Without pain, without sacrifice, we would have nothing, Fight Club
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).
Understanding the Problem. We are given a plane z = 6 -3x -2y and asked to find the volume of the solid region that lies between this plane and the xy-plane over the unit square. The xy-plane corresponds to z = 0, and the unit square (x^{2} + y^{2} = 1) is the region in the xy-plane where both x and y range from 0 to 1.
Set up the integral. In Cartesian coordinates, the volume is given by the triple integral $\int \int \int 1dV$
In Cartesian coordinates, the volume element dV is dxdydz. We need to set up the bounds of the integral based on the given region and the plane. The unit square is described by: 0 ≤ x ≤ 1, 0 ≤ y ≤ 1. The plane z = 6 −3x −2y is bounded above by this plane and below by the xy-plane, where z = 0 (Figure C).
So the integral in Cartesian coordinates is set up as: $= \int_{0}^{1} \int_{0}^{1} \int_{0}^{6-3x-2y} dz·dy·dx$
Compute the inner integral: $\int_{0}^{6-3x-2y} dz = z\bigg|_{0}^{6-3x-2y} = 6-3x-2y$
Middle integral with respect to y: $\int_{0}^{1} (6-3x-2y)dy = 6y -3xy -y^2\bigg|_{0}^{1} = 6 -3x -1 -(6(0)−3x(0)−0^2)= 5 -3x.$
Outer integral with respect to x: $\int_{0}^{1} (5 -3x)dx = 5x -\frac{3x^2}{2}\bigg|_{1}^{0} = 5 -\frac{3}{2} = \frac{7}{2}$. So the volume of the region between the xy-plane and the plane z = 6 −3x −2y over the unit square R is: $V = \frac{7}{2}$.
Understanding the Problem. We want to find the volume of the region bounded by:
Set up the integral. In Cartesian coordinates, the volume V of a region is given by the triple integral $\int \int \int_{R} 1dV$
In Cartesian coordinates, the volume element dV is dxdydz (Figure D). We need to set up the bounds of the integral based on the given region and the planes.
The bounds for x, y, and z are determined by the given planes: z ranges from 0 to 4 (since z = 4 is the upper boundary and the xy-plane is the lower boundary, where z = 0); y ranges from 0 to 6 −3x (since y = 6 −3x is the upper boundary and y = 0 is the lower boundary); x ranges from 0 to 2 (since when y=0, solving 6−3x=0 gives x = 2).
So the integral in Cartesian coordinates is set up as: $= \int_{0}^{2} \int_{0}^{6-3x} \int_{0}^{4} dz·dy·dx$
Compute the inner integral with respect to z: $\int_{0}^{4} dz = z\bigg|_{0}^{4} = 4$
Middle integral with respect to y: $\int_{0}^{6-3x} 4dy = 4y\bigg|_{0}^{6-3x} = 24-12x$
Outer integral with respect to x: $\int_{0}^{2} (24-12x)dx = 24x -6x^2\bigg|_{0}^{2} = 48 -6·4 -(0)=24$. So the volume of the prism bounded by the planes z = 4 and y = 6 −3x is: V = 24 cubic units.
Solution:
We first need to determine the limits of integration for the triple integral in Cartesian coordinates, where the volume element dV is given by dx·dy·dz. These limits are determined by the intersection of the two surfaces.
Determine the bounds. The surfaces intersect where z = x^{2} + y^{2} and z = 4 - x^{2} - y^{2} ↭[Setting these equal gives:] x^{2} + y^{2} < 4 -x^{2} -y^{2} ⇒2(x^{2} + y^{2}) = 4 ⇒x^{2} + y^{2} = 2, this equation describes a circle of radius $\sqrt{2}$ in the xy-plane.
Set up the integral. The volume V between the two paraboloids can be found using a triple integral in Cartesian coordinates. In Cartesian coordinates, the volume is given by $\int \int \int 1dV = \int_{-\sqrt{2}}^{\sqrt{2}} \int_{-\sqrt{2-x^2}}^{+\sqrt{2-x^2}} \int_{x^2+y^2}^{4-x^2-y^2} dz·dy·dx$
The lower bound for z is z = x^{2} + y^{2}, and the upper bound is z = 4 - x^{2} - y^{2}. For a fixed x, y ranges for $-\sqrt{2-x^2}$ to $\sqrt{2-x^2}$. Since the region is circular, x^{2} + y^{2} = 2, x ranges from $-\sqrt{2}$ to $\sqrt{2}$.
Compute the inner integral: $\int_{4-x^2-y^2}^{x^2+y^2} dz = z\bigg|_{4-x^2-y^2}^{x^2+y^2} = 4-2x^2-2y^2$
$\int_{-\sqrt{2}}^{\sqrt{2}} \int_{-\sqrt{2-x^2}}^{+\sqrt{2-x^2}} \int_{x^2+y^2}^{4-x^2-y^2} dz·dy·dx 1dV = \int_{-\sqrt{2}}^{\sqrt{2}} \int_{-\sqrt{2-x^2}}^{+\sqrt{2-x^2}} 4-2x^2-2y^2 dydx$
Evaluating triple integrals can be quite complex, so we are better off figuring out some techniques to simplify calculations.
Convert to cylindrical coordinates. Cylindrical coordinates are a simple extension of the two-dimensional polar coordinates to three dimensions. A Point (x, y, z) in a three-dimensional space is described using three coordinates (r, θ, z) (Figure 4 and 5 where r = a is a cylindrical surface that lies at a fixed radial distance “a” from the z-axis), where
Recall z = z, x = rcos(θ), y = rsin(θ) ⇒ Combining these contributions together, we get the volume element dV in cylindrical coordinates: dV = rdrdθdz (see Figure 6).
