Surface Integrals, Heat Flow & Divergence Theorem Intuition

1) What does a flux integral physically represent?

The total mass of a surface

The net flow of a vector field through a surface

The average value of a function on a surface

The surface area of the region

A flux integral measures how much of a vector field passes through a given surface. Imagine the field as fluid velocity and the surface as a net: the integral gives the net amount of fluid crossing the surface per unit time. If vectors align with the chosen normal, flux is positive; opposite gives negative; parallel gives zero.

Explanation

A flux integral measures how much of a vector field passes through a given surface. Imagine the field as fluid velocity and the surface as a net: the integral gives the net amount of fluid crossing the surface per unit time. If vectors align with the chosen normal, flux is positive; opposite gives negative; parallel gives zero.

2) Let S be the portion of the plane z = 4 - x - y that lies above the triangular region in the xy-plane with vertices (0,0), (1,0), and (0,1). Evaluate ∬_S dS (surface area).

√2/2

√3/2

√5/2

2

For z=g(x,y)=4-x-y, g_x=-1 and g_y=-1, so dS=√(1+(-1)^2+(-1)^2)dA=√3 dA. The projection D is a right triangle with area 1/2, so ∬_S dS=√3·(1/2)=√3/2.

Explanation

For z=g(x,y)=4-x-y, g_x=-1 and g_y=-1, so dS=√(1+(-1)^2+(-1)^2)dA=√3 dA. The projection D is a right triangle with area 1/2, so ∬_S dS=√3·(1/2)=√3/2.

3) Let F(x,y,z)=⟨x, y, z⟩. Find the flux of F through the portion of the plane x + y + z = 1 in the first octant, with downward orientation.

1/2

-1/2

√3/2

-√3/2

On the plane x+y+z=1, we have x+y+z=1. A unit normal with downward orientation is n=-(1,1,1)/√3. Then F·n=-(x+y+z)/√3=-1/√3. Also, for z=1-x-y, dS=√3 dA, so F·n dS = (-1/√3)(√3)dA = -dA. The projected triangle has area 1/2, hence flux = -1/2.

Explanation

On the plane x+y+z=1, we have x+y+z=1. A unit normal with downward orientation is n=-(1,1,1)/√3. Then F·n=-(x+y+z)/√3=-1/√3. Also, for z=1-x-y, dS=√3 dA, so F·n dS = (-1/√3)(√3)dA = -dA. The projected triangle has area 1/2, hence flux = -1/2.

4) What is the relationship between surface integrals of vector fields and flux integrals?

They are unrelated concepts

Flux integrals are a special type of surface integral for vector fields

Surface integrals only apply to scalar fields

Flux only applies to closed surfaces

A surface integral of a vector field is commonly called a flux integral: ∬_S F·n dS measures the normal component of the field passing through the surface.

Explanation

A surface integral of a vector field is commonly called a flux integral: ∬_S F·n dS measures the normal component of the field passing through the surface.

5) Consider the temperature field T(x,y,z)=x²+y²+z². Find the rate of heat flow through the sphere x²+y²+z²=4, using Fourier’s law: heat flow = -k ∬_S ∇T·n dS.

-64πk

64πk

-32πk

32πk

∇T=⟨2x,2y,2z⟩. On the radius-2 sphere, the outward unit normal is n=⟨x,y,z⟩/2, so ∇T·n=(2x,2y,2z)·(x,y,z)/2 = x²+y²+z² = 4. Thus ∬_S ∇T·n dS = 4·Area(S)=4·(4π·2²)=4·16π=64π. Heat flow = -k·64π = -64πk.

Explanation

∇T=⟨2x,2y,2z⟩. On the radius-2 sphere, the outward unit normal is n=⟨x,y,z⟩/2, so ∇T·n=(2x,2y,2z)·(x,y,z)/2 = x²+y²+z² = 4. Thus ∬_S ∇T·n dS = 4·Area(S)=4·(4π·2²)=4·16π=64π. Heat flow = -k·64π = -64πk.

