Advanced Surface & Flux Integrals Quiz

1) Evaluate ∬ₛ 1 dS for the surface S given by z = x + y, 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 - x.

√3

√3 / 2

√3 * 1/2

2√3

The domain D is the triangle with area 1/2. The surface is z = x + y, g_x = 1, g_y = 1, so dS = √(1+1+1) dA = √3 dA. The surface area is √3 times the area of D, giving √3 * (1/2).

Explanation

The domain D is the triangle with area 1/2. The surface is z = x + y, g_x = 1, g_y = 1, so dS = √(1+1+1) dA = √3 dA. The surface area is √3 times the area of D, giving √3 * (1/2).

2) Compute the surface area ∬ₛ dS where S is the part of the plane x + y + z = 1 in the first octant.

√2/2

√3/2

1/2

√3

The projection of S onto the xy-plane is the triangle with vertices (0,0), (1,0), (0,1), which has area 1/2. For the graph z = 1 − x − y we have ∂z/∂x = −1 and ∂z/∂y = −1, so dS = √(1 + (−1)² + (−1)²) dA = √3 dA. Therefore the surface area is √3 times the projected area: √3 × (1/2) = √3/2.

Explanation

The projection of S onto the xy-plane is the triangle with vertices (0,0), (1,0), (0,1), which has area 1/2. For the graph z = 1 − x − y we have ∂z/∂x = −1 and ∂z/∂y = −1, so dS = √(1 + (−1)² + (−1)²) dA = √3 dA. Therefore the surface area is √3 times the projected area: √3 × (1/2) = √3/2.

3) Compute ∬ₛ z dS where S is the part of the plane x + y + z = 1 in the first octant.

√3/12

√3/6

√3/4

√3/3

For z = 1 - x - y, dS = √3 dA. The integral is √3 ∬_D (1 - x - y) dA, where D is the triangle with vertices (0,0), (1,0), (0,1). Compute ∬_D (1 - x - y) dA = ∫_0^1 ∫_0^{1-x} (1 - x - y) dy dx = ∫_0^1 [ (1-x)y - xy - y²/2 ]_0^{1-x} dx = ∫_0^1 [ (1-x)² - (1-x)²/2 ] dx = ∫_0^1 (1-x)²/2 dx = (1/2) ∫_0^1 (1-x)² dx. Let u = 1-x, then dx = -du, and the integral becomes (1/2) ∫_0^1 u² du = (1/2) * (1/3) = 1/6. So the surface integral is √3 * (1/6) = √3/6.

Explanation

For z = 1 - x - y, dS = √3 dA. The integral is √3 ∬_D (1 - x - y) dA, where D is the triangle with vertices (0,0), (1,0), (0,1). Compute ∬_D (1 - x - y) dA = ∫_0^1 ∫_0^{1-x} (1 - x - y) dy dx = ∫_0^1 [ (1-x)y - xy - y²/2 ]_0^{1-x} dx = ∫_0^1 [ (1-x)² - (1-x)²/2 ] dx = ∫_0^1 (1-x)²/2 dx = (1/2) ∫_0^1 (1-x)² dx. Let u = 1-x, then dx = -du, and the integral becomes (1/2) ∫_0^1 u² du = (1/2) * (1/3) = 1/6. So the surface integral is √3 * (1/6) = √3/6.

4) Consider a sphere centered at the origin. The vector field F is given by F = <y, -x, 0>. Without performing any integration, determine the flux of F through the sphere and the geometric reason for it.

Positive, because the field magnitude increases with distance.

Negative, because the field circulates clockwise.

Zero, because the field vectors are everywhere tangent to the surface.

Zero, because the field is constant in the z-direction.

The vector field <y, -x, 0> represents a rotation around the z-axis. At any point (x, y, z) on the sphere, the normal vector is radial, proportional to <x, y, z>. The dot product of the field and the normal is <y, -x, 0> · <x, y, z> = xy - yx + 0 = 0. Since the field is perpendicular to the normal vector at every point on the surface (the field "swirls" around the sphere without entering or exiting), the flux is zero.

