Life is a math equation. In order to gain the most, you have to know how to convert negatives into positives, Anonymous.
Pure mathematics is, in its way, the poetry of logical ideas, Albert Einstein.
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).
Spherical coordinates are a system used to represent points in three-dimensional space using three parameters: the radial distance r (also denoted as ρ), the polar angle θ, and the azimuthal angle ϕ. In other words, it describes or specifies the position of a point in a three-dimensional space by three numbers (r, θ, φ) (Figure i). This system is particularly effective for problems involving spherical symmetry, such as those involving spheres or radial distributions. Understanding spherical coordinates provides a powerful tool for solving complex problems in multivariable calculus and physics.
They make it simple to describe a sphere (ρ = a are spheres of radius a centered at the origin), just as cylindrical coordinates make it easy to describe a cylinder (Figure ii).
Let’s consider the plane of the visual field as being a representation of the xy plane, Figure 3, and we get z = ρcos(φ), r = ρsin(φ) ⇒[Cylindrical conversion formulas, x = rcos(θ), y = rsin(θ), z = z] Spherical coordinates can be converted to Cartesian coordinates using the following formulas: x = rcos(θ) = ρsin(φ)cos(θ), y = rsin(θ) = ρsin(φ)sin(θ), and z = ρcos(φ).
These relationships describe how a point in spherical coordinates maps onto the x, y, and z axes of Cartesian coordinates. The inverse ransformations are given by, ρ = $\sqrt{x^2+y^2+z^2}, tan(θ) = \frac{y}{x}, cos(φ) = \frac{z}{\sqrt{x^2+y^2+z^2}}$
φ = ^{π}⁄_{4} is a cone where the points on the surface have a constant angle φ from the z-axis.(Figure iv), φ = ^{π}⁄_{2} is the xy plane, where all points lie equidistantly from the z-axis.
When performing triple integrals in spherical coordinates, it’s crucial to use the correct volume element dV. The volume element in spherical coordinates is: dV = ρ^{2}sin(φ)dρdφdθ. It can be derived from the Jacobian determinant of the transformation from Cartesian to spherical coordinates or graphically (Figure v).
Understanding spherical coordinates and their relationship to Cartesian coordinates is crucial for tackling complex problems in multivariable calculus, particularly those involving radial symmetry.
Understanding the problem We need to integrate the density function over the volume of the sphere in spherical coordinates. The volume element in spherical coordinates is dV = ρ^{2}sin(φ)dρdφdθ.
Boundaries in Spherical Coordinates:
Setting Up the Integral The integral to find the mass M is:
$\int_{0}^{2π} \int_{0}^{π} \int_{0}^{4} (5 -ρ)ρ^2sin(φ)dρdφdθ = \int_{0}^{2π} \int_{0}^{π} \int_{0}^{4} (5ρ^2-ρ^3)sin(φ)dρdφdθ$
Evaluating the problem
Calculating the inner integral with respect to ρ: $\int_{0}^{4} (5ρ^2-ρ^3)dρ = (\frac{5}{3}ρ^3-\frac{1}{4}ρ^4)\bigg|_{0}^{4} = (\frac{5}{3}64-\frac{1}{4}256) = \frac{320}{3}-64=\frac{128}{3}$
Calculating the middle integral with respect to φ: $\int_{0}^{π} \frac{128}{3}sin(φ)dφ = -\frac{128}{3}cos(φ)\bigg|_{0}^{π} = -\frac{128}{3}(-1-1)=\frac{256}{3}$
Calculating the outer integral: $M = \int_{0}^{2π} \frac{256}{3} dθ = \frac{256}{3}θ\bigg|_{0}^{2π} = \frac{512}{3}π$. This result represents the total mass of the sphere with the given variable density, computed by integrating over the entire volume using spherical coordinates.
Understanding the problem: The unit sphere is defined by the equation: x^{2} + y^{2} + z^{2} = 1. In spherical coordinates, where ρ is the radial distance, ϕ is the polar angle, and θ is the azimuthal angle, this equation becomes: ρ=1. The plane is given by: z = $\frac{1}{\sqrt{2}}$. In spherical In spherical coordinates, z = ρcos(ϕ), so the plane equation becomes: ρcos(ϕ)= $\frac{1}{\sqrt{2}}$
Setting Up the Integral To set up the integral, we need to determine the limits of integration in spherical coordinates. The volume element in spherical coordinates is given by dV = ρ^{2} sin(ϕ)dρdϕdθ.
