Surface Integrals of Scalar Fields: Surface Area & Basic Applications

The surface is the unit disk in the xy-plane, defined by z=0 and x² + y² ≤ 1. For a surface given by z=g(x,y)=0, the surface area element is dS = √( (∂z/∂x)² + (∂z/∂y)² + 1 ) dA. Since ∂z/∂x=0 and ∂z/∂y=0, this simplifies to dS = √(0+0+1) dA = 1 dA. The scalar function is f(x,y,z)=2. The surface integral becomes ∫∫ₛ 2 dS = ∫∫_D 2 * (1) dA, where D is the unit disk (x²+y² ≤ 1). This is 2 times the area of D. The area of the unit disk is π(1)² = π. Therefore, the integral equals 2 * π = 2π.

For a surface given by z = g(x,y) = 4 - 2x - y, we compute the partial derivatives: ∂z/∂x = -2 and ∂z/∂y = -1. The formula for the surface area element is dS = √( (∂z/∂x)² + (∂z/∂y)² + 1 ) dA. Plugging in the values, we get √( (-2)² + (-1)² + 1 ) = √(4 + 1 + 1) = √(6). Therefore, dS = √(6) dA.

The surface is given by z = g(x,y) = x² + y². The partial derivatives are ∂z/∂x = 2x and ∂z/∂y = 2y. The surface area element is dS = √( (∂z/∂x)² + (∂z/∂y)² + 1 ) dA = √( (2x)² + (2y)² + 1 ) dA = √(4x² + 4y² + 1) dA. The function to integrate over the surface is f(x,y,z) = x² + y². Since on the surface, z = x² + y², we have f(x,y,z) = x² + y². Therefore, the surface integral becomes the double integral over the projection D (the disk x² + y² ≤ 4) of [ (x² + y²) * √(4x² + 4y² + 1) ] dA.

When the scalar function f(x, y, z) is identically equal to 1, the surface integral ∫∫ₛ 1 dS simplifies to ∫∫ₛ dS. This integral sums up the infinitesimal surface area elements dS over the entire surface S. The total sum of these area elements is precisely the total surface area of S. Therefore, the integral gives the area of the surface S.

The cylinder x² + y² = 9 has radius r = 3. A standard parameterization for the lateral surface is: x = 3 cos θ, y = 3 sin θ, z = z, with 0 ≤ θ ≤ 2π and 0 ≤ z ≤ 5. We compute the fundamental vector cross product. The parameterization is r(θ, z) = (3 cos θ, 3 sin θ, z). The partial derivatives are r_θ = (-3 sin θ, 3 cos θ, 0) and r_z = (0, 0, 1). Their cross product is r_θ × r_z = (3 cos θ, 3 sin θ, 0). The magnitude of this vector is √( (3 cos θ)² + (3 sin θ)² + 0² ) = √(9 cos² θ + 9 sin² θ) = √(9) = 3. Therefore, the surface area element is dS = ||r_θ × r_z|| dz dθ = 3 dz dθ.

First, compute the gradient of T: ∇T = (∂T/∂x, ∂T/∂y, ∂T/∂z) = (2x, 2y, 0). Multiply by -k = -2: -k∇T = -2*(2x, 2y, 0) = (-4x, -4y, 0). The surface S is the plane z=0, with the unit normal vector n = (0, 0, 1) (pointing in positive z direction). The dot product is (-4x, -4y, 0) • (0, 0, 1) = 0. Therefore, the integrand (-k ∇T) • n simplifies to 0. This means the heat flux across this particular surface, given this temperature field and orientation, is zero.

The surface is defined by z = g(x,y) = xy. We compute the partial derivatives: ∂z/∂x = y and ∂z/∂y = x. The surface area element is dS = √( (∂z/∂x)² + (∂z/∂y)² + 1 ) dA = √( y² + x² + 1 ) dA. The function to integrate over the surface is f(x,y,z)=z. On the surface, z = xy, so f(x,y,z) = xy. Therefore, the surface integral ∫∫S z dS becomes the double integral over the rectangular region D (0≤x≤1, 0≤y≤2) of the integrand: (xy) * √( y² + x² + 1 ) dA. This is written as ∫{y=0}² ∫_{x=0}¹ (xy) * √(x² + y² + 1) dx dy.

First, find the equation of the plane through the three points. The points are not collinear. The plane equation can be found by noting the sum of coordinates: x + y + z = 1. So, the surface S is the part of the plane z = 1 - x - y that lies over the triangular region D in the xy-plane bounded by x=0, y=0, and x+y=1. Compute dS: For z = 1 - x - y, ∂z/∂x = -1, ∂z/∂y = -1. Then dS = √( (-1)² + (-1)² + 1 ) dA = √(1+1+1) dA = √3 dA. The function is f(x,y,z)=x. On the surface, this is just x. So the integral is ∫∫ₛ x dS = ∫∫D x * (√3) dA = √3 ∫∫D x dA where the region D is defined as 0 ≤ x ≤ 1, 0 ≤ y ≤ 1-x. Compute the double integral: √3 ∫x=01 ∫y=0 (1-x) x dy dx = √3 ∫x=01  [ x * (1-x) ] dx = √3 ∫₀¹ (x - x²) dx = √3 [ (x²/2) - (x³/3) ]01 = √3 [ (1/2) - (⅓) ] = √3 [ (3/6) - (2/6) ] = √3 * (1/6) = √3/6.