6) Let S be the parametric surface r(u,v)=⟨u, v, u+v⟩ for 0 ≤ u ≤ 1, 0 ≤ v ≤ 1. Evaluate ∬_S (x + y) dS.

√2

√3

2√3

3

r_u=⟨1,0,1⟩ and r_v=⟨0,1,1⟩, so r_u×r_v=⟨-1,-1,1⟩ with magnitude √3. Also x+y=u+v. Hence ∬_S (x+y)dS = ∫_0^1∫_0^1 (u+v)√3 du dv = √3·1 = √3.

Explanation

r_u=⟨1,0,1⟩ and r_v=⟨0,1,1⟩, so r_u×r_v=⟨-1,-1,1⟩ with magnitude √3. Also x+y=u+v. Hence ∬_S (x+y)dS = ∫_0^1∫_0^1 (u+v)√3 du dv = √3·1 = √3.

7) What does the orientation of a surface affect in a flux integral?

Only the magnitude of the result

Only the sign of the result

Both magnitude and sign

Neither magnitude nor sign

Reversing orientation changes n to -n, so F·n changes sign. The magnitude (absolute value) is unaffected; only the sign flips.

Explanation

Reversing orientation changes n to -n, so F·n changes sign. The magnitude (absolute value) is unaffected; only the sign flips.

8) A surface is parameterized by r(u,v)=⟨u+v, u-v, uv⟩. What is r_u × r_v?

⟨u, v, 1⟩

⟨u+v, v-u, -2⟩

⟨v, u, -2⟩

⟨1, 1, 1⟩

r_u=⟨1,1,v⟩ and r_v=⟨1,-1,u⟩. Cross product: i: 1·u - v·(-1)=u+v; j: v·1 - 1·u = v-u; k: 1·(-1)-1·1=-2. So r_u×r_v=⟨u+v, v-u, -2⟩.

Explanation

r_u=⟨1,1,v⟩ and r_v=⟨1,-1,u⟩. Cross product: i: 1·u - v·(-1)=u+v; j: v·1 - 1·u = v-u; k: 1·(-1)-1·1=-2. So r_u×r_v=⟨u+v, v-u, -2⟩.

9) A temperature field is T(x,y,z)=2x+3y+4z. Find the rate of heat flow through the portion of the plane x+y+z=1 in the first octant using Fourier’s law: heat flow = -k ∬_S ∇T·n dS.

-9k/2

9k/2

-3k/2

3k/2

∇T=⟨2,3,4⟩. For the plane x+y+z=1 with unit normal n=⟨1,1,1⟩/√3, ∇T·n=(2+3+4)/√3=9/√3=3√3. For z=1-x-y, dS=√3 dA, so ∇T·n dS = (3√3)(√3)dA = 9 dA. Over the projected triangle of area 1/2, ∬ 9 dA = 9/2. Multiply by -k gives -9k/2.

Explanation

∇T=⟨2,3,4⟩. For the plane x+y+z=1 with unit normal n=⟨1,1,1⟩/√3, ∇T·n=(2+3+4)/√3=9/√3=3√3. For z=1-x-y, dS=√3 dA, so ∇T·n dS = (3√3)(√3)dA = 9 dA. Over the projected triangle of area 1/2, ∬ 9 dA = 9/2. Multiply by -k gives -9k/2.

10) When is it appropriate to use the divergence theorem instead of directly evaluating a surface integral of a vector field?

When the surface is open

When the surface is closed and you can compute the divergence more easily

When the vector field is constant

When the surface is not orientable

The divergence theorem applies to closed surfaces and converts ∬_S F·n dS into ∭_V div(F) dV, often simplifying the computation when div(F) is easier to integrate.

Explanation

The divergence theorem applies to closed surfaces and converts ∬_S F·n dS into ∭_V div(F) dV, often simplifying the computation when div(F) is easier to integrate.

11) Evaluate ∬_S (x² + y² + z²) dS over the sphere x² + y² + z² = R².

4πR²

4πR³

4πR⁴

(4/3)πR³

On the sphere, x²+y²+z²=R² is constant, so ∬_S (x²+y²+z²)dS = ∬_S R² dS = R²·Area(S)=R²·(4πR²)=4πR⁴.

