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A vector field is an assignment of a vector $\vec{F}$ to each point (x, y) in a space, i.e., $\vec{F} = M\vec{i}+N\vec{j}$ where M and N are functions of x and y.
A vector field on a plane can be visualized as a collection of arrows, each attached to a point on the plane. These arrows represent vectors with specific magnitudes and directions.
Work is defined as the energy transferred when a force acts on an object and displaces it along a path. In the context of vector fields, we calculate the work done by a force field along a curve or trajectory C using a line integral. The work done by a force field $\vec{F}$ along a curve C is: W = $\int_{C} \vec{F}·d\vec{r} = \int_{C} Mdx + Ndy = \int_{C} \vec{F}·\hat{\mathbf{T}}ds$, where $\hat{\mathbf{T}}$ is the unit tangent vector.
A vector field is conservative if there exist a scalar function such that $\vec{F}$ = ∇f (the vector field is its gradient). This scalar function is known or referred to as the potential function associated with the vector field.
Theorem. Fundamental theorem of calculus for line integral. If $\vec{F}$ is a conservative vector field in a simply connected region of space (i.e., a region with no holes), and if f is a scalar potential function for $\vec{F}$ in that region, then $\int_{C} \vec{F}·d\vec{r} = \int_{C} ∇f·d\vec{r} = f(P_1)-f(P_0)$ where P_{0} and P_{1} are the initial and final points of the curve C, respectively.
Conservative force. A force $\vec{F}$ is considered conservative if the work done by the force around any closed curve C is zero. Mathematically, this is expressed as $\int_{C} \vec{F}·d\vec{r} = 0$.
Path independence. A force field is path-independent, meaning the work done by the force in moving an object from one point to another is the same, regardless of the path taken between the two points.
Gradient Field. A vector field $\vec{F}$ is a gradient field if it can be expressed as the gradient of a scalar potential function f. In mathematical terms, this means $\vec{F} = ∇f$, where f is a scalar function and the vector field $\vec{F}$ has components $\vec{F} = ⟨M, N⟩ = ⟨\frac{∂f}{∂x}, \frac{∂f}{∂y}⟩ = ⟨f_x, f_y⟩$. Here, f is the potential function associated with the vector field $\vec{F}$. The lineal integral of $\vec{F}$ along a path C measures the work done by the vector field in moving an object along the path C. If $\vec{F}$ is a gradient field, then: $\int_{C} \vec{F}·d\vec{r} = f(P_1)-f(P_0).$
Exact differential. In the context of differential forms, a differential expression Mdx+Ndy is called an exact differential if there exist a scalar function f(x, y) such that df = $\frac{∂f}{∂x}dx+\frac{∂f}{∂y}dy$ = Mdx + Ndy. This implies that the vector field $\vec{F} = ⟨M, N⟩$ is conservative, and there exists a potential function f such that $\vec{F} = ∇f$.
If a differential is exact, then the line integral of $\vec{F}$ over any path C can be evaluated by simply finding the difference in the potential function values at the endpoints of the path.
These four properties —conservative force, path independence, gradient field, and exact differential— are different perspectives of the same fundamental concept: the conservativeness of a vector field.
The criterion for checking whether a vector field $\vec{F}$ is conservative can be summarized as follows: If $\frac{∂M}{∂y} = \frac{∂N}{∂x}$, and if the functions M and N have continuous first partial derivatives across the entire domain D, and the domain D is open and simply connected, then $\vec{F}$ is a gradient field, meaning it is conservative.
Green’s Theorem. If C is a positively oriented (counterclockwise) simple closed curve enclosing a region R, and $\vec{F} = ⟨M(x, y), N(x,y)$ is a vector field that is defined and has continuous partial derivatives on an open region containing R, then Green's Theorem states that: $\oint_C \vec{F} \cdot{} d\vec{r} = \iint_R curl(\vec{F}) dA ↭ \oint_C Mdx + N dy = \iint_R (N_x-M_y)dA = \iint_R (\frac{∂N}{∂x}-\frac{∂M}{∂y})dA$ where
Green’s Theorem for Flux. If C is a positively oriented (counterclockwise), simple closed curve that encloses a region R and if $\vec{F} = ⟨P, Q⟩$ is a continuously differentiable vector field defined on an open region that contains R, then the flux of $\vec{F}$ across C is equal to the double integral of the divergence of $\vec{F}$ over R. Mathematically, this is expressed as: $\oint_C \vec{F}·\vec{n}d\vec{s} = \int \int_{R} div \vec{F}dA$ where the divergence of $\vec{F}$ is given by $div \vec{F} = P_x + Q_y = \frac{∂P}{∂x}+\frac{∂Q}{∂y}$.
