Flux Integrals of Vector Fields: Definition, Setup & Physical Meaning

To find the surface area element for a graph defined by a function of two variables, we treat the surface as a level surface or parameterize it using x and y. We begin by finding the partial derivatives of the function g with respect to x and y. We then form the tangent vectors in the x and y directions. The cross product of these tangent vectors gives a normal vector, the magnitude of which represents the scaling factor between the area in the domain and the area on the surface. Calculating this magnitude results in the square root of the sum of one plus the squares of the partial derivatives. Therefore, the correct differential element is the square root of 1 plus the partial derivative with respect to x squared plus the partial derivative with respect to y squared, multiplied by the area element of the domain.

First, we identify the function to be integrated, which is f(x, y, z) = z. Since the surface is given by z = 2 - x - y, we substitute this expression into the function, replacing z with (2 - x - y). Next, we calculate the surface area element dS. We take the partial derivative of z with respect to x, which is -1, and the partial derivative of z with respect to y, which is also -1. We square these values and add them to 1, giving us 1 + (-1)² + (-1)², which equals 3. Taking the square root gives us √(3). Finally, we multiply the substituted function by this factor to get the integrand (2 - x - y) * √(3).

The flux integral measures how much of a vector field passes through a given surface. We calculate the dot product of the vector field F and the unit normal vector n, which gives the component of the field perpendicular to the surface at any point. Integrating this normal component over the entire surface area sums up the total flow crossing the boundary. If the field represents fluid velocity, this integral calculates the volume of fluid passing through the surface per unit of time. Therefore, the correct interpretation is the net amount of flow passing through the surface.

To find the normal vector for a surface defined by z = g(x, y), we can rewrite the equation as a level surface G(x, y, z) = z - g(x, y) = 0. In this case, we define G(x, y, z) = z - x² - y². We then compute the gradient of G, which gives us the vector <-2x, -2y, 1>. For an "upward" orientation, the k-component (the z-component) of the normal vector must be positive. Since the k-component of our gradient vector is +1, this vector points upward. Thus, the correct non-normalized normal vector is <-2x, -2y, 1>.

We start by identifying the unit normal vector for the surface. Since the surface lies in the plane z = 0 and is oriented in the positive z-direction, the unit normal vector n is <0, 0, 1>. Next, we calculate the dot product of the vector field F and the normal vector n. Computing <0, 0, 4> · <0, 0, 1> results in the scalar value 4. We then integrate this constant value over the area of the surface. The surface is a square with side length 1, so its area is 1 times 1, which equals 1. Multiplying the dot product result by the area gives 4 times 1, which equals 4.

Heat flow is governed by Fourier's Law, which states that the heat flux density vector is proportional to the negative gradient of the temperature. Specifically, the heat flux vector J is equal to -k times the gradient of T, because heat flows from regions of higher temperature to regions of lower temperature. To find the total heat flow across a surface, we must sum the component of this flux vector that is normal to the surface. This requires taking the dot product of the heat flux vector J with the unit normal vector n and integrating over the surface. Substituting the expression for J, we obtain the integral of -k * (gradient of T) · n dS.

We begin by determining the partial derivatives of the function z = √(x² + y²). The partial derivative with respect to x is x / √(x² + y²), and the partial derivative with respect to y is y / √(x² + y²). Next, we calculate the surface area element dS by squaring these derivatives, adding 1, and taking the square root. The sum of the squares simplifies to (x² + y²) / (x² + y²), which equals 1. Adding the initial 1 gives 2, so dS equals √(2) dA. We then integrate this constant √(2) over the domain D, which is a disk of radius 3. The area of the disk isπ * 3² = 9pi. Multiplying the area by the constant √(2) yields 9 *π * √(2).

We must analyze the relationship between the vector field and the normal vectors of the top and bottom surfaces. For the top disk at z = 5, the outward normal vector is k = <0, 0, 1>. For the bottom disk at z = 0, the outward normal vector is -k = <0, 0, -1>. The vector field is given as F = <x, y, 0>, which has a z-component of 0. When we take the dot product of F with either the top normal <0, 0, 1> or the bottom normal <0, 0, -1>, the result is 0 because the vectors are orthogonal. Since the integrand is zero on both disks, the flux contribution from these surfaces is zero.

