I have yet to see any problem, however complicated, which, when you looked at it in the right way, did not become still more complicated, Poul Anderson.
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 $div \vec{F} = P_x + Q_y = \frac{∂P}{∂x}+\frac{∂Q}{∂y}$.
Consider a vector field $\vec{F}$ illustrated in Figure A. C′ is an outer boundary, C′′ is an inner boundary enclosing a region where $\vec{F}$ might not be defined at certain points within it, and R is the region between these two boundaries C′ and C′′.
We aim to compute the total line integral of the vector field $\vec{F}$ around the composite curve C, which is formed by combining both C′ and C′′. he total line integral around C involves considering both the outer boundary C′ and the inner boundary C′′.
Green’s Theorem is a powerful tool that relates the line integral around a simple, closed curve C to a double integral over the region R enclosed by C. However, in a non-simply connected region, where C is not a single simple curve but consists of multiple curves (like C′ and C′′ ), Green’s Theorem must be applied carefully.
Apply Green’s Theorem to the outer and inner boundaries C′ and C′′. To find the total line integral around the composite curve C, we subtract the integral around C′′ because C′′ is taken in the clockwise direction (which is opposite to the counterclockwise direction). $\oint_{C} \vec{F}d\vec{r} = \oint_{C’} \vec{F}d\vec{r}-\oint_{C’’} \vec{F}d\vec{r} = \int \int_R curl \vec{F}dA =$[In our previous case, Counterexample. Consider the vector field $\vec{F} = \frac{-y\vec{i}+x\vec{j}}{x^2+y^2}, curl(\vec{F}) = \frac{∂}{∂x}(\frac{x}{x^2+y^2})-\frac{∂}{∂y}(\frac{-y}{x^2+y^2}) = 0$] 0.
In some cases the vector field $\vec{F}$ is not defined at certain points within the region R, such as within the inner boundary C’’. This could occur if there is a singularity or a point where $\vec{F}$ becomes undefined, e.g., the origin.
To handle this situation, we modify our approach:
Composite curve. We define C as the combination if C’ and C’’. The total line integral around C can then be expressed as the difference between the line integrals around C′ and C′′.
Well-defined Regions. Even if $\vec{F}$ is not defined within the hole enclosed by C′′, Green’s Theorem can still be applied to the well-defined region between C′ and C′′, R, Figure B. In other words,we can create a curve that will enclose this region counterclockwise and is well defined, hence the total line integral $\oint_C \vec{F}·\hat{\mathbf{T}}·ds = \oint_{C’}\vec{F}d\vec{r} -\oint_{C’’}\vec{F}d\vec{r}$[C’’ is clockwise, and the interior blue segments cancel out] = $\int \int_R curl \vec{F}dA$
Definition. A region R in the plane is simply connected if, for any closed curve C within R, the entire interior of C is also contained within R. In simpler terms, a simply connected region does not have any holes or gaps; it’s a continuous and unbroken area.
Imagine drawing any loop or curve within the region R. If you can always shrink that loop to a single point without leaving the region, then R is simply connected.
For example, in Figure C, C_{1} is simply connected because any closed curve you draw within C_{1} can be contracted to a point within the region. On the other hand, C_{2} is not simply connected because it contains a hole (indicated by the red curve).
Understanding whether a region is simply connected is crucial when applying Green’s Theorem. Green’s Theorem is a fundamental result in vector calculus that relates the line integral around a closed curve C to a double integral over the region R enclosed by C.
Green’s Theorem states: $\oint_C \vec{F}·\hat{\mathbf{T}}·ds = \int \int_R curl \vec{F}dA$ where
More importantly, this holds true if the region R is simply connected and the vector field $\vec{F}$ is well-defined and differentiable throughout R. This ensures that the interior of any curve C does not encounter any discontinuities or undefined points in $\vec{F}$.
A vector field $\vec{F}$ is called conservative (or a gradient field) if it can be expressed as the gradient of some scalar potential function, f. In other words, $\vec{F} = ∇f$ where ∇f denotes the gradient of f.
For a vector field to be conservative, two main conditions must be satisfied:
When these conditions are met, the line integral of $\vec{F}$ around any closed curve C in R is zero: $\vec{F}$ is conservative, $\oint_C \vec{F}·\hat{\mathbf{T}}·ds = \int \int_R curl\vec{F}dA = \int \int_R 0·dA$ = 0.
This result is significant because it tells us that the work done by a conservative force field around any closed loop is zero. In practical terms, if you move a particle in a conservative force field, the energy you spend moving it along a closed path will be recovered when you return to the starting point.
The unit circle is defined by x^{2}+y^{2} = 1, and “upper half” means that we are considering the part of the circle where y ≥ 0.