Determining the Limits of Integration
The lower bound for ϕ (Figure B) is the point of the intersection between the sphere ρ = 1 and the plane $z = \frac{1}{\sqrt{2}} = ρcos(φ)⇒cos(φ) = \frac{1}{\sqrt{2}} ⇒ φ = \frac{π}{4}$
V = $\int_{0}^{2π} \int_{0}^{\frac{π}{4}} \int_{\frac{sec(φ)}{\sqrt{2}}}^{1} ρ^2sin(φ)dρdφdθ = \frac{2π}{3}-\frac{5π}{6\sqrt{2}}$
Evaluating the Integral
Integrating with respect to ρ: $\int_{\frac{\sec(\phi)}{\sqrt{2}}}^{1} \rho^2 = \frac{1}{3} \rho^3 \bigg|_{\frac{\sec(\phi)}{\sqrt{2}}}^{1}$
Integrate wit respect to φ:
$\int_{0}^{\frac{\pi}{4}} \frac{1}{3} \sin(φ) - \frac{1}{3·2·\sqrt{2}} \sec^3(φ) \sin(φ)dφ$
Let’s break this down into two parts:
$\frac{1}{3} \int_{0}^{\frac{\pi}{4}} \sin(φ) dφ = \frac{1}{3}[-cos(φ)]\bigg|_{0}^{\frac{\pi}{4}} = \frac{1}{3}[-cos(\frac{\pi}{4})+1]=\frac{1}{3}(\frac{-\sqrt{2}}{2}+1)$
$- \frac{1}{3·2·\sqrt{2}} \int_{0}^{\frac{\pi}{4}} \sec^3(φ) \sin(φ)dφ$
$\int_{0}^{\frac{\pi}{4}} \sec^3(φ) \sin(φ)dφ = \int_{0}^{\frac{\pi}{4}} \frac{sin(φ)}{cos^3(φ)} = \int_{0}^{\frac{\pi}{4}} tan(φ)sec^2(φ)dφ$
Substitution. Let u = sec(φ), du = sec(φ)tan(φ)dφ
d/dx [sec(x)] = d/dx [1/cos(x)] = [(cos(x) * 0) - (1 * -sin(x))] / cos²(x) = sin(x) / cos²(x) = (1/cos(x)) * (sin(x)/cos(x)) = sec(x)tan(x)
$\int_{0}^{\frac{\pi}{4}} tan(φ)sec^2(φ) = \int_{0}^{\frac{\pi}{4}} udu = \frac{u^2}{2} = \frac{sec^2(φ)}{2}\bigg|_{0}^{\frac{\pi}{4}} = 1-\frac{1}{2} = \frac{1}{2}$.
Finally, the outer integral is:
$\int_{0}^{2π} \frac{1}{3}(\frac{-\sqrt{2}}{2}+1)- \frac{1}{3·2^2·\sqrt{2}} dθ = 2π[ \frac{1}{3}(\frac{-\sqrt{2}}{2}+1)- \frac{1}{12·\sqrt{2}} ] = 2π[\frac{-\sqrt{2}+2}{3}- \frac{1}{12·\sqrt{2}}] = 2π[\frac{-2\sqrt{2}+4-1}{12\sqrt{2}}] = 2π[\frac{-2\sqrt{2}+3}{12\sqrt{2}}] = \frac{-4π\sqrt{2}+6π}{12\sqrt{2}} = \frac{-π}{3} + \frac{π}{2\sqrt{2}}$
We are asked to find the volume of the region that lies inside both a cone and a sphere. The cone has an angle φ = ^{π}/_{3} with the z-axis, and the sphere has a radius ρ = 6. The spherical coordinates (ρ, φ, θ) are used to describe this region.
φ = ^{π}⁄_{3} describes the region above a cone opening upwards from the origin, with the cone’s surface making an angle of ^{π}⁄_{3} with the z-axis.
Setting Up the Integral We need to integrate the density function over the volume of the sphere in spherical coordinates. The volume element in spherical coordinates is dV = ρ^{2}sin(φ)dρdφdθ.
Boundaries in Spherical Coordinates:
The volume V of the region can be found by integrating the volume element over the specified ranges of ρ, φ, and θ: $\int_{0}^{2π}\int_{0}^{\frac{π}{3}}\int_{0}^{6} ρ^2sin(φ)dρdφdθ$
Calculate the inner integral with respect to ρ: $\int_{0}^{6} ρ^2dρ = \frac{ρ^3}{3}\bigg|_{0}^{6} = \frac{216}{3} = 72$
Calculate the middle integral with respect to φ: $\int_{0}^{\frac{π}{3}} 72sin(φ)dφ = 72(-cos(φ))\bigg|_{0}^{\frac{π}{3}} = 72(-cos(\frac{π}{3})+cos(0)) = 72(-\frac{1}{2}+1) = 72·\frac{1}{2} = 36$
Compute the outer integral: $\int_{0}^{2π} 36dθ = 36θ\bigg|_{0}^{2π} = 72π$
The volume V of the region inside both the cone φ = π/3 and the sphere ρ = 6 is: V=72π.