We can approach this using the Divergence Theorem, which relates the flux across a closed surface to the triple integral of the divergence over the enclosed volume. We calculate the divergence of F by taking the partial derivative of the x-component (y) with respect to x, which is 0, adding the partial derivative of the y-component (-x) with respect to y, which is 0, and adding the partial derivative of the z-component (2) with respect to z, which is also 0. Since the divergence is exactly zero everywhere, the triple integral of the divergence is zero. Consequently, the total net flux across the closed sphere must be zero.

First, we compute the tangent vector with respect to u, denoted as r_u. Differentiating the components of r with respect to u gives <1, 0, 2u>. Next, we compute the tangent vector with respect to v, denoted as rᵥ. Differentiating the components with respect to v gives <0, 1, -2v>. We then calculate the cross product r_u x rᵥ. The i-component is (0)(-2v) - (2u)(1) = -2u. The j-component is -((1)(-2v) - (2u)(0)) = 2v. The k-component is (1)(1) - (0)(0) = 1. Combining these components, we obtain the vector <-2u, 2v, 1>.

For a graph z = g(x,y) the tangent vectors are (1,0,g_x) and (0,1,g_y); their cross product has magnitude sqrt(1 + g_x² + g_y²), which converts the area element dA in the xy‑plane into the actual surface area element dS.

We can solve this by using symmetry arguments rather than a direct complex parameterization. The surface integral of x² over the sphere must be equal to the surface integral of y² and the surface integral of z² due to the spherical symmetry of the surface. Let I be the value of the integral we want to find. Then 3I = Integral(x² + y² + z²) dS. On the unit sphere, x² + y² + z² equals 1. Therefore, 3I is equal to the integral of 1 dS, which is simply the surface area of the unit sphere. The surface area of a unit sphere is 4 π. So, we have the equation 3I = 4 π. Dividing by 3, we find that the integral I equals 4 *π / 3.

We begin by parameterizing the cylindrical surface using cylindrical coordinates: x = 2cos(theta), y = 2sin(theta), z = z. The bounds are 0  ≤  theta  ≤  2π and 0  ≤  z  ≤  3. The outward normal vector for a cylinder of radius r is <x, y, 0> / r, but for the flux integral dS calculation, the vector element n dS simplifies to <x, y, 0> dz dtheta (using the non-normalized normal from the cross product of partials, which is <2cos(theta), 2sin(theta), 0>). The vector field F on the surface is <2cos(theta), 2sin(theta), z>. We take the dot product of F and the normal vector <2cos(theta), 2sin(theta), 0>. This gives 4cos²(theta) + 4sin²(theta) + 0, which simplifies to 4. We integrate this value over the bounds: the integral of 4 dz dtheta. Integrating with respect to z from 0 to 3 gives 12. Integrating 12 with respect to theta from 0 to 2π gives 24 π.

First, we determine the surface area element dS for the function z = x² + y². The partial derivatives are dz/dx = 2x and dz/dy = 2y. The formula for dS is √(1 + (2x)² + (2y)²) dA, which simplifies to √(1 + 4x² + 4y²) dA. Next, we convert this to polar coordinates because the domain is a disk. In polar coordinates, x² + y² becomes r², so the square root term becomes √(1 + 4r²). The area element dA in polar coordinates is r dr dtheta. Therefore, combining these parts, the integrand becomes r * √(1 + 4r²). The limits for the cylinder x² + y² = 4 correspond to r going from 0 to 2, and theta going from 0 to 2π.

We calculate the flux by projecting the surface onto the xy-plane. The surface z = 1 - x² - y² intersects the xy-plane (z=0) at the circle x² + y² = 1, so our domain D is the unit disk. The upward normal vector for a graph z = g(x,y) is given by <-dg/dx, -dg/dy, 1>. For our function, the partial derivatives are -2x and -2y, so the normal vector is <2x, 2y, 1>. We take the dot product of the vector field F = <y, -x, 1> with this normal vector. The dot product is (y)(2x) + (-x)(2y) + (1)(1), which simplifies to 2xy - 2xy + 1 = 1. We then integrate this result over the domain D. The integral of 1 over the unit disk is simply the area of the disk, which is π * 1² = pi.