Set up the integral in polar coordinates. In polar coordinates: x = rcos(θ), y = rsin(θ), and x^{2} + y^{2} = r^{2}, the differential area element dA in polar coordinates is rdrdθ.
The transformation to polar coordinates simplifies the bounds of integration and the integrand, making the problem more manageable.
The region R is the upper half of the unit circle, so θ ranges from 0 to π (since θ = 0 corresponds to the positive x-axis and θ = π corresponds to the negative x-axis, covering the upper semicircle) and r ranges from 0 to 1 (the equation of the unit circle x^{2}+y^{2}=1, becomes r^{2}=1 ⇒ r = ± 1).
Thus, the integral in polar coordinates becomes:
$\int \int_R (1-r^2)dA = \int_{0}^{π}\int_{0}^{1} (1-r^2)rdrdθ$ =[Simplify the integrand] $\int_{0}^{π}\int_{0}^{1} (r-r^3)drdθ.$
Calculate the inner integral: $\int_{0}^{1} (r-r^3)dr = \frac{r^2}{2}-\frac{r^4}{4}\bigg|_{0}^{1} = (\frac{1}{2}-\frac{1}{4}-(0-0)) = \frac{1}{4}$.
Calculate the outer integral: $\int_{0}^{1} \frac{1}{4}dθ = \frac{1}{4}θ\bigg|_{0}^{1} = \frac{1}{4}π-0 = \frac{π}{4}$
Visualizing the Problem. The paraboloid z = 4 -x^{2} -y^{2} opens downward, and its vertex is at the point (0, 0, 4) on the z-axis. The xy-plane is where z = 0, which forms the base of the solid. We are interested in the region above this plane and below the surface of the paraboloid.
Set up the integral in polar coordinates. The equation of the paraboloid z = 4 -x^{2} -y^{2} is given in Cartesian coordinates. To simplify the integration, we convert this equation into polar coordinates. In polar coordinates: x = rcos(θ), y = rsin(θ), and the paraboloid equation becomes z = 4 - (x^{2} + y^{2}) = 4 - r^{2}, the differential area element dA in polar coordinates is rdrdθ.
Determine the region of integration: The paraboloid intersects the xy-plane when z = 0, $4 -r^2 = 0⇒r^2 = 4 ⇒ r = 2.$ So, r ranges from 0 to 2 (Another way of thinking about it is this: The region of integration R corresponds to the projection of the paraboloid onto the xy-plane. This projection is a circle, which we find by setting z = 0 in the equation of the paraboloid) and θ ranges from 0 to 2π.
Set Up the Volume Integral. The volume V under the paraboloid and above the xy-plane can be calculated using a double integral over the region R: $V = \int \int_R zdA = \int \int_R (4-r^2)dA = \int_{0}^{2π}\int_{0}^{2} (4-r^2)rdrdθ$ =[Simplify the integrand] $\int_{0}^{2π}\int_{0}^{2} (4r-r^3)drdθ.$
Evaluate the inner integral with respect to r: $\int_{0}^{2} (4r-r^3)dr = 2r^2-\frac{r^4}{4}\bigg|_{0}^{2} = (8-4)-(0-0)=4.$
Evaluate the outer integral with respect to θ: $\int_{0}^{2π} 4dθ = 4θ\bigg|_{0}^{2π} = 8π.$
The circulation of a vector field around a closed curve C is given by the line integral: $\oint_C \vec{F}·\hat{\mathbf{T}}ds = \oint_C \vec{F}·d\vec{r} = \oint_C Mdx + Ndy.$ Here, M = -y and N = x.
Parameterize the Circle. The circle of radius 2 centered at the origin can be parameterized using a vector function: $\vec{r}(t) = ⟨2cos(t), 2sin(t)⟩, 0 ≤ t ≤ 2π.$
This parameterization describes a point $\vec{r}(t)$ on the circle as t varies from 0 to 2π, tracing out the entire circle in the counterclockwise direction.
Compute the Components of $\vec{F}$ and the differentials dx and dy. Given $\vec{F} = ⟨-y, x⟩, M = -y =[y=2sin(t)] -2sin(t), N = x = 2cos(t)⇒ dx = -2sin(t)dt, dy = 2cos(t)dt.$
Set Up the Line Integral. The line integral for circulation around the curve C is given by: $\oint_C \vec{F}·\hat{\mathbf{T}}ds = \oint_C \vec{F}·d\vec{r} = \oint_C Mdx + Ndy =[\text{Substituting the values of M, N, dx, and dy into this expression}] \int_{0}^{2π} (-2sin(t))(-2sin(t)dt)+(2cos(t))(2cos(t))dt = \int_{0}^{2π} (4sin^2(t)+4cos^2(t))dt =[\text{Notice that } sin^2(t)+cos^2(t) = 1] \int_{0}^{2π} 4dt = 4t\bigg|_{0}^{2π} = 8π - 0 = 8π$
Determine the region of integration. Before we can reverse the order of integration, we need to understand the region over which we’re integrating. The given limits for y are from 0 to 4. For a fixed value of y, x ranges from $\sqrt{y}$ to 2. These limits describe a region in the xy-plane.
