar{F} \cdot \hat{n}}{|\hat{n} \cdot \hat{i}|} \, dy dz \] \[ \bar{F} \cdot \hat{n} = (z\hat{i} + x\hat{j} - 3y^{2}z\hat{k}) \cdot \frac{x\hat{i} + y\hat{j}}{4} = \frac{xz + xy}{4} \] \[ \hat{n} \cdot \hat{i} = \frac{x}{4} \] \[ \iint\limits_{S} \bar{F} \cdot \hat{n} \, dS = \iint\limits_{R} \frac{\frac{xz + xy}{4}}{\frac{x}{4}} \, dy dz = \iint\limits_{R} (y + z) \, dy dz \]For the first octant portion of the cylinder, \(y\) ranges from 0 to 4 and \(z\) from 0 to 5.
\[ \int_{z=0}^{5} \int_{y=0}^{4} (y + z) \, dy dz = \int_{z=0}^{5} \left[ \frac{y^{2}}{2} + yz \right]_{0}^{4} dz \] \[ = \int_{z=0}^{5} (8 + 4z) \, dz = \left[ 8z + 2z^{2} \right]_{0}^{5} = 40 + 50 = 90 \] \[ \iint\limits_{S} \bar{F} \cdot \hat{n} \, dS = \boxed{90} \]Note on Projection Choice
The \(yz\)-plane projection was chosen because it simplifies the integral by eliminating the \(x\) dependence (since \(x = \sqrt{16-y^2}\) on the cylinder surface). Other projections would require more complex integration.
Example 2: Evaluate \(\iint\limits_{S} \bar{F} \cdot \hat{n} \, dS\) where: \[ \bar{F} = (x + y^2)\hat{i} - 2x\hat{j} + 2yz\hat{k} \] and \(S\) is the surface of the plane \(2x + y + 2z = 6\) in the first octant.
Solution: For the plane \(2x + y + 2z = 6\), the normal vector is:
\[ \nabla f = 2\hat{i} + \hat{j} + 2\hat{k} \] \[ \hat{n} = \frac{2\hat{i} + \hat{j} + 2\hat{k}}{\sqrt{4 + 1 + 4}} = \frac{2}{3}\hat{i} + \frac{1}{3}\hat{j} + \frac{2}{3}\hat{k} \] \[ \bar{F} \cdot \hat{n} = (x + y^2)\left(\frac{2}{3}\right) - 2x\left(\frac{1}{3}\right) + 2yz\left(\frac{2}{3}\right) \] \[ = \frac{2}{3}x + \frac{2}{3}y^2 - \frac{2}{3}x + \frac{4}{3}yz = \frac{2}{3}y^2 + \frac{4}{3}yz \]The plane intersects the axes at (3,0,0), (0,6,0), and (0,0,3). Projecting onto xy-plane:
\[ \iint\limits_{S} \bar{F} \cdot \hat{n} \, dS = \iint\limits_{R} \frac{\frac{2}{3}y^2 + \frac{4}{3}yz}{\frac{2}{3}} \, dx dy = \iint\limits_{R} (y^2 + 2yz) \, dx dy \]Substitute \(z = \frac{6 - 2x - y}{2}\):
\[ = \iint\limits_{R} \left[y^2 + y(6 - 2x - y)\right] \, dx dy = \iint\limits_{R} (6y - 2xy) \, dx dy \]Limits: \(x\) from 0 to 3, \(y\) from 0 to \(6-2x\):
\[ \int_{x=0}^{3} \int_{y=0}^{6-2x} (6y - 2xy) \, dy dx \] \[ = \int_{x=0}^{3} \left[3y^2 - xy^2\right]_{0}^{6-2x} dx = \int_{x=0}^{3} [3(6-2x)^2 - x(6-2x)^2] dx \] \[ = \int_{0}^{3} (3 - x)(6 - 2x)^2 dx = \cdots = 81 \] \[ \iint\limits_{S} \bar{F} \cdot \hat{n} \, dS = \boxed{81} \]Example 3: Evaluate \(\iint\limits_{S} \bar{F} \cdot \hat{n} \, dS\) where \(\bar{F} = z\hat{i} + x\hat{j} - 3y^2z\hat{k}\) and \(S\) is the surface of the cylinder \(x^2 + y^2 = 1\) in the first octant between \(z = 0\) and \(z = 2\).
Solution
For the cylinder \(x^2 + y^2 = 1\), the normal vector is \(\hat{n} = x\hat{i} + y\hat{j}\).
\[\bar{F} \cdot \hat{n} = zx + xy\].
The integral becomes \(\iint\limits_{R} (z + y) \, dy dz\), where \(x = \sqrt{1 - y^2}\), \(y\) from 0 to 1, \(z\) from 0 to 2: \[\int_{y=0}^{1} \int_{z=0}^{2} (z + y) \, dz dy\].
\[\int_{y=0}^{1} \left[\frac{z^2}{2} + yz\right]_{0}^{2} dy = \int_{y=0}^{1} (2 + 2y) dy = \left[2y + y^2\right]_{0}^{1} = 3\].
\[\iint\limits_{S} \bar{F} \cdot \hat{n} \, dS = \boxed{3}\].