Compute the integral: $\int \int \int 1dV = \int_{0}^{2π} \int_{0}^{\sqrt{2}} \int_{r^2}^{4-r^2} dz·r·dr·dθ = $
Inner integral with respect to z: $\int_{r^2}^{4-r^2} dz·r = r\int_{r^2}^{4-r^2}dz = r·z\bigg|_{r^2}^{4-r^2} = r(4-r^2-r^2)=r(4-2r^2)$
Middle integral with respect to r: $\int_{0}^{\sqrt{2}} r(4-2r^2)dr = \int_{0}^{\sqrt{2}}4rdr -\int_{0}^{\sqrt{2}}2r^3dr = 2r^2-\frac{1}{4}·2·r^4 = 2r^2-\frac{1}{2}r^4\bigg|_{0}^{\sqrt{2}} = 2·2-\frac{1}{2}·2^2 = 4-2 =2$
Outer integral with respect to θ: $\int_{0}^{2π} \int_{0}^{\sqrt{2}} \int_{r^2}^{4-r^2} dz·r·dr·dθ = \int_{0}^{2π} 2dθ = 2θ\bigg|_{0}^{2π}=4π$
Therefore, the volume of the region between the two paraboloids is 4π.
Understanding the problem. x varies from -1 to 1, which corresponds to the range of the unit circle in the x-direction. y varies from $-\sqrt{1-x^2}$ to $\sqrt{1-x^2}$, both x and y describing the unit circle in the xy-plane.
Besides, z ranges from $\sqrt{x^2+y^2}$ to 1. The lower bound $z = \sqrt{x^2+y^2}$ is the equation of a cone (with its vertex at the origin and opening upward). The upper bound z = 1 is a horizontal plane (which form the upper “cap” of the solid) (Refer to Figure ii for a visual representation and aid in understanding it). In simpler terms, this solid is the part of the cone $z = \sqrt{x^2+y^2}$ that lies below the plane z = 1 and within the unit circle.
Convert to cylindrical coordinates. Cylindrical coordinates are a simple extension of the two-dimensional polar coordinates to three dimensions. A Point (x, y, z) in a three-dimensional space is described using three coordinates (r, θ, z) (Figure 4 and 5 where r = a is a cylindrical surface that lies at a fixed radial distance “a” from the z-axis), where
Recall z = z, x = rcos(θ), y = rsin(θ) ⇒ Combining these contributions together, we get the volume element dV in cylindrical coordinates: dV = rdrdθdz.
V = $\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{\sqrt{x^2+y^2}}^{1} dzdydx = \int_{0}^{2π}\int_{0}^{1}\int_{r}^{1} rdzdrdθ$
$\int_{r}^{1} rdz = rz\bigg|_{r}^{1} = r(1-r)$
$\int_{0}^{1} r(1-r)dr = \frac{r^2}{2}-\frac{r^3}{3}\bigg|_{0}^{1} = \frac{1}{2}-\frac{1}{3} = \frac{3}{6}-\frac{2}{6} =\frac{1}{6}$
V = $\int_{0}^{2π} \frac{1}{6}dθ = \frac{1}{6}θ\bigg|_{0}^{2π} = \frac{2π}{6} = \frac{π}{3}$. The Volume of the solid is π/3, that is expected because V_{cone} = $\frac{1}{3}πr^2h = \frac{1}{3}π1^2·1 = \frac{π}{3}$ (r = h = 1).
Understanding the problem. In the xy-plane, the intersection of the paraboloid with the plane z = 0 occurs when 9 -x^{2} -y^{2} = 0 ⇒ x^{2} + y^{2} = 9. This describes a circle centered at the origin with a radius of 3. Therefore, in the xy-plane, the region E is a disk with radius 3. z varies from 0 (xy-plane) and the paraboloid (z = 9 -x^{2} -y^{2}), which opens downwards with its vertex at (0, 0, 9) and intersects the xy-plane at z = 0 (Refer to Figure A for a visual representation and aid in understanding it).
Convert to cylindrical coordinates. Cylindrical coordinates are a simple extension of the two-dimensional polar coordinates to three dimensions. A Point (x, y, z) in a three-dimensional space is described using three coordinates (r, θ, z) (Figure 4 and 5 where r = a is a cylindrical surface that lies at a fixed radial distance “a” from the z-axis), where
Recall z = z, x = rcos(θ), y = rsin(θ) ⇒ Combining these contributions together, we get the volume element dV in cylindrical coordinates: dV = rdrdθdz.
Set up the Integral: V = $\int_{0}^{2π}\int_{0}^{3}\int_{0}^{9-r^2} r·rdzdrdθ$
$\int_{0}^{9-r^2} r^2dz = r^2z\bigg|_{0}^{9-r^2} = r^2(9-r^2)$
$\int_{0}^{3} r^2(9-r^2)dr = \int_{0}^{3} 9r^2-r^4dr = 3r^3-\frac{r^5}{5}\bigg|_{0}^{3} = 3·27-\frac{243}{5} = 81 -\frac{243}{5} = \frac{405}{5}-\frac{243}{5}= \frac{162}{5}$
V = $\int_{0}^{2π}\int_{0}^{3}\int_{0}^{9-r^2} r·rdzdrdθ = \int_{0}^{2π} \frac{162}{5}dθ = \frac{162}{5}θ\bigg|_{0}^{2π} = \frac{162·2π}{5} =\frac{324π}{5}$