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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).
Triple integrals are a powerful tool in multivariable calculus with a wide range of applications in physics, engineering, and other fields. These integrals allow us to calculate properties of three-dimensional objects such as mass, moments of inertia, electric charge, and volumes.
The mass of a three-dimensional object can be calculated using triple integrals when the object’s density is known. Density, denoted as δ, is a function that describes how mass is distributed within the object. The relationship between mass m, volume V, and density δ is given by: $δ = \frac{m}{V}⇒dm = δ·dV.$
To calculate the mass of a three-dimensional solid object occupying a region R in space, we use the triple integral $Mass = \int \int \int_{R} δ·dV$ where δ is the density of the object and dV represents an infinitesimal volume.
The integral sums the contributions of mass from each infinitesimal volume element over the entire region R.
If the density δ is constant and equal to 1, the mass is simply the volume of the region, $V = \int \int \int_{R} dV$.
The average value of a function of three variables f(x, y, z) in a region R, say occupied by a three-dimensional solid, can be calculated using a triple integral. This average value represents the “mean” value of the function across the entire region. It is given by: $\overline {f} = \frac{1}{Vol(R)}·\int \int \int_{R} f·dV$ where Vol(R) is the volume of the region R.
If the solid has a varying density and we can calculate the weighted average of the function f(x, y, z) by incorporating the density into the integral, so the calculation becomes, $\overline {f} = \frac{1}{Mass(R)}·\int \int \int_{R} f·δ·dV$. This weighted average takes into account how the density varies throughout the region.
The center of mass of a three-dimensional body is the point where the entire mass of the body can be considered to be concentrated. It represents the average position of all the mass in the object, weighted by the mass density. The coordinates $(\overline {x}, \overline {y}, \overline {z})$ of the center of mass are calculated by the following formulas $\overline {x} = \frac{1}{Mass(R)}·\int \int \int_{R} x·δ·dV, \overline {y} = \frac{1}{Mass(R)}·\int \int \int_{R} y·δ·dV, \overline {z} = \frac{1}{Mass(R)}·\int \int \int_{R} z·δ·dV$.
These integrals calculate the average position of the mass along each coordinate axis. This allow us to describe the position of the object as a whole, such as the point where the object would balance if placed on a perfectly frictionless support, e.g., the point where you would put your finger to balance your ruler. For a body with uniform density, the center of mass coincides with its geometric center.
Moment of Inertia is the tendency of a body to maintain its state of rotation or rest. If the body is in rest, it tries to remain at rest and resist any force trying to bring it into motion. If the body is in rotational motion, it is the tendency which opposes the change in its rotational motion due to external forces, the body tries to continue in motion and resist any force trying to bring it to change its motion.
The moment of inertia of a body about a given axis is a measure of a body’s resistance to change in its rotational motion about that axis. It depends on both the mass distribution of the object and the distance of each mass element from the axis of rotation. The general formula for the moment of inertia is I = mr^{2} where m is the mass of the object and r is the distance from the axis of rotation. The general formula for the moment of inertia I about an axis is: I = $\int \int \int_{R} (\text{distance to axis})^2·δ·dV$.
Higher moment of inertia implies that it is harder to change the body’s rotational motion about the axis (more torque is required to change the rotational speed of the object).
The moment of inertia of a body about the z-axis, $I_z = \int \int \int_{R} r^2·δ·dV$ =[Cartesian coordinates] $\int \int \int_{R} (x^2+y^2)·δ·dV$. By symmetry, $I_y = \int \int \int_{R} (x^2+z^2)·δ·dV, I_x = \int \int \int_{R} (y^2+z^2)·δ·dV$
These integrals calculate the distribution of mass relative to each axis, providing a comprehensive understanding of the body’s resistance to rotational changes.
Understanding the problem. The goal is to set up the integral for z over the region E using Cartesian coordinates. We need to understand the boundaries of E in three-dimensional space:
Limits of Integration
Set up the limit (Cartesian Coordinates). Given the ranges for x, y, and z, the triple integral in Cartesian coordinates is: $\int\int\int_{E} zdV = \int_{0}^{1} \int_{3x}^{3} \int_{0}^{\sqrt{9-y^2}} zdzdydx$
Understanding the problem. We need to understand the boundaries of our solid in three-dimensional space:
Limits of Integration
Setting Up the Triple Integral. The volume V of the solid can be found using the following triple integral in Cartesian coordinates:
V = $\int_{-1}^{1} \int_{-\sqrt{4-4x^2}}^{\sqrt{4-4x^2}} \int_{0}^{z+2} dydzdx.$
Understanding the Problem. The volume we seek is bounded
z = 4: 4 = 6 - x² - y² => x² + y² = 2. This is a circle of radius √2 centered at the origin in the plane z = 4. z = 2: 2 = 6 - x² - y² => x² + y² = 4. This is a circle of radius 2 centered at the origin in the plane z = 2.