Explanation

On the sphere, x²+y²+z²=R² is constant, so ∬_S (x²+y²+z²)dS = ∬_S R² dS = R²·Area(S)=R²·(4πR²)=4πR⁴.

12) Find the upward flux of F(x,y,z)=⟨0,0,z⟩ through the paraboloid z=x²+y² bounded by the plane z=1.

π/2

π

2π

0

Use the graph form: for z=g(x,y)=x²+y², upward dS-vector is ⟨-g_x,-g_y,1⟩ dA = ⟨-2x,-2y,1⟩ dA. F=⟨0,0,z⟩ gives F·⟨-2x,-2y,1⟩ = z = x²+y². Over the disk x²+y²≤1, flux = ∬_D (x²+y²) dA = ∫_0^{2π}∫_0^1 r²·r dr dθ = 2π·(1/4)=π/2.

Explanation

Use the graph form: for z=g(x,y)=x²+y², upward dS-vector is ⟨-g_x,-g_y,1⟩ dA = ⟨-2x,-2y,1⟩ dA. F=⟨0,0,z⟩ gives F·⟨-2x,-2y,1⟩ = z = x²+y². Over the disk x²+y²≤1, flux = ∬_D (x²+y²) dA = ∫_0^{2π}∫_0^1 r²·r dr dθ = 2π·(1/4)=π/2.

13) What does a negative value of a flux integral indicate?

The vector field is zero everywhere

More field lines exit than enter the surface

More field lines enter than exit the surface

The surface is not properly oriented

With a chosen orientation, negative flux means the net flow is opposite the chosen normal direction (e.g., for an outward normal on a closed surface, net flow is into the volume).

Explanation

With a chosen orientation, negative flux means the net flow is opposite the chosen normal direction (e.g., for an outward normal on a closed surface, net flow is into the volume).

14) Consider the temperature field T(x,y,z)=x²+y². Find the rate of heat flow through the curved surface of the cylinder x²+y²=4 between z=0 and z=2 using Fourier’s law: heat flow = -k ∬_S ∇T·n dS.

-32πk

32πk

-16πk

16πk

∇T=⟨2x,2y,0⟩. On the radius-2 cylinder, outward unit normal is n=⟨x/2,y/2,0⟩, so ∇T·n = (2x,2y,0)·(x/2,y/2,0)=x²+y²=4. Curved surface area is (circumference)(height)=(2π·2)·2=8π, so ∬_S ∇T·n dS = 4·8π=32π and heat flow = -k·32π = -32πk.

Explanation

∇T=⟨2x,2y,0⟩. On the radius-2 cylinder, outward unit normal is n=⟨x/2,y/2,0⟩, so ∇T·n = (2x,2y,0)·(x/2,y/2,0)=x²+y²=4. Curved surface area is (circumference)(height)=(2π·2)·2=8π, so ∬_S ∇T·n dS = 4·8π=32π and heat flow = -k·32π = -32πk.

15) Use the divergence theorem to find the flux of F(x,y,z)=⟨x³, y³, z³⟩ through the sphere x²+y²+z²=9.

2916π/5

972π/5

972π

2916π

div(F)=3x²+3y²+3z²=3(x²+y²+z²)=3ρ². Over the radius-3 ball: flux=∭ 3ρ² dV = ∫_0^{2π}∫_0^{π}∫_0^3 3ρ²·ρ² sinφ dρ dφ dθ = 3(2π)(2)(∫_0^3 ρ⁴ dρ)=12π·(3⁵/5)=12π·(243/5)=2916π/5.

Explanation

div(F)=3x²+3y²+3z²=3(x²+y²+z²)=3ρ². Over the radius-3 ball: flux=∭ 3ρ² dV = ∫_0^{2π}∫_0^{π}∫_0^3 3ρ²·ρ² sinφ dρ dφ dθ = 3(2π)(2)(∫_0^3 ρ⁴ dρ)=12π·(3⁵/5)=12π·(243/5)=2916π/5.