Understanding the boundaries
$\sqrt{3(x^2+y^2)} ≤ z ≤ \sqrt{16-x^2-y^2}$ represents the region between a cone $z=\sqrt{3(x^2+y^2)}$ with its vertex at the origin and opening upwards and a sphere of radius 4 (x^{2}+y^{2}+z^{2} = 16) (Figure 2).
$z = \sqrt{3(x^2+y^2)}$ is a cone with its vertex at the origin and its axis along the z-axis. The constant 3 under the square root affects the “steepness” of the cone. (1) z = √3: √3 = √(3(x² + y²)) => x² + y² = 1. This is a circle of radius 1 centered at the origin in the plane z = √3. (2) z = 2√3: 2√3 = √(3(x² + y²)) => x² + y² = 4. This is a circle of radius 2 centered at the origin in the plane z = 2√3.
$0 ≤ x ≤ \sqrt{4-y^2}$ and 0 ≤ y ≤ 2, is a quarter circle of radius 2 (positive x and y).
Spherical coordinates: x=ρsin(φ)cos(θ), y=ρsin(φ)sin(θ), z=ρcos(φ).
Spherical Coordinates Boundaries
Setting Up the Integral in Spherical Coordinates. The volume element in spherical coordinates is: dV = ρ^2sin(φ)dρdθdφ, the integrand $x^2+y^2+z^2$ becomes ρ^{2} in spherical coordinates. Therefore, the integral becomes,
$\int_{0}^{2} \int_{0}^{\sqrt{4-y^2}} \int_{\sqrt{3(x^2+y^2)}}^{\sqrt{16-x^2-y^2}} (x^2+y^2+z^2)dzdxdy =[dxdydz = dV = ρ^2sin(φ)dρdθdφ] \int_{0}^{\sqrt{\frac{π}{6}}} \int_{0}^{\frac{π}{2}} \int_{0}^{4} (ρ^2)ρ^2sin(φ)dρdθdφ$
Evaluating the Integral
Inner integral with respect to ρ: $\int_{0}^{4} (ρ^4)dρ = \frac{ρ^5}{5}\bigg|_{0}^{4} = \frac{4^5}{5} = \frac{1024}{5}$
Middle integral with respect to θ: $\int_{0}^{\frac{π}{2}} \frac{1024}{5}θ\bigg|_{0}^{\frac{π}{2}} = \frac{1024}{5}·\frac{π}{2} = \frac{1024π}{10} = \frac{512π}{5}$
Outer integral with respect to φ: $\int_{0}^{\sqrt{\frac{π}{6}}}·\frac{512π}{5}sin(φ)dφ = \frac{512π}{5}\int_{0}^{\sqrt{\frac{π}{6}}}sin(φ)dφ$
$-cos(φ)\bigg|_{0}^{\frac{π}{6}} = -cos(\frac{π}{6})+cos(0)= -\frac{\sqrt{3}}{2}+1$
The final result for the volume is: $\frac{512π}{5}(1-\frac{\sqrt{3}}{2})$
We are asked to evaluate the triple integral $\int\int\int_{E} (x^2+y^2)dV$ where E is the solid region below the sphere defined by x^{2} + y^{2} + z^{2} = 1 and above the xy-plane (Refer to Figure i for a visual representation and aid in understanding it).
Spherical coordinates: x=ρsin(φ)cos(θ), y=ρsin(φ)sin(θ), z=ρcos(φ). The integrand $x^2+y^2 = (ρsin(φ)cos(θ))^2+(ρsin(φ)sin(θ))^2 = ρ^2sin^2(φ)(cos^2(θ)+sin^2(θ))= ρ^2sin^2(φ) = (ρsin(φ))^2$, dV = ρ^{2}sin(φ)dρdθdφ
Spherical Coordinates Boundaries
$\int_{0}^{2π} \int_{0}^{\frac{π}{2}} \int_{0}^{1} (ρsin(φ))^2 ρ^2sin(φ)dρdφdθ$
Evaluate the Integral
Integrating with respect to ρ: $\int_{0}^{1} ρ^4dρ = \frac{ρ^5}{5}\bigg|_{0}^{1} = \frac{1}{5}$
Integrating with respect to ρ: $\int_{0}^{\frac{π}{2}} sin^3(φ)dφ =\int_{0}^{\frac{π}{2}} sin(φ)(1-cos^2(φ))dφ = \int_{0}^{\frac{π}{2}} sin(φ)dφ - \int_{0}^{\frac{π}{2}} sin(φ)cos^2(φ)dφ$
The first integral is straightforward: $\int_{0}^{\frac{π}{2}} sin(φ)dφ = -cos(φ)\bigg|_{0}^{\frac{π}{2}}= 1$
For the second integral, using the substitution φ = cos(φ), du=−sin(φ)dφ: $\int_{0}^{\frac{π}{2}} sin(φ)cos^2(φ)dφ = \int_{0}^{1} -u^2du = \frac{u^3}{3} = \frac{1}{3}$
Thus, $\int_{0}^{\frac{π}{2}} sin^3(φ)dφ = 1 - \frac{1}{3} = \frac{2}{3}$.