Determining the Limits of Integration
Setting up the triple integral. The volume V can be expressed as a triple integral in cylindrical coordinates: V = $\int_{0}^{2π} \int_{0}^{2} \int_{r}^{6-r^2} dzrdrdθ$
Evaluating the Triple Integral
Evaluating the inner integral with respect to z: $\int_{r}^{6-r^2} dz = z\bigg|_{r}^{6-r^2}=6-r^2-r$
Evaluating the middle integral with respect to r: $\int_{0}^{2} (6-r^2-r)rdr = \int_{0}^{2} (6r-r^3-r^2)dr = 3r^2-\frac{r^4}{4}-\frac{r^3}{3}\bigg|_{0}^{2} = 3·4-\frac{16}{4}-\frac{8}{3} = 12 -4 -\frac{8}{3} = 8 -\frac{8}{3} = \frac{24}{3} -\frac{8}{3} = \frac{16}{3}$
Evaluating the outer integral with respect to θ: $\int_{0}^{2π} \frac{16}{3}dθ = \frac{16}{3}θ\bigg|_{0}^{2π} = \frac{16}{3}·2π = \frac{32π}{3}$.
Thus, the volume of the region between the cone and the paraboloid is: V = $\frac{32π}{3}$.
Understanding the problem. The region of interest is bounded by
z = 7: 7 = 9 - x² - y² => x² + y² = 2. This is a circle of radius √2 centered at the origin in the plane z = 7. z = 5: 5 = 9 - x² - y² => x² + y² = 4. This is a circle of radius 2 centered at the origin in the plane z = 5.
The density function varies with z, specifically f(r, θ, z) = 12 −z, which means the density decreases as z increases.
Setting up the integral. We will use cylindrical coordinates (r, θ, z) where x = rcos(θ), y = rsin(θ), z = z, r is the radial distance from the z-axis, and θ is the angle in the xy-plane measured counterclockwise from the positive x-axis.
The bounds are:
Given these bounds, the mass M of the solid is given by the triple integral: $\int_{0}^{2π}\int_{0}^{3}\int_{0}^{9-r^2} (12 -z)dzrdrdθ$
Evaluating the integral.
Evaluating the inner integral with respect to z: $\int_{0}^{9-r^2} (12 -z)dz = 12z -\frac{z^2}{2}\bigg|_{0}^{9-r^2}= 12(9-r^2)-\frac{(9-r^2)^2}{2} = 108-12r^2-\frac{1}{2}(81-18r^2+r^4) = 108-12r^2-\frac{81}{2}+9r^2-\frac{1}{2}r^4 = \frac{135}{2}-3r^2-\frac{1}{2}r^4$
Evaluating the middle integral with respect to r: $\int_{0}^{3} (\frac{135}{2}r-3r^3-\frac{1}{2}r^5)dr = \frac{135}{4}r^2-\frac{3}{4}r^4-\frac{1}{12}r^6\bigg|_{0}^{3} = \frac{135}{4}·9-\frac{3}{4}·81-\frac{1}{12}·729 = \frac{1215}{4}-\frac{243}{4}-\frac{243}{4} = \frac{729}{4}$
Evaluate the outer integral with respect to θ: $\int_{0}^{2π} \frac{729}{4}dθ = \frac{729}{4}θ\bigg|_{0}^{2π} = \frac{729π}{2}$
Thus, the mass of the solid is M = $\frac{729π}{2}$.
The mass M of the solid is: M=364.5π.
Understanding the Problem.
z = ar. The value of ‘a’ determines the slope of the surface. If ‘a’ is positive, the surface will slope upwards as ‘r’ increases. Besides, the height ‘z’ is directly proportional (increases linearly) to the radial distance ‘r’. This means that for a fixed angle θ, the surface will form a straight line. As you rotate this plane around the z-axis (changing θ), the straight line will sweep out a cone-like shape. The surface is a cone with its vertex at the origin (0, 0, 0).
The moment of inertia I_{z} about the z-axis is a measure of how the mass of the cone is distributed related to the axis. It is given by the triple integral: $I_z = \int \int \int_{R} r^2·δ·dV$ where r is the radial distance from the z-axis, δ is the density function, and dV is the differential volume element.
Given that the density δ = 1, this simplifies to: $I_z = \int \int \int_{R} r^2·dV = \int \int \int_{R} r^2dV$
In cylindrical coordinates, where dV = rdrdθdz, the integral becomes: $\int \int \int_{R} r^3drdθdz$.