Explanation

The vector field <y, -x, 0> represents a rotation around the z-axis. At any point (x, y, z) on the sphere, the normal vector is radial, proportional to <x, y, z>. The dot product of the field and the normal is <y, -x, 0> · <x, y, z> = xy - yx + 0 = 0. Since the field is perpendicular to the normal vector at every point on the surface (the field "swirls" around the sphere without entering or exiting), the flux is zero.

5) We wish to evaluate the surface integral of a function f(x, y, z) over the surface S defined by x = 1 - y² - z² (a paraboloid opening along the x-axis). If we project this surface onto the yz-plane to set up the integral, what is the correct expression for the surface area element dS in terms of y and z?

√(1 + 4y² + 4z²) dy dz

√(1 + 2y + 2z) dy dz

√(1 + y² + z²) dy dz

Dy dz

When a surface is defined as x = g(y, z), the surface area element is calculated analogously to the z = g(x, y) case, but using partial derivatives with respect to y and z. The formula is dS = √(1 + (dx/dy)² + (dx/dz)²) dA. Here, x = 1 - y² - z². The partial derivative dx/dy is -2y and dx/dz is -2z. Squaring these gives 4y² and 4z². Therefore, dS = √(1 + 4y² + 4z²) dy dz.

Explanation

When a surface is defined as x = g(y, z), the surface area element is calculated analogously to the z = g(x, y) case, but using partial derivatives with respect to y and z. The formula is dS = √(1 + (dx/dy)² + (dx/dz)²) dA. Here, x = 1 - y² - z². The partial derivative dx/dy is -2y and dx/dz is -2z. Squaring these gives 4y² and 4z². Therefore, dS = √(1 + 4y² + 4z²) dy dz.

6) For the surface S given by g(x,y,z) = x² + y² − z = 0 with upward orientation, the correct vector surface element when projecting onto the xy-plane is

<2x, 2y, 1> dA

<−2x, −2y, 1> dA

<2x, 2y, −1> dA

<−2x, −2y, −1> dA

For a surface z = g(x,y), the upward orientation (positive z-direction) uses the vector surface element <−g_x, −g_y, 1> dA. Here g(x,y) = x² + y², so g_x = 2x and g_y = 2y, giving <−2x, −2y, 1> dA. The positive z-component guarantees the upward direction.

Explanation

For a surface z = g(x,y), the upward orientation (positive z-direction) uses the vector surface element <−g_x, −g_y, 1> dA. Here g(x,y) = x² + y², so g_x = 2x and g_y = 2y, giving <−2x, −2y, 1> dA. The positive z-component guarantees the upward direction.

7) The rate of heat flow out of a surface S is given by ∬ₛ −κ ∇T · dS (κ > 0 constant, outward normal). If ∇T = <1, 1, 1> and S is the triangular surface x + y + z = 1 in the first octant oriented outward from the tetrahedron, the heat flow is

−√3κ

−3κ/2

0

√3κ/2

The flux of ∇T through S (outward) is ∇T · (vector area of S). The vector area is (1/2)<1,1,1> (computed from cross product of two sides). Then ∇T · (1/2)<1,1,1> = (1/2)(1+1+1) = 3/2. Hence the Heat flow = −κ (3/2).

Explanation

The flux of ∇T through S (outward) is ∇T · (vector area of S). The vector area is (1/2)<1,1,1> (computed from cross product of two sides). Then ∇T · (1/2)<1,1,1> = (1/2)(1+1+1) = 3/2. Hence the Heat flow = −κ (3/2).

8) Compute ∬ₛ (x² + y² + z²) dS where S is the part of the sphere x² + y² + z² = 4 above z = 0 (upper hemisphere of radius 2).