Triple integrals are powerful tools in calculus, particularly useful for calculating various physical quantities over a region in three-dimensional space. These integrals allow us to compute volume, mass, and other physical quantities over a solid region bounded by surfaces in 3d space.
A triple integral extends the concept of integration to three dimensions. When you have a function f(x, y, z) that varies over a 3D region R, you can use a triple integral to sum up the values of f(x, y, z) over that region.
The triple integral of a function f(x, y, z) over a region R in three-dimensional space is expressed as $\int \int \int_{R} f(x, y, z)dV$ where
To evaluate a triple integral, you need to set up the bounds of integration, which describe the region R over which you’re integrating. The process involves: (1) Identifying the region. (2) Choosing the order of integration: In Cartesian coordinates, you typically integrate with respect to x, y, and z. However, the order can be rearranged. (3) Setting up the limits: These limits are the ranges for x, y, and z.
Triple integrals can be interpreted or computed in various coordinate systems depending on the symmetry of the region R. Choosing the appropriate coordinate system is crucial as it can either simplify or complicate the calculation by aligning the integration limits with the geometry of the region.
Conversion to Cartesian Coordinates: z = z, x = rcos(θ), y = rsin(θ), and the differential volume element is: dV = rdrdθdz.
They are ideal for regions with spherical symmetry such as spheres, hemispheres, or cones. In this system, the coordinates or parameters are:
The differential volume element dV = ρ^{2}sin(ϕ)dρdθdϕ. Conversion to Cartesian Coordinates: x = rcos(θ) = ρsin(φ)cos(θ), y = rsin(θ) = ρsin(φ)sin(θ), and z = ρcos(φ).
We want to calculate the flux of the vector field $\vec{F} = z\vec{k}$ through the portion of the paraboloid z = f(x, y) = x^{2}+y^{2} that lies above the unit disk in the xy-plane. It is a circular paraboloid that opens upwards, has its vertex at the origin, and has circular cross-sections perpendicular to the z-axis. The normal vector to the surface is pointing upwards (Figure 03). The vector field $\vec{F} = z\hat{\mathbf{k}}$ represents a field that points in the positive z-direction, with a magnitude that increases with z. In other words, the higher the point on the surface, the stronger the vector field at that point.
Surface Element and Normal Vector For a surface defined by z = f(x,y), the surface element vector is given by: $\hat{\mathbf{n}}dS = ±⟨-f_x, -f_y, 1⟩dxdy$
For the paraboloid z = f(x, y) = x^{2}+y^{2}, f_x = $\frac{∂}{∂x}(x^2+y^2) = 2x$, f_y = $\frac{∂}{∂y}(x^2+y^2) = 2y$
Therefore, $\hat{\mathbf{n}}·dS = ⟨-2x, -2y, 1⟩dxdy$. Since the surface is oriented upwards, we choose the positive sign for the normal vector.
Flux integral. The flux through the surface S is given by the surface integral: Flux = $\int \int_{S} \vec{F}·\hat{\mathbf{n}}·dS$ =[Substitute $\vec{F} = z\vec{k} = ⟨0, 0, z⟩$ and $\hat{\mathbf{n}}·dS = ⟨-2x, -2y, 1⟩dxdy$, we have:] Flux = $\int \int_{S}⟨0, 0, z⟩·⟨-2x, -2y, 1⟩dxdy $
Simplifying the dot product:
$Flux = \int \int_{S} zdxdy$ =[Given z = x^{2}+y^{2} on the surface] $\int \int_{S} (x^2+y^2)dxdy$
Converting to Polar Coordinates
The integral is really in the shadow region below the surface as it only involves x and y. To evaluate this double integral, it’s convenient to switch to polar coordinates because the region of integration (the unit disk) is circular. The relationships between Cartesian and polar coordinates are: x = r·cos(θ), y = r·sin(θ), x^{2}+y^{2} = r^{2}, dA = dxdy= rdrdθ.