Integrating with respect to θ: $\int_{0}^{2π} \frac{1}{5}\frac{2}{3}dθ = \frac{2}{15}θ\bigg|_{0}^{2π} = \frac{2}{15}·2π = \frac{4π}{15}$.
In rectangular (Cartesian) coordinates, the cylinder is aligned with the z-axis, and its equation is defined by the following constraints:
Explanation of the limits:
Thus, the volume of the cylinder can be calculated using the following triple integral: $\int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \int_{0}^{3} dzdydx$
Cylindrical coordinates (r, θ, z) are naturally suited for problems involving cylinders. The volume element in cylindrical coordinates is dV = rdzdrdθ, and the equations of the cylinder in these coordinates are:
The volume of the cylinder can be calculated using the following triple integral in cylindrical coordinates: $\int_{0}^{2π} \int_{0}^{2} \int_{0}^{3} rdzdrdθ$
In spherical coordinates (ρ, φ, θ), the integral setup for the volume of a cylinder is more complex because spherical coordinates are not inherently suited for cylindrical shapes. The volume element in spherical coordinates is dV = ρ^{2}sin(φ)dρdφdθ.
To compute the volume of the cylinder, we divide the integral into two parts: one for the top of the cylinder and one for the side of the cylinder.
Part 1: Volume Contribution from the Top of the Cylinder
The integral representing the volume from this part is: $\int_{0}^{2π} \int_{0}^{tan^{-1}(\frac{2}{3})} \int_{0}^{3sec(φ)} ρ^2sin(φ)dρdφdθ$
Part 2: Volume Contribution from the Side of the Cylinder
$\int_{0}^{2π} \int_{tan^{-1}(\frac{2}{3})}^{\frac{π}{2}} \int_{0}^{2csc(φ)} ρ^2sin(φ)dρdφdθ$
The total volume of the cylinder is the sum of these two integrals:
$\int_{0}^{2π} \int_{0}^{tan^{-1}(\frac{2}{3})} \int_{0}^{3sec(φ)} ρ^2sin(φ)dρdφdθ + \int_{0}^{2π} \int_{tan^{-1}(\frac{2}{3})}^{\frac{π}{2}} \int_{0}^{2csc(φ)} ρ^2sin(φ)dρdφdθ$
The gravitational force is a fundamental force that attracts any two objects with mass. According to Newton’s law of universal gravitation, the force between two point masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. The formula for the gravitational force is: $F = G\frac{m_1m_2}{r^2}$
where:
The goal is to calculate the gravitational force $\vec{F}$ exerted by an object on a mass m located at the origin.
To calculate the gravitational force exerted by a small differential mass ΔM located at coordinates (x, y, z) on a mass m at the origin, we use the formula:
$|\vec{F}| = G\frac{ΔM·m}{ρ^2},\text{ and the direction is given by } dir(\vec{F}) = \frac{⟨x, y, z⟩}{ρ}$ where:
Thus, the vector form of the gravitational force is: $\vec{F} = \frac{G·ΔM·m}{ρ^3}⟨x, y, z⟩$ (Figure C).
To find the gravitational force exerted by the entire object, we integrate this gravitational force over the entire volume occupied by the object.
Given that ΔM = δ· ΔV, where δ is the density and ΔV is the volume element, we can express the total gravitational force as: $\vec{F} = \int \int \int \frac{G·m·⟨x, y, z⟩}{ρ^3}δdV$
This integral can be complex, so it is useful to simplify by using spherical coordinates. In spherical coordinates, z = ρcos(φ), and the volume element is dV = ρ^{2}sin(φ)dρdφdθ. If we place the solid in a way that the z-axis is an axis of symmetry, the gravitational force will be directed along the z-axis -Figure D-, that is, it will reduce the force vector to $\vec{F}=⟨0, 0, F_z⟩$
Therefore, the integral simplifies to: $F_z = \int \int \int \frac{G·m·z}{ρ^3}δdV = Gm\int \int \int \frac{ρcos(φ)}{ρ^3}δ·ρ^2sin(φ)dρdφdθ = Gm \int \int \int δ·cos(φ)sin(φ)dρdφdθ.$