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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 on a plane is an assignment of a vector $\vec{F}$ to every point $(x, y)$ in a given space. Mathematically, it is expressed as: $\vec{F}(x, y) = \langle M(x, y), N(x, y) \rangle = M\hat{i} + N\hat{j}$ where $M$ and $N$ are scalar 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. In physics, vector fields often represent forces (like gravity or electromagnetism). If we want to calculate the work done by this force field in moving an object along a specific trajectory $C$, we use a line integral.
The work $W$ done by a force field $\vec{F}$ along a curve $C$ is the integral of the tangential component of the force along the path: $W = \int_C \vec{F} \cdot d\vec{r} = \int_C (M \, dx + N \, dy)$
A vector field is considered conservative if it is the gradient of a scalar potential function $f(x,y)$, meaning $\boxed{\vec{F} = \nabla f}$.
If a vector field is continuous on an open, simply-connected region (a region with no “holes”), the following four properties are mathematically equivalent. If one is true, they are all true:
For these conservative fields, we can bypass complex line integrals using the Fundamental Theorem of Line Integrals. The work is simply the difference in potential energy between the endpoints: $\boxed{\int_C \nabla f \cdot d\vec{r} = f(B) - f(A)}$
Before diving into flux, it is crucial to understand conservative vector fields. Let $\vec{F}(x, y) = \langle P(x, y), Q(x, y) \rangle$ be a 2D vector field.
The criterion for checking whether a vector field $\vec{F}$ is conservative can be summarized as follows: If the cross-partials are equal, meaning $\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}$, and if the scalar functions $P$ and $Q$ have continuous first partial derivatives across an open, simply connected domain $D$ (a region with no holes), then $\vec{F}$ is a gradient field, meaning it is perfectly conservative.
This conservative property is deeply linked to Green’s Theorem for Circulation (Work).
Green’s Theorem establishes the fundamental relationship between a macroscopic line integral around a boundary and the microscopic rotation (curl) within the region it encloses.
Green’s Theorem. If $C$ is a positively oriented (counterclockwise) simple closed curve enclosing a region $R$, and $\vec{F} = \langle P(x, y), Q(x,y) \rangle$ is a vector field that is defined and has continuous partial derivatives on an open region containing $R$, then Green's Theorem states that: $\oint_C \vec{F} \cdot{} dr = \iint_R (\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dA$ where:
If the field is conservative, the cross-partials are identical ($\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}$). Consequently, the integrand of the double integral becomes zero. This mathematically proves that the work done around any closed loop in a conservative field is always zero.
Consider a vector field $\vec{F}$ operating in a region with a “hole,” as illustrated in Figure A. Let $C'$ be an outer boundary and $C''$ be an inner boundary enclosing a space where $\vec{F}$ is either undefined or singular (e.g., division by zero at the origin). The valid domain $R$ is the annular region between $C'$ and $C''$.
Green’s Theorem requires a simple, closed curve bounding a continuous region. In a non-simply connected region containing multiple curves, the theorem must be adapted by creating a composite boundary and applied carefully.
To evaluate the total line integral of $\vec{F}$ around the composite curve $C$, we must account for both the outer boundary $C'$ (oriented counterclockwise) and the inner boundary $C''$ (oriented clockwise to keep the valid region $R$ to the left of the path).
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). $\oint_{C} \vec{F}\cdot dr = \oint_{C'} \vec{F}\cdot dr -\oint_{C''} \vec{F}\cdot dr = \iint_R \operatorname{curl} \vec{F}dA$
As an example, consider the classic vortex field $\vec{F} = \left\langle \frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2} \right\rangle$. While its macroscopic circulation around the origin is non-zero, calculating the curl yields: $curl(\vec{F}) = \frac{∂}{∂x}(\frac{x}{x^2+y^2})-\frac{∂}{∂y}(\frac{-y}{x^2+y^2}) = 0$.
Because the curl is zero everywhere except the undefined origin, the integral over the valid region $R$ evaluates to $0$.
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:
Definition. A region $R$ in the plane is simply connected if, for any closed curve $C$ completely contained within R, the entire interior of $C$ is also strictly contained within $R$. In simpler terms, a simply connected region does not have any holes, gaps, or missing points; it’s a continuous and unbroken area.