The region of integration is 0 ≤ r ≤1 and 0 ≤ θ ≤2π. The integral becomes:
$\int_{0}^{2π} \int_{0}^{1} r^2·rdrdθ = \int_{0}^{2π} \int_{0}^{1} r^3drdθ = \int_{0}^{2π}\frac{r^4}{4}\bigg|_{0}^{1} = $
$ = \int_{0}^{2π} \frac{1}{4}dθ = \frac{1}{4}θ\bigg|_{0}^{2π} = \frac{π}{2}.$
The flux of the vector field $\vec{F} = z\hat{\mathbf{k}}$ through the paraboloid z = x^{2} + y^{2} over the unit disk is ^{π}⁄_{2}.
Suppose a surface S is parametrized by two parameters u and u, which vary over a region D in the uv-plane. The position of any point on the surface can be described by the position vector: $\vec{r}(u, v) = ⟨x(u, v), y(u, v), z(u, v)⟩$. Here x(u, v), y(u, v), and (u, v) are functions that describe the coordinates of the surface in terms of the parameters u and v.
Consider a small section of the surface dS corresponding to small changes Δu and Δv in the parameters u and v. This small section can be approximated by a parallelogram whose sides are given by the partial derivatives of $\vec{r}$ with respect to u and v: $\frac{∂\vec{r}}{∂u}Δu = ⟨\frac{∂\vec{x}}{∂u}Δu, \frac{∂\vec{y}}{∂u}Δu, \frac{∂\vec{z}}{∂u}Δu⟩$ and $\frac{∂\vec{r}}{∂v}Δv = ⟨\frac{∂\vec{x}}{∂v}Δv, \frac{∂\vec{y}}{∂v}Δv, \frac{∂\vec{z}}{∂v}Δv⟩$
The partial derivative $\frac{\partial \vec{r}}{\partial u}$ represents how the position vector $\vec{r}$ changes as the parameter u changes, giving us the tangent vector to the surface in the direction of increasing u. Similarly, Δu corresponds to the small change in the parameter u. The product $\frac{\partial \vec{r}}{\partial u}Δu$ is a vector describing the change in position along the surface S in the direction of u. This vector has x y, and z components, representing the changes in the coordinates of a point on the surface as “u” changes.
The area of the parallelogram formed by these two tangent vectors is given by the magnitude of their cross product. The cross product itself gives a vector that is perpendicular to the surface at that point, with a magnitude equal to the area of the parallelogram.
$±\hat{\mathbf{n}}dS = (\frac{∂\vec{r}}{∂u}Δu)x(\frac{∂\vec{r}}{∂v}Δv) =(\frac{∂\vec{r}}{∂u}x\frac{∂\vec{r}}{∂v})ΔuΔv$
Conclusion. Thus, the surface element vector can be expressed as: $±\hat{\mathbf{n}}dS = (\frac{∂\vec{r}}{∂u}x\frac{∂\vec{r}}{∂v})ΔuΔv$. To calculate the Flux of a vector field $\vec{F}$ over the entire surface S = $\int \int_{D} \vec{F}·d\vec{S} = \int \int_{D} \vec{F}·\hat{\mathbf{n}}dS = \int \int_{D} \vec{F}·(\frac{∂\vec{r}}{∂u}x\frac{∂\vec{r}}{∂v})dudv$
This integral sums up the contributions of the flux through all the infinitesimal surface elements.
The flux of a vector field $\vec{F}$ across a surface S measures how much of the vector field passes through the surface in the direction of the outward normal vector. Specifically, it calculates the component of the vector field that is perpendicular to the surface at each point, summed (integrated) over the entire surface.
Parametrization of the Surface. To compute the flux, we first need to parametrize the surface z = 1 -x^{2} -y^{2}, z ≥ 0. The surface is the upper half of a circular paraboloid opening downwards, with its vertex at (0, 0, 1). The condition z ≥ 0 ensures we are only considering the upper portion of this paraboloid.