16π

32π

64π/3

80π/3

Use spherical coordinates: ρ = 2, 0 ≤ φ ≤ π/2, 0 ≤ θ ≤ 2π. The integrand is ρ² = 4. dS = ρ² sinφ dφ dθ = 4 sinφ dφ dθ. The integral is ∫_0^{2π} dθ ∫_0^{π/2} 4 · 4 sinφ dφ = 16 ∫_0^{2π} dθ ∫_0^{π/2} sinφ dφ = 16 × 2π × [−cosφ]_0^{π/2} = 32π (0 − (−1)) = 32π.

Explanation

Use spherical coordinates: ρ = 2, 0 ≤ φ ≤ π/2, 0 ≤ θ ≤ 2π. The integrand is ρ² = 4. dS = ρ² sinφ dφ dθ = 4 sinφ dφ dθ. The integral is ∫_0^{2π} dθ ∫_0^{π/2} 4 · 4 sinφ dφ = 16 ∫_0^{2π} dθ ∫_0^{π/2} sinφ dφ = 16 × 2π × [−cosφ]_0^{π/2} = 32π (0 − (−1)) = 32π.

9) A surface S is parametrized by r(u,v) = <u, v, u² − v²>, 0 ≤ u ≤ 1, 0 ≤ v ≤ 1. Which parametrization gives the orientation with positive k component for the normal?

Given r(u,v)

R(v,u)

R(u,1−v)

Negative orientation

Compute r_u = <1, 0, 2u>, rᵥ = <0, 1, −2v>. Then r_u × rᵥ = <(0)(−2v) − (1)(2u), (2u)(0) − (1)(−2v), (1)(1) − (0)(0)> = <−2u, 2v, 1>. The k component is 1 > 0, so the given parametrization already has a positive k component.

Explanation

Compute r_u = <1, 0, 2u>, rᵥ = <0, 1, −2v>. Then r_u × rᵥ = <(0)(−2v) − (1)(2u), (2u)(0) − (1)(−2v), (1)(1) − (0)(0)> = <−2u, 2v, 1>. The k component is 1 > 0, so the given parametrization already has a positive k component.

10) A hot object is bounded by the closed surface consisting of the upper hemisphere x² + y² + z² = 1 (z ≥ 0) and the disk in the xy-plane. The temperature gradient is ∇T = <x, y, z>. The net heat flow out of the object is

0

π

2π

4π/3

By the divergence theorem the net heat flow out is ∭ᵥ div(∇T) dV = ∭ᵥ ∇²T dV. Here ∇²T = div ∇T = 3, volume of the hemispherical object is 2π/3, so net heat flow = 3 × (2π/3) = 2π.

Explanation

By the divergence theorem the net heat flow out is ∭ᵥ div(∇T) dV = ∭ᵥ ∇²T dV. Here ∇²T = div ∇T = 3, volume of the hemispherical object is 2π/3, so net heat flow = 3 × (2π/3) = 2π.

11) Compute the flux of F = <x, y, 0> through the part of the paraboloid z = x² + y² − 1 for z ≤ 0, oriented upward.

−π

−π/2

0

π

Projection D: x² + y² ≤ 1. Upward dS = <−2x, −2y, 1> dA. F · dS = x(−2x) + y(−2y) + 0 = −2(x² + y²) dA. Flux = ∬_D −2(x² + y²) dA = −2 × 2π∫₀¹ r³ dr = −4π (1/4) = −π.

Explanation

Projection D: x² + y² ≤ 1. Upward dS = <−2x, −2y, 1> dA. F · dS = x(−2x) + y(−2y) + 0 = −2(x² + y²) dA. Flux = ∬_D −2(x² + y²) dA = −2 × 2π∫₀¹ r³ dr = −4π (1/4) = −π.

12) A closed surface is the boundary of a solid region (sphere). To apply the divergence theorem correctly, the orientation must be

Inward on the bottom half and outward on the top half

Outward everywhere

Upward everywhere

Arbitrary as long as it is consistent

The divergence theorem states that for a closed surface bounding a region V, ∬ₛ F · dS = ∭ᵥ div F dV when the surface is oriented with the outward normal (positive orientation). Any other consistent orientation would give the negative of the triple integral.