Determining the Limits of Integration
The region R is bounded by the cone z = ar and the plane z = b. We need to set up our limits for r, θ, and z. Fixing z at a particular value, we get a circle with radius r = z/a.
Setting Up the Triple Integral. With these limits, the triple integral in cylindrical coordinates is: $I_z = \int_{0}^{b} \int_{0}^{2π} \int_{0}^{\frac{z}{a}} r^3drdθdz$
Evaluating the Integral
Evaluating the Inner Integral. First, we integrate with respect to r: $\int_{0}^{\frac{z}{a}} r^3dr = \frac{r^4}{4}\bigg|_{0}^{\frac{z}{a}} = \frac{(\frac{z}{a})^4}{4} = \frac{z^4}{4a^4}$
Evaluating the middle integral. Next, we integrate with respect to θ: $\int_{0}^{2π} \frac{z^4}{4a^4} dθ = \frac{z^4}{4a^4}\int_{0}^{2π} dθ = \frac{z^4}{4a^4}(θ\bigg|_{0}^{2π}) = \frac{z^4π}{2a^4}$
Evaluating the Outer Integral. Finally, we integrate with respect to z: $I_z = \int_{0}^{b} \frac{z^4π}{2a^4}dz = \frac{z^5π}{5·2·a^4}\bigg|_{0}^{b} = \frac{πb^5}{10a^4}$
Therefore, the moment of inertia I_{z} of the solid cone about the z-axis is: $I_z = \frac{πb^5}{10a^4}$∎
Understanding the problem. The solid is above the xy-plane (z = 0), under the cone z = $\sqrt{x^2+y^2}$ which open upwards from the origin. This cone has its vertex at the origin and its axis along the z-axis (z = 1: 1 = √(x² + y²) => x² + y² = 1. This is a circle of radius 1 centered at the origin in the plane z = 1. z = 2: 2 = √(x² + y²) => x² + y² = 4. This is a circle of radius 2 centered at the origin in the plane z = 2) and inside the sphere centered at the origin with a radius r = 2, x² + y² + z² = 4.
Intersection: To find the intersection between the cone and the sphere, z^{2} = x^{2} + y^{2}, 2x^{2} + 2y^{2} = 4 ⇒ x^{2} + y^{2} = 2 ↭ r = √2, this is a circle of radius √2 in the xy-plane (Refer to Figure B for a visual representation and aid in understanding it)
Setting Up the Triple Integral. In cylindrical coordinates, where x = rcosθ, y = rsinθ, and z = z, the volume element dV is expressed as dV = rdzdrdθ. The function to be integrated f(x, y, z) in cylindrical coordinates becomes f(r, θ, z).
The region E can be split into two parts:
Region below the cone and inside the sphere (green area):
The integral for this part is: $\int_{0}^{2π}\int_{0}^{\sqrt{2}}\int_{0}^{r} f(r, θ, z) rdzdrdθ$
Region inside and below the sphere (blue area):
Combining both parts, the total volume integral over the region E is: $\int\int\int_{E} f(x, y, z)dV = \int_{0}^{2π}\int_{0}^{\sqrt{2}}\int_{0}^{r} f(r, θ, z) rdzdrdθ + \int_{0}^{2π}\int_{0}^{\sqrt{2}}\int_{0}^{r} f(r, θ, z) rdzdrdθ$
Understanding the problem. The unit sphere x^{2}+ y^{2}+z^{2} = 1 is a sphere centered at the origin with a radius of 1. The region of interest lies above the plane z = 1 −y and inside the sphere.
Plane z = 1 -y. z-intercept: When x = 0 and y = 0, z = 1. This gives us the point (0, 0, 1). y-intercept: When x = 0 and z = 0, we have 0 = 1 - y, so y = 1. This gives us the point (0, 1, 0). The equation shows that z decreases by 1 unit for every 1 unit increase in y. This means the plane has a slope of -1 in the y-direction.
Determining the Limits of Integration:
The integral in Cartesian coordinates becomes:
$I_z = \int_{0}^{1} \int_{-\sqrt{2y -2y^2}}^{\sqrt{2y -2y^2}} \int_{1-y}^{\sqrt{1-x^2-y^2}} dzdxdy$, this is not an easy integral, so it will be wise to use symmetry and it is left to the reader to do so.
Given the complexity of evaluating these integrals directly, using symmetry and potentially switching to cylindrical coordinates (x = rcos(θ), y = rsin(θ), z = z) simplifies the problem.
Limits in Cylindrical Coordinates:
Set Up the New Integral: $\int_{0}^{2π}\int_{0}^{1}\int_{1-rsin(θ)}^{\sqrt{1-r^2}} rdzdrdθ$. It is left for the reader as a difficult exercise.