Since the surface is radially symmetric, it’s convenient to express the coordinates (x, y, z) in terms of polar coordinates (r, θ), where: x = rcos(θ), y =rsin(θ), z = 1 −r^{2}.
The “shadow” of this bowl-shaped surface with its vertex at (0, 0, 1) is a unit disk, meaning the radial coordinate r (corresponding to the radius in the polar coordinate system) ranges from 0 to 1, and the angular coordinate θ (corresponding to the angle in the polar coordinate system) ranges from 0 to 2π.
The position vector $\vec{r}(r,θ)$ of a point on the surface in terms of the parameters r and θ is given by: $\vec{r}(u,v)=⟨rcos(θ), rsin(θ), 1-r^2⟩$
Plug these values in the vector field. $\vec{F}(r, θ) = ⟨rcos(θ), rsin(θ), 1-r^2⟩$
Compute the Partial Derivatives. To find the vector area element $d\vec{S}$, we need to compute the partial derivatives of $\vec{r}(r, θ)$ with respect to the parameters r and θ: $\frac{∂\vec{r}}{∂r} = ⟨cos(θ), sin(θ), -2r⟩, \frac{∂\vec{r}}{∂θ} = ⟨-rsin(θ, rcos(θ), 0⟩$.
Compute the Cross Product
The cross product of $\frac{∂\vec{r}}{∂u}x\frac{∂\vec{r}}{∂v}$ gives a vector perpendicular to the surface, which represents the surface element vector $d\vec{S}$:
$\frac{∂\vec{r}}{∂r}x\frac{∂\vec{r}}{∂θ} = |\begin{smallmatrix}\vec{i} & \vec{j} & \vec{k}\\ cos(θ) & sin(θ) & -2r\\ -rsin(θ) & rcos(θ) & 0\end{smallmatrix}| = ⟨2r^2cos(θ), 2r^2sin(θ), r⟩$.
Calculate the flux
The flux is calculated by integrating the dot product of the vector field $\vec{F}$ and the vector area element $d\vec{S}$:
Flux = $\int \int_{D} \vec{F}·d\vec{S} = \int \int_{D} \vec{F}·\hat{\mathbf{n}}dS = \int \int_{D} \vec{F}·(\frac{∂\vec{r}}{∂r}x\frac{∂\vec{r}}{∂θ})drdθ = \int \int_{D} ⟨rcos(θ), rsin(θ), 1-r^2⟩⟨2r^2cos(θ), 2r^2sin(θ), r⟩drdθ = \int_{0}^{2π} \int_{0}^{1} (2r^3cos^2(θ)+2r^3sin^2(θ)+r-r^3) drdθ = [\text{Simplifying}] = \int_{0}^{2π} \int_{0}^{1} (2r^3+r-r^3) drdθ = \int_{0}^{2π} \int_{0}^{1} (r^3+r) drdθ$
Evaluate the inner integral (over r): $\int_{0}^{1} (r^3+r) dr = \frac{r^4}{4}+\frac{r^2}{2}\bigg|_{0}^{1} = \frac{1}{4}+\frac{1}{2} = \frac{3}{4}$
Outer Integral (over θ): Flux = $\int_{0}^{2π} \frac{3}{4}dθ = \frac{3}{4}θ\bigg|_{0}^{2π} = \frac{3·2π}{4} = \frac{3π}{2}$.
Thus, the flux of the vector field $\vec{F}=⟨x, y, z⟩$ across the surface z = 1 -x^{2} -y^{2}, where z ≥ 0 is ^{3π}⁄_{2}.
Solution:
The flux measures how much of the vector field passes through the surface in the direction of the outward normal vector.
Parametrization of the Surface. To calculate the flux, we first need to parametrize the surface of the upper hemisphere. The upper hemisphere can be conveniently described using polar coordinates because of its symmetry.
Domain: The “shadow” of the hemisphere on the xy-plane is a unit disk, meaning the radial coordinate u (corresponding to the radius in the polar coordinate system) ranges from 0 to 1, and the angular coordinate v (corresponding to the angle in the polar coordinate system) ranges from 0 to 2π.