Explanation

The divergence theorem states that for a closed surface bounding a region V, ∬ₛ F · dS = ∭ᵥ div F dV when the surface is oriented with the outward normal (positive orientation). Any other consistent orientation would give the negative of the triple integral.

13) Calculate the flux of the vector field F = <1, 0, 0> through the portion of the plane x + 2y + 3z = 6 that lies in the first octant, oriented with the normal pointing away from the origin.

3

6

9

12

While this can be done by projecting onto the xy-plane, it is much faster to project onto the yz-plane because the vector field F = <1, 0, 0> only has an x-component. The surface is x = 6 - 2y - 3z. The normal vector for x = g(y,z) is <1, -dg/dy, -dg/dz> (oriented positively in x). Here, g_y = -2 and g_z = -3, so the normal vector is <1, 2, 3>. The normalized condition isn't needed if we use the correct dS mapping. The flux integral is the double integral over the projection D (in yz-plane) of F dot <1, 2, 3> dy dz. F dot <1, 2, 3> = 1*1 + 0 + 0 = 1. So we just need the area of the projection D in the yz-plane. The plane x + 2y + 3z = 6 intersects the yz-plane (x=0) at 2y + 3z = 6. The intercepts are y=3 and z=2. The area of this triangle is (1/2) * base * height = (1/2) * 3 * 2 = 3. The flux is 1 * Area = 3.

Explanation

While this can be done by projecting onto the xy-plane, it is much faster to project onto the yz-plane because the vector field F = <1, 0, 0> only has an x-component. The surface is x = 6 - 2y - 3z. The normal vector for x = g(y,z) is <1, -dg/dy, -dg/dz> (oriented positively in x). Here, g_y = -2 and g_z = -3, so the normal vector is <1, 2, 3>. The normalized condition isn't needed if we use the correct dS mapping. The flux integral is the double integral over the projection D (in yz-plane) of F dot <1, 2, 3> dy dz. F dot <1, 2, 3> = 1*1 + 0 + 0 = 1. So we just need the area of the projection D in the yz-plane. The plane x + 2y + 3z = 6 intersects the yz-plane (x=0) at 2y + 3z = 6. The intercepts are y=3 and z=2. The area of this triangle is (1/2) * base * height = (1/2) * 3 * 2 = 3. The flux is 1 * Area = 3.

14) Let S be the unit disk in the xy-plane oriented upward. Which of the following vector fields yields the largest positive flux through S?

F = <5, 5, 0>

F = <0, 0, 1 - x² - y²>

F = <x, y, 1>

F = <0, 0, 2>

Flux is the integral of (F · n) over the surface. For the unit disk in the xy-plane, the upward normal n is <0, 0, 1>.

For A: F · n = 0. Flux is 0.

For B: F · n = 1 - x² - y². This is a paraboloid shape over the disk. The volume under 1 - r² over the unit disk is integral(0 to 1) of (1-r²) * 2π*r dr = 2π * [r²/2 - r⁴/4] = 2π(1/4) = π/2 (approx 1.57).

For C: F · n = 1. Flux is Area * 1 =π (approx 3.14).

For D: F · n = 2. Flux is Area * 2 = 2π (approx 6.28).

Therefore, field D yields the largest flux. This requires evaluating the "shape" of the normal component over the domain.

Explanation

Flux is the integral of (F · n) over the surface. For the unit disk in the xy-plane, the upward normal n is <0, 0, 1>.

For A: F · n = 0. Flux is 0.

For B: F · n = 1 - x² - y². This is a paraboloid shape over the disk. The volume under 1 - r² over the unit disk is integral(0 to 1) of (1-r²) * 2π*r dr = 2π * [r²/2 - r⁴/4] = 2π(1/4) = π/2 (approx 1.57).

For C: F · n = 1. Flux is Area * 1 =π (approx 3.14).

For D: F · n = 2. Flux is Area * 2 = 2π (approx 6.28).

Therefore, field D yields the largest flux. This requires evaluating the "shape" of the normal component over the domain.

Advanced Surface & Flux Integrals: Symmetry, Projections & Mixed Applications