Sketch the region (Figure B). We get the region bounded by the curve y = x^{2} from x = 0 to x = 2 and the line x = 2. Alternatively, the region is bound by 0 ≤ x ≤ 2 and for a fixed x, 0 ≤ y ≤ x^{2} (y varies from 0 to x^{2}).
Set up the integral with reversed limits: $\int_{0}^{4}\int_{\sqrt{y}}^{2} \sqrt{x^3+1}dxdy = \int_{0}^{2}\int_{0}^{x^2} \sqrt{x^3+1}dydx$
Evaluate the inner integrate: $\int_{0}^{x^2} \sqrt{x^3+1}dy = y\sqrt{x^3+1}\bigg|_{0}^{x^2} = x^2\sqrt{x^3+1}$
Evaluate the outer integral: $\int_{0}^{2} x^2\sqrt{x^3+1}dx =$[To evaluate this integral, we use a substitution. Let: u = x^{3}+1, du = 3x^{2}dx ⇒x^{2}dx = ^{1}⁄_{3}du] $\frac{1}{3}\int_{0}^{2} \sqrt{u}du = \frac{1}{3}·\frac{2}{3}u^{\frac{3}{2}} = \frac{2}{9}(x^3+1)^{\frac{3}{2}}\bigg|_{0}^{2} = \frac{2}{9}(9^{\frac{3}{2}}-1^{\frac{3}{2}}) = \frac{2}{9}(27-1) = \frac{52}{9}$
Understanding the region of integration. Before we can reverse the order of integration, we need to understand the region over which we’re integrating. The inner integral $\int_{x}^{2x} f(x,y)dy$ indicates that for a fixed x, y ranges from x to 2x. The outer integral $\int_{0}^{1} f(x,y)dx$ indicated that x ranges from 0 to 1.
Sketch the region: The lower bound is y = x, a line with slope 1 passing through the origin. The upper bound is y = 2x, a line with slope 2 passing through the origin. x ranges from 0 to 1.
Set up the new integral with reversed limits: The new region of integration is: y ranges from 0 to 2, x ranges from y/2 to y or y/2 to 1, depending on y ([0, 1] or [1, 2]). $\int_{0}^{1}\int_{x}^{2x} f(x,y)dydx = \int_{0}^{1}\int_{\frac{y}{2}}^{y} f(x,y)dxdy + \int_{1}^{2}\int_{\frac{y}{2}}^{1} f(x,y)dxdy$
The problem asks us to find the volume of a tetrahedron bounded by the plane 4x +2y +z =8 and the coordinate planes (the xy-, xz-, and yz-planes). This tetrahedron is in the first octant, meaning all coordinates (x, y, z) are non-negative.
Understanding the Geometry. To understand the shape of the tetrahedron, let’s determine where the plane intersects the coordinate axes:
Setting Up the Triple Integral in Cartesian Coordinates. The volume V can be calculated using a triple integral over the region defined by the tetrahedron. We integrate the function f(x, y, z) = 1 over this region because the volume is simply the integral of 1 over the region.
V = $\int_{0}^{2}\int_{0}^{-2x+4} \int_{0}^{8-4x-2y} 1·dzdydx$ and a double integral by integrating over the projection of the tetrahedron onto the xy-plane.
Simplifying the Integral to a Double Integral. The innermost integral with respect to z is straightforward since we are integrating a constant 1: $\int_{0}^{8-4x-2y} 1·dz = (8-4x-2y) -0 = 8-4x-2y$
Set up the integral in Cartesian coordinates: $V = \int_{0}^{2} \int_{0}^{-2x +4} (8 -4x-2y)dydx$
Compute the inner integral with respect to y: $\int_{0}^{-2x +4} (8 -4x-2y)dy = 8y -4xy -y^2\bigg|_{0}^{-2x +4} = 8(-2x+4)-4x(-2x+4)-(-2x+4)^2 = -16x +32 +8x^2 -16x -(4x^2-16x+16) = 4x^2-16x+16$
Compute the outer integral with respect to x: $\int_{0}^{2} (4x^2-16x+16)dx = \frac{4}{3}x^3-8x^2+16x\bigg|_{0}^{2} = \frac{4}{3}8-8·4+16·2=\frac{32}{3}-32+32 = \frac{32}{3}$.
Conclusion: The volume of the tetrahedron in the first octant bounded by the plane 4x +2y +z = 8 is $\frac{32}{3}$ cubic units.