Parametrization: The position vector $\vec{r}(u,v)$ of a point on the surface in terms of the parameters u and v is given by: $\vec{r}(u,v)=⟨ucos(v), usin(v), \sqrt{1-u^2}⟩$ where:
Plug these values in the vector field. $\vec{F}(\vec{r}(u, v)) = ⟨-y, -x, 1⟩ = ⟨-usin(v), -ucos(v), 1⟩$.
Compute the Partial Derivatives. We need to find the partial derivatives of $\vec{r}(u,v)=⟨ucos(v), usin(v), \sqrt{1-u^2}⟩$ with respect to u and v:
$\frac{∂\vec{r}}{∂u} = ⟨cos(v), sin(v), \frac{-u}{\sqrt{1-u^2}}⟩, \frac{∂\vec{r}}{∂v} = ⟨-usin(v), ucos(v), 0⟩$. $\frac{∂\vec{r}}{∂u}$ This vector points in the direction of increasing u, with v held constant.
Compute the Cross Product
The cross product of $\frac{∂\vec{r}}{∂u}x\frac{∂\vec{r}}{∂v}$ gives a vector perpendicular to the surface, which represents the surface element vector $d\vec{S}$:
$\frac{∂\vec{r}}{∂u}x\frac{∂\vec{r}}{∂v} = |\begin{smallmatrix}\vec{i} & \vec{j} & \vec{k}\\ cos(v) & sin(v) & \frac{-u}{\sqrt{1-u^2}}\\ -usin(v) & ucos(v) & 0\end{smallmatrix}| = ⟨\frac{u^2cos(v)}{\sqrt{1-u^2}}, \frac{u^2sin(v)}{\sqrt{1-u^2}}, u⟩$. Observe the z-component is u, u is positive, that will point upwards & out ($\hat{\mathbf{n}}$), so it is well-oriented.
Calculate the flux
S = $\int \int_{D} \vec{F}·d\vec{S} = \int \int_{D} \vec{F}·\hat{\mathbf{n}}dS = \int \int_{D} \vec{F}·(\frac{∂\vec{r}}{∂u}x\frac{∂\vec{r}}{∂v})dudv = \int_{0}^{2π}\int_{0}^{1} ⟨-usinv, -ucosv, 1⟩·⟨\frac{u^2cos(v)}{\sqrt{1-u^2}}, \frac{u^2sin(v)}{\sqrt{1-u^2}}, u⟩dudv = \int_{0}^{2π}\int_{0}^{1} (-usin(v)·\frac{u^2cos(v)}{\sqrt{1-u^2}}+-ucosv·\frac{u^2sin(v)}{\sqrt{1-u^2}}+1·u)dudv = \int_{0}^{2π}\int_{0}^{1} (\frac{-2u^3sin(v)cos(v)}{\sqrt{1-u^2}}+u)dudv = \int_{0}^{2π} sin(v)cos(v)dv\int_{0}^{1}\frac{-2u^3}{\sqrt{1-u^2}}du + \int_{0}^{2π}\int_{0}^{1}ududv$
Since sin(v)cos(v) is odd over [0, 2π], the integral of this term over v is zero.
$\int_{0}^{2π}\int_{0}^{1} ududv =\int_{0}^{2π} \frac{u^2}{2}\bigg|_{0}^{1}dv =$
$\int_{0}^{2π} \frac{1}{2}dv = \frac{1}{2}v\bigg|_{0}^{2π} = \frac{1}{2}2π=π$
The flux of the vector field $\vec{F}$ = ⟨−y, −x, 1⟩ across the upper hemisphere of the unit sphere with an outward-pointing normal is π.
Description of the problem.
We are tasked with finding the flux of the vector field $\vec{F}= ⟨x, z, 0⟩$ across the paraboloid z = x^{2} + y^{2}, a circular paraboloid that opens upwards, has its vertex at the origin, and has circular cross-sections perpendicular to the z-axis. As you move upwards along the z-axis, the cross-sections become larger circles. The paraboloid extends infinitely upwards.
Parametrization of the Surface.
To calculate the flux, we first need to parametrize the surface of the paraboloid. A natural parametrization in this case uses cylindrical coordinates because of the symmetry of the paraboloid.