Imagine drawing any closed loop or curve within the region $R$. If you can always continuously shrink that loop down to a single infinitesimal point without ever 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 single point within the region. On the other hand, $C_2$ is not simply connected because it contains a hole (indicates by the red curve) -a loop encompassing the central hole cannot be contracted to a point without leaving the valid domain.
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: $\oint_C (\vec{F} \cdot \hat{\mathbf{T}})ds = \int \int_R curl(\vec{F})dA$ where:
More importantly, this holds true only if the region $R$ is simply connected and the vector field $\vec{F}$ is well-defined and differentiable everywhere inside $R$. This ensures that the interior of any curve $C$ does not encounter any discontinuities or undefined points in $\vec{F}$ invalidating the double integral.
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, $\vec{F} = \nabla f$ where $\nabla f$ denotes the gradient of $f$.
For a vector field to be conservative, two main conditions must be simultaneously satisfied:
When both conditions are met, the line integral of $\vec{F}$ around any closed curve $C$ in $R$ is zero: $\vec{F}$ is conservative, $\oint_C (\vec{F} \cdot \hat{\mathbf{T}})ds = \int \int_R curl(\vec{F})dA = \int \int_R 0dA = 0$. The field is irrotational and path-independent.
This result is significant because it tells us that the work done by a conservative force field (like gravity or electrostatics) around any closed loop is zero (it requires zero net energy). 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 only considering the part of the circle where $y \ge 0$.
Set up the integral in polar coordinates. In polar coordinates: $x = r\cos(\theta), y = r\sin(\theta)$, and $x^2 + y^2 = r^2$ the differential area element $dA$ in polar coordinates is $rdrd\theta$. The transformation to polar coordinates simplifies the bounds of integration and the integrand, making the problem more manageable.
Determine the bounds. The region $R$ is the upper half of the unit circle, so the angle $\theta$ ranges or sweeps from $0$ to $\pi$ (since $\theta = 0$ corresponds to the positive x-axis and $\theta = \pi$ corresponds to the negative x-axis, covering the upper semicircle) and $r$ ranges from the origin $0$ to the boundary $1$ (the equation of the unit circle $x^2 + y^2 = 1$, becomes $r^2 = 1 \implies r = \pm 1$, $r = 1$ -polar radius is non-negative).
Thus, the integral in polar coordinates becomes: $\int \int_R (1-r^2)dA = \int_{0}^{\pi}\int_{0}^{1} (1-r^2)rdrd\theta$ =[Simplify the integrand] $\int_{0}^{\pi}\int_{0}^{1} (r-r^3)drd\theta.$
Evaluate the integral.
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: $\int_{0}^{\pi} \frac{1}{4}d\theta = \frac{1}{4}\theta \bigg|_{0}^{\pi} = \frac{1}{4}\pi -0 = \frac{\pi}{4}$
Visualizing the Problem.
The paraboloid $z = 4 -x^2 -y^2$ opens downward from its peak at $(0, 0, 4)$ 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: $x = r\cos(\theta), y = r\sin(\theta)$, and the paraboloid equation becomes $z = 4 -x^2 -y^2 = 4 -r^2$, the differential area element $dA$ in polar coordinates is $rdrd\theta$.
Determine the region of integration: The paraboloid intersects the xy-plane when $z = 0, 4 -r^2 = 0 \implies r^2 = 4 \implies r = 2.$ 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 centered at the origin with radius $2$, which we find by setting $z = 0$ in the equation of the paraboloid) and $\theta$ ranges from 0 to $2\pi$.
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 $R$: $V = \int \int_R zdA = \int \int_R (4-r^2)dA = \int_{0}^{2\pi}\int_{0}^{2} (4-r^2)rdrd\theta$ =[Simplify the integrand] $\int_{0}^{2\pi}\int_{0}^{2} (4r-r^3)dr\,d\theta.$
Evaluate the integral.
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: $\int_{0}^{2\pi} 4d\theta = 4\theta \bigg|_{0}^{2\pi} = 8\pi.$
Recall. The circulation of a vector field around a closed curve $C$ is given by the line integral: $\oint_C (\vec{F}\cdot \hat{\mathbf{T}})ds = \oint_C (\vec{F} \cdot dr) = \oint_C Pdx + Qdy$ where, $P = -y$ and $Q = x$.
Parameterize the Circle. The circle of radius 2 centered at the origin can be parameterized using a vector function: $\vec{r}(t) = \langle 2\cos(t), 2\sin(t) \rangle, 0 \le t \le 2\pi.$
This parameterization describes a point $\vec{r}(t)$ on the circle as t varies from $0$ to $2\pi$, tracing out the entire circle in the counterclockwise direction.