For a surface S given explicitly as the graph of a function z = g(x, y) over some region D in the xy-plane, the surface integral of a scalar function f(x, y, z) is evaluated using the formula: ∫∫ₛ f(x, y, z) dS = ∫∫_D f(x, y, g(x, y)) * √( (∂g/∂x)² + (∂g/∂y)² + 1 ) dA. Here, D is the projection of the surface S onto the xy-plane. Therefore, the surface integral is computed as a double integral over this two-dimensional region D.

For a sphere of radius ρ = 2, a standard parameterization is: x = 2 sin φ cos θ, y = 2 sin φ sin θ, z = 2 cos φ, where φ goes from 0 to π/2 (since z≥0) and θ from 0 to 2π. Compute the fundamental vector cross product. The parameterization is r(θ, φ) = (2 sin φ cos θ, 2 sin φ sin θ, 2 cos φ). Compute partial derivatives: r_θ = (-2 sin φ sin θ, 2 sin φ cos θ, 0) and r_φ = (2 cos φ cos θ, 2 cos φ sin θ, -2 sin φ). Their cross product r_θ × r_φ has magnitude: ||r_θ × r_φ|| = ρ² sin φ = (2)² sin φ = 4 sin φ. Therefore, the surface area element is dS = ||r_θ × r_φ|| dθ dφ = 4 sin φ dθ dφ.

The surface S is the disk x² + y² ≤ 1 in the plane z=0. For this flat surface, dS = dA (since z=0 constant). The function is T(x,y) = 100 - x² - y². So the surface integral is ∫∫S T dS = ∫∫D (100 - x² - y²) dA, where D is the unit disk. Convert to polar coordinates: x = r cos θ, y = r sin θ, dA = r dr dθ, with 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π. The integrand becomes 100 - r². The integral is ∫θ=02π ∫r=01 (100 - r²) r dr dθ. Compute the inner integral: ∫₀¹ (100r - r³) dr = [50r² - (r⁴/4)]01 = 50 - 1/4 = (200/4 - 1/4) = 199/4. Then multiply by the θ-integral range 2π: (199/4) * 2π = (199/2)π.

For a parameterized surface r(u,v), the vectors r_u and rᵥ are tangent to the surface at a point. The cross product r_u × rᵥ yields a vector that is normal to the surface. The magnitude of this cross product, ||r_u × rᵥ||, is equal to the area of the parallelogram whose sides are the vectors r_u and rᵥ. This area scaling factor is used to convert the area element in the parameter domain (dudv or dA) to the area element dS on the surface.

If the temperature T is constant over the surface S, then its gradient ∇T is the zero vector (0,0,0). Therefore, the heat flux vector -k ∇T is also the zero vector. The dot product of the zero vector with any unit normal n is 0. Thus, the integrand (-k ∇T) • n is identically zero over the entire surface. The surface integral of zero is zero, regardless of the surface shape or area. So the heat flow through S is zero.

The cube has 6 faces. On each face, one coordinate is constant (either 0 or 1). By symmetry, the integral of (x+y+z) over each face will be the same if we consider opposite faces together. Alternatively, we can compute the integral over all six faces and add. However, a simpler method: For each face, the integral ∫∫face (x+y+z) dS can be computed. For example, take the face where x=0, with 0≤y≤1, 0≤z≤1. On this face, x=0, so the integrand is y+z. dS = dA = dy dz. The integral over this face is ∫{z=0}¹ ∫_{y=0}¹ (y+z) dy dz = ∫₀¹ [ (y²/2 + yz) from y=0 to 1 ] dz = ∫₀¹ (1/2 + z) dz = [ (z/2) + (z²/2) ] from 0 to 1 = (1/2 + 1/2) = 1. Similarly, the opposite face x=1 gives integrand 1+y+z, and its integral is ∫₀¹∫₀¹ (1+y+z) dy dz = ∫₀¹ [ (y + y²/2 + yz) from 0 to 1 ] dz = ∫₀¹ (1 + 1/2 + z) dz = ∫₀¹ (3/2 + z) dz = [ (3z/2) + (z²/2) ] from 0 to 1 = (3/2 + 1/2) = 2. So the pair of faces x=0 and x=1 contribute 1+2=3. By symmetry, the pairs y=0 and y=1 also contribute 3, and the pairs z=0 and z=1 also contribute 3. Total = 3+3+3 = 9.

The parameterization is r(u,v) = (u cos v, u sin v, u). Compute partial derivatives: r_u = (cos v, sin v, 1) and rᵥ = (-u sin v, u cos v, 0). Compute the cross product: r_u × rᵥ = determinant |i j k; cos v sin v 1; -u sin v u cos v 0| = i(sin v0 - 1u cos v) - j(cos v*0 - 1*(-u sin v)) + k(cos v(u cos v) - sin v(-u sin v)) = (-u cos v) i - j(0 + u sin v) + k*(u cos² v + u sin² v) = (-u cos v, -u sin v, u(cos² v+ sin² v)) = (-u cos v, -u sin v, u). The magnitude is √( (-u cos v)² + (-u sin v)² + u² ) = √( u² cos² v + u² sin² v + u² ) = √( u² (cos² v+ sin² v+1) ) = √( u² * (1+1) ) = √(2u²) = u √(2) (since u ≥ 0). The function f(x,y,z)=y. In terms of parameters, y = u sin v. So the surface integral becomes: ∫∫S y dS = ∫{v=0}^{π} ∫{u=0}^{2} (u sin v) * (||r_u × rᵥ||) du dv = ∫{v=0}^{π} ∫{u=0}^{2} (u sin v) * (u √(2)) du dv = ∫{v=0}^{π} ∫_{u=0}^{2} u² √(2) sin v du dv.