Triple Integrals

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Vector Fields, Line Integrals, and the Theorems of Calculus

Understanding how quantities flow, circulate, and accumulate through space requires the mathematical machinery of vector fields and multiple integrals. These concepts form the bedrock of fluid dynamics, electromagnetism, and classical mechanics.

The Anatomy of a Vector Field

A vector field is a mathematical assignment of a vector to every point in a given space. Unlike a scalar field, which assigns a single magnitude (like temperature) to each point, a vector field assigns both a magnitude and a direction (like wind velocity).

In two-dimensional space ($\mathbb{R}^2$), a vector field is expressed as: $\vec{F}(x,y) = M(x,y)\mathbf{\hat{i}} + N(x,y)\mathbf{\hat{j}} = \langle M(x,y), N(x,y) \rangle$

In three-dimensional space ($\mathbb{R}^3$), it takes the form: $\vec{F}(x,y,z) = P(x,y,z)\mathbf{\hat{i}} + Q(x,y,z)\mathbf{\hat{j}} + R(x,y,z)\mathbf{\hat{k}}$

Visual Intuition: Imagine a flat plane covered in tiny arrows. At every coordinate $(x,y)$, the arrow points in a specific direction with a specific length. If the vector field models an ocean current, the arrows show which way the water is moving and how fast.

Work and Line Integrals

In physics, work is defined as the energy transferred when a force acts on an object and displaces it along a path. If the force is constant and the path is straight, Work is simply Force times Distance. But what if the object moves along a curved path $C$ through a varying force field $\vec{F}$?

We calculate this using a line integral. The work $W$ done by a force field $\vec{F}$ in moving a particle along a parameterized curve $C$ is the integral of the tangential component of the force along the curve: $W = \int_C \vec{F} \cdot d\vec{r}$

Because $d\vec{r} = \langle dx, dy \rangle$, we can expand this in two dimensions: $\int_C \vec{F} \cdot d\vec{r} = \int_C (M dx + N dy)$.

Alternatively, using the unit tangent vector $\mathbf{\hat{T}}$ and the arc length differential $ds$: $\int_C (\vec{F} \cdot \mathbf{\hat{T}}) ds$

Conservative Vector Fields and the Potential Function

A vector field $\vec{F}$ is called conservative if it can be expressed as the gradient of a scalar function $f$. $\vec{F} = \nabla f = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right\rangle$

This scalar function $f$ is known as the potential function associated with the vector field. Gravitational and electrostatic fields are prime examples of conservative fields.

The Fundamental Theorem for Line Integrals

If $\vec{F}$ is a conservative vector field with a continuously differentiable potential function $f$, evaluating the line integral of $\vec{F}$ along a curve $C$ becomes remarkably simple. The path taken no longer matters; only the starting point $P_0$ and the ending point $P_1$ matter.

$\boxed{\int_C \vec{F} \cdot d\vec{r} = \int_C \nabla f \cdot d\vec{r} = f(P_1) - f(P_0)}$ where $P_0$ and $P_1$ are the initial and final points of the curve C, respectively. This is the multi-dimensional analog to the First Fundamental Theorem of Calculus.

Equivalent Properties of Conservative Fields

The concept of a conservative field is so fundamental that it can be defined through four mathematically equivalent properties (assuming the domain is well-behaved):

1. Path Independence: The work done by the force field in moving an object from point $A$ to point $B$ is identical, regardless of the path taken between the two points.
2. Zero Circulation (Conservative Force): The work done by the force around any closed loop $C$ is always exactly zero. Mathematically, this is expressed as $\oint_C \vec{F} \cdot d\vec{r} = 0$. If you walk in a circle on a mountain and return to your exact starting point, your net change in elevation is zero.
3. 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 $\vec{F} = \nabla f$, where $f$ is a scalar function and the vector field $\vec{F}$ has components $\vec{F} = \langle M, N \rangle = \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \rangle = \langle f_x, f_y \rangle$. Here, $f$ is the potential function associated with the vector field $\vec{F}$.
   The line 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: $\int_{C} \vec{F} \cdot d\vec{r} = f(P_1)-f(P_0).$
4. Exact Differential: The differential form $M dx + N dy$ is exact, meaning there exists a scalar function $f$ such that $df = M dx + N dy$.

The Cross-Partial Test: How to Check for Conservativeness

How do you know if a vector field $\vec{F} = \langle M, N \rangle$ is conservative without guessing its potential function? You test its partial derivatives.

The criterion for checking whether a vector field $\vec{F}$ is conservative can be summarized as follows: If $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, 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.

This test only guarantees a conservative field if the domain is simply connected—meaning the region consists of one piece and has absolutely no “holes” in it. Consider the classic vortex field $\vec{F} = \langle \frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2} \rangle$. If you apply the cross-partial test, you will find that $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$. However, this field is not conservative over $\mathbb{R}^2$ because the domain has a singularity (a hole) at the origin $(0,0)$. Integrating this field around the unit circle yields $2\pi$, not $0$.

Green’s Theorem: Connecting Curves to Areas

Green’s Theorem provides a profound relationship between a line integral around a closed boundary curve $C$ and a double integral over the plane region $R$ bounded by $C$. It comes in two distinct forms: Circulation and Flux.

Green’s Theorem. If $C$ is a positively oriented (counterclockwise) simple closed curve enclosing a region $R$, and $\vec{F} = \langle M(x, y), N(x,y) \rangle$ is a vector field that is defined and has continuous partial derivatives on an open region containing $R$, then Green's Theorem states that: $\oint_C \vec{F} \cdot{} d\vec{r} = \iint_R \operatorname{curl}(\vec{F}) dA ↭ \oint_C Mdx + N dy = \iint_R (N_x-M_y)dA = \iint_R (\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y})dA$ where:

• $\oint_C \vec{F} \cdot{} d\vec{r}$ represents the work or line integral of the vector field $\vec{F}$ around the curve $C$.
• $\iint_R (\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y})dA$ represents the double integral over the region $R$, with the integrand being the difference between the partial derivatives of $N$ and $M$.
• Physical Intuition: The integrand $\left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right)$ is the 2D scalar curl of the field. Imagine dropping a microscopic paddlewheel into the field at a specific point. The curl measures how fast that paddlewheel spins. Green’s Theorem states that the macroscopic circulation around the boundary $C$ is exactly equal to the sum of all the microscopic spins inside the region $R$.

Instead of measuring how much of the field flows along the curve, we can measure how much of the field flows across the curve (outward).

Green’s Theorem for Flux. If $C$ is a positively oriented (counterclockwise), simple closed curve that encloses a region $R$ and if $\vec{F} = \langle M, N \rangle$ is a continuously differentiable vector field defined on an open region that contains $R$, then the outward flux of the vector field across the curve $C$ is equal to the double integral of the divergence of $\vec{F}$ over $R$. Mathematically, this is expressed as