Compute the Components of $\vec{F}$ and the differentials $dx$ and $dy$. Given $\vec{F} = \langle -y, x \rangle, P = -y = -2\sin(t), Q = x = 2\cos(t) \implies dx = -2\sin(t)dt, dy = 2\cos(t)dt.$
Set Up the Line Integral. The line integral for circulation around the curve $C$ is given by: $\oint_C (\vec{F}\cdot \hat{\mathbf{T}})ds = \oint_C (\vec{F} \cdot dr) = \oint_C Pdx + Qdy =[\text{Substituting the values of P, Q, dx, and dy into this expression}] \int_{0}^{2\pi} (-2\sin(t))(-2\sin(t)dt)+(2\cos(t))(2\cos(t))dt = \int_{0}^{2\pi} (4\sin^2(t)+4\cos^2(t))dt$.
Factor out the 4 and apply the Pythagorean identity ($\sin^2(t) + \cos^2(t) = 1$): $= \int_{0}^{2\pi} 4(1) \, dt$
Integrate with respect to $t$: $= \left[ 4t \right]_{0}^{2\pi} = 8\pi$
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 $0 \le x \le 2$ and for a fixed $x, 0 \le y \le x^2$.
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 integral: $\int_{0}^{x^2} \sqrt{x^3+1}dy = y\sqrt{x^3+1}\bigg|_{0}^{x^2} = x^2\sqrt{x^3+1}$
Evaluate the outer integral: $\int_{0}^{2} x^2\sqrt{x^3+1}dx$
To evaluate this integral, we use a substitution. Let: $u = x^3+1$ which means $du = 3x^2dx$ or $x^2dx = \frac{1}{3}du$.
Update the limits of integration for $u$: When $x = 0$, $u = 0^3 + 1 = 1$. When $x = 2$, $u = 2^3 + 1 = 9$.
Substitute and solve: $\frac{1}{3}\int_{1}^{9} \sqrt{u} \, du = \frac{1}{3} \left[ \frac{2}{3}u^{\frac{3}{2}} \right]_{1}^{9} = \frac{2}{9} \left( 9^{\frac{3}{2}} - 1^{\frac{3}{2}} \right)$
$= \frac{2}{9}(27 - 1) = \frac{52}{9}$
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:
To integrate with respect to $x$ first ($dx \, dy$), we use horizontal slices. Because the left boundary of the region changes behavior at $y = 1$, we must split the integral into two separate regions along the y-axis:
$\star$ Lower region ($0 \le y \le 1$): A horizontal slice enters at the line $y = 2x$ (or $x = \frac{y}{2}$) and exits at the line $y = x$ (or $x = y$).
$\star$ Upper region ($1 \le y \le 2$): A horizontal slice enters at the line $y = 2x$ (or $x = \frac{y}{2}$) and exits at the vertical boundary $x = 1$.
Combining these gives the reversed setup: $\int_{0}^{1}\int_{x}^{2x} f(x,y) \, dy \, dx = \int_{0}^{1}\int_{\frac{y}{2}}^{y} f(x,y) \, dx \, dy + \int_{1}^{2}\int_{\frac{y}{2}}^{1} f(x,y) \, dx \, dy$
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:
Thus, the vertices of the tetrahedron are $(0,0,0)$, $(2,0,0)$, $(0,4,0)$, and $(0,0,8)$.
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.
Projecting the tetrahedron onto the xy-plane.
First, ignore $z$ and look at the “shadow” of the tetrahedron on the $xy$-plane ($z = 0$), the trace is $4x+2y = 8$
$y$ ranges from $0$ to $4 −2x$. When $z = 0$, solving $4x +2y = 8$ gives $y = 4 −2x$.
$x$ ranges from the origin ($x = 0$) to the x-intercept ($x = 2$).
Limits for z.
For each fixed point $(x, y)$ in that triangular region:
$\star$ The bottom of the tetrahedron is the $xy$-plane: $z = 0$.
$\star$ The top is the plane: $4x + 2y + z = 0 \implies z = 8 -4x -2y$
Putting it all together. $\boxed{V = \int_{0}^{2}\int_{0}^{-2x+4} \int_{0}^{8-4x-2y} 1dzdydx}$
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.
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