Domain: The “shadow” of the paraboloid on the xy-plane is a circular region.
$\vec{r}(u,v)=⟨ucos(v), usin(v), u^2⟩$.
We calculate the flux using the surface integral: $\int \int_{D} \vec{F}·d\vec{S} = \int \int_{D} \vec{F}·\hat{\mathbf{n}}dS = \int \int_{D} \vec{F}·(\frac{∂\vec{r}}{∂u}x\frac{∂\vec{r}}{∂v})dudv$
Plug these values in the vector field. $\vec{F}(\vec{r}(u, v)) = ⟨x, z, 0⟩ = ⟨ucos(v), u^2, 0⟩$.
Compute the Partial Derivatives. We need to find the partial derivatives of $\vec{r}$ with respect to u and v:
$\frac{∂\vec{r}}{∂u} = ⟨cos(v), sin(v), 2u⟩, \frac{∂\vec{r}}{∂v} = ⟨-usin(v), ucos(v), 0⟩$
Compute the Cross Product
$\frac{∂\vec{r}}{∂u}x\frac{∂\vec{r}}{∂v} = |\begin{smallmatrix}\vec{i} & \vec{j} & \vec{k}\\ cos(v) & sin(v) & 2u\\ -usin(v) & ucos(v) & 0\end{smallmatrix}| = ⟨-2u^2cos(v), -2u^2sin(v), u⟩$. Since 1 ≤ u ≤ 3, the z-component of the cross product u is positive, indicating that the vector $⟨-2u^2cos(v), -2u^2sin(v), u⟩$ points inward (towards the inside of the paraboloid). Because we need an outward-pointing normal vector, we multiply the vector by −1.
Calculate the flux
We calculate the flux using the formula:
S = $\int \int_{D} \vec{F}·d\vec{S} = \int \int_{D} \vec{F}·\hat{\mathbf{n}}dS = \int \int_{D} \vec{F}·(\frac{∂\vec{r}}{∂u}x\frac{∂\vec{r}}{∂v})dudv$ =[We are going to introduce a minus because the orientation] $-\int_{0}^{2π}\int_{1}^{3} ⟨ucos(v), u^2, 0⟩·⟨-2u^2cos(v), -2u^2sin(v), u⟩dudv = -\int_{0}^{2π}\int_{1}^{3} (-2u^3cos^2(v)-2u^4sin(v))dudv = 2\int_{0}^{2π} (\frac{u^4}{4}cos^2(v)+\frac{u^5}{5}sin(v))\bigg|_{1}^{3}dv =$
=$2\int_{0}^{2π} (\frac{81}{4}-\frac{1}{4})cos^2(v)+(\frac{3^5}{5}-\frac{1}{5})sin(v)dv$
Recall that $\int_{0}^{2π} sin(v)dv = 0, \frac{81}{4}-\frac{1}{4} = 20$
= $40\int_{0}^{2π} cos^2(v) = 40\int_{0}^{2π} cos^2(v)dv =$ [Use the identity $cos^2(v)=\frac{1+cos(2v)}{2}] 40(\frac{1}{2}\int_{0}^{2π}dv +\frac{1}{2}\int_{0}^{2π}cos(2v)dv) = $
Recall that $\int_{0}^{2π} cos(2v)dv = 0$
= $20\int_{0}^{2π}dv = 20v\bigg|_{0}^{2π} = 20·2π = 40π.$ Thus, the flux of $\vec{F}= ⟨x, z, 0⟩$ across the paraboloid z = x^{2} + y^{2}, 1 ≤ z ≤ 9 with outward-pointing normal is 40π.
Problem Recap and Surface Description We are tasked with calculating the flux of $\vec{F} = ⟨x, y, 5⟩$ cross the surface of a solid bounded by:
The objective is to compute the flux of the vector field through these surfaces, taking into account that the normal vector must point outward.
Flux Integral. The flux of the vector field $\vec{F}$ across a surface S is given by the surface integral: $\int \int_{D} \vec{F}·d\vec{S} = \int \int_{D} \vec{F}·\hat{\mathbf{n}}dS = \int \int_{D} \vec{F}·(\frac{∂\vec{r}}{∂u}x\frac{∂\vec{r}}{∂v})dudv$
Parametrization of the Cylinder Surface
To compute the flux across the curved surface of the cylinder, we parameterize this surface using cylindrical coordinates.
We parameterize the curved surface of the cylinder. Given x^{2} + z^{2} = 1, we can use $\vec{r}(θ,y)=⟨cos(θ), y, sin(θ)⟩$ where: θ varies between 0 and 2π (it represents the angle around the y-axis), y varies along the height of the cylinder, that is, between 0 (the bottom plane, y = 0) and 2 -x = 2 - cos(θ) (the top plane, z + y = 2).
Compute the Partial Derivatives. We need to find the partial derivatives of $\vec{r}$ with respect to θ and y:
$\frac{∂\vec{r}}{∂θ} = ⟨-sin(θ), 0, cos(θ)⟩, \frac{∂\vec{r}}{∂y} = ⟨0, 1, 0⟩$
Compute the Cross Product
$\frac{∂\vec{r}}{∂θ}x\frac{∂\vec{r}}{∂y} = |\begin{smallmatrix}\vec{i} & \vec{j} & \vec{k}\\ -sin(θ) & 0 & cos(θ)\\ 0 & 1 & 0\end{smallmatrix}| = ⟨-cos(θ), 0, -sin(θ)⟩$. This normal vector points inward, e.g., θ = 0, ⟨-cos(θ), 0, -sin(θ)⟩ = ⟨-1, 0, 0⟩, toward the axis of the cylinder. Since we require an outward-pointing normal, we reverse the direction by multiplying by −1: ⟨cos(θ), 0, sin(θ)⟩.
Calculate the flux
We calculate the flux using the formula:
S_{1} = $\int \int_{D} \vec{F}·d\vec{S} = \int \int_{D} \vec{F}·\hat{\mathbf{n}}dS = \int \int_{D} \vec{F}·(\frac{∂\vec{r}}{∂θ}x\frac{∂\vec{r}}{∂y})dA = \int \int_{D} ⟨cos(θ), y, 5⟩·⟨cos(θ), 0, sin(θ)⟩dA = \int_{0}^{2π}\int_{0}^{2-cos(θ)} (cos^2(θ)+5sin(θ))dydθ$
Evaluate the Inner Integrals:
$\int_{0}^{2-cos(θ)} (cos^2(θ)+5sin(θ))dy = cos^2(θ)y+5sin(θ)y\bigg|_{0}^{2-cos(θ)}=cos^2(θ)(2-cos(θ))+5sin(θ)(2-cos(θ))$
Evaluate the Outer Integrals:
Flux_{S1} = 2π + 0 + 0 + 0 = 2π.
S_{2} is the cap of the cylinder, y = 0. The vector field in this surface is $\vec{F} = ⟨x, 0, 5⟩$, and the normal vector $\hat{\mathbf{n}} = ⟨0, -1, 0⟩$ (the outward-pointing normal vector for this surface, $⟨\frac{∂y}{∂x}, -1, \frac{∂y}{∂z}⟩$).
Flux_{S2} = $\int \int_{S_2} \vec{F}·\hat{\mathbf{n}}dS = \int \int_{S_2} ⟨x, 0, 5⟩·⟨0, -1, 0⟩dS = \int \int_{S_2} 0 = 0.$
S_{3} is the plane x + y = 2 which intersects the cylinder. y = 2 -x. The vector field in this surface is $\vec{F} = ⟨x, y, 5⟩ = ⟨x, 2 - x, 5⟩$, and the normal vector $\hat{\mathbf{n}} = ⟨-\frac{∂(2-x)}{∂x}, 1, -\frac{∂(2-x)}{∂z}⟩ = ⟨1, 1, 0⟩$ derived from the gradient of the plane x +y −2 = 0.
Flux_{S3} = $\int \int_{S_3} \vec{F}·\hat{\mathbf{n}}dS = \int \int_{S_3} ⟨x, 2 -x, 5⟩·⟨1, 1, 0⟩dS = \int \int_{S_3} (x +2 -x)dA = 2\int \int_{D_3} dA = $ 2·Area(D_{3}) =[D_{3} is the projection of the surface onto the xz plane, which is a circle of radius 1, giving an area of π] 2π
Thus, the flux of $\vec{F} = ⟨x, y, 5⟩$ over the surface of the solid enclosed by x^{2}+z^{2} = 1, y = 0, and x + y = 2 is 2π + 0 + 2π = 4π.
When dealing with surfaces in three-dimensional space, it’s often necessary to compute surface area elements, which are infinitesimal areas on the surface. These surface elements can be determined using the normal vector to the surface.
Normal vector to the Surface. A normal vector to a surface is a vector that is perpendicular (orthogonal) to the surface at a given point. When we know a normal vector $\vec{N}$ to a surface (not necessarily a unit vector), we can use it to determine the surface area element. For example:
The gradient ∇g points in the direction of the steepest ascent of the function g, and hence is perpendicular to the surface.
Relation Between ΔS and ΔA
To understand how to calculate the surface area element ΔS using the normal vector, let’s consider a surface S that is slanted relative to a reference plane, such as the xy-plane. We want to relate ΔS (a small patch of the actual surface) to the area in the xy-plane ΔA. ΔA is a projection of ΔS onto the xy-plane, which is easier to calculate directly.
The area ΔA and the actual surface area ΔS are related by the angle α between the normal vector $\vec{N}$ to the surface and the normal vector to the xy-plane, which is the $\hat{\mathbf{k}}$ vector (the unit vector in the z-direction).
This relationship is given by: ΔA = ΔS⋅cos(α). Here cos(α) is the cosine of the angle between the normal vector $\vec{N}$ to the surface and the z-axis $\hat{\mathbf{k}}$. This is illustrated in Figure A.
The angle α is the angle between the two vectors perpendiculars to and can be computed using the dot product between the normal vector $\vec{N}$ and the unit vector $\hat{\mathbf{k}}$. Recall that $\vec{u}·\vec{v}=|\vec{u}||\vec{v}|cos(θ)$ where θ is the angle between the vectors u and v. For our case, cos(α) = $\frac{\vec{N}\hat{\mathbf{k}}}{|\vec{N}||\mathbf{k}|} = \frac{\vec{N}\hat{\mathbf{k}}}{|\vec{N}|}$, hence ΔS = $\frac{1}{cos(α)}ΔA = \frac{|\vec{N}|}{\vec{N}\hat{\mathbf{k}}}ΔA$
Area Element with Unit Normal Vector. $\hat{\mathbf{n}}dS = \hat{\mathbf{n}}\frac{|\vec{N}|}{\vec{N}\hat{\mathbf{k}}}dA$ [Since $\hat{\mathbf{n}}$ is a unit vector perpendicular to the surface, i.e., $\hat{\mathbf{n}} = \frac{\vec{N}}{|\vec{N}|}$, the expression simplifies to] ±$\frac{\vec{N}}{\vec{N}\hat{\mathbf{k}}}dA$. Finally, we obtain $\hat{\mathbf{n}}dS = ±\frac{\vec{N}}{\vec{N}\hat{\mathbf{k}}}dxdy$. The sign depends on the orientation of the surface.
Consider a specific surface S defined explicitly as z = f(x, y) or can also be written as an implicit function g(x, y, z) = z - f(x, y) = 0.
The normal vector to this surface is given by the gradient of g: $\vec{N} = ∇g = ⟨-\frac{∂f}{∂x}, -\frac{∂f}{∂y}, 1⟩ = ⟨-f_x, -f_y, 1⟩$.
We compute the surface area element, $\hat{\mathbf{n}}dS = \frac{\vec{N}}{\vec{N}\hat{\mathbf{k}}}dxdy = \frac{⟨-f_x, -f_y, 1⟩}{⟨-f_x, -f_y, 1⟩\hat{\mathbf{k}}}dxdy = \frac{⟨-f_x, -f_y, 1⟩}{⟨-f_x, -f_y, 1⟩⟨0, 0, 1⟩}dxdy = \frac{⟨-f_x, -f_y, 1⟩}{1}dxdy = ⟨-f_x, -f_y, 1⟩dxdy.$
Therefore, the formula for the surface area element for the surface z = f(x, y) is: dS = ⟨-f_{x}, -f_{y}, 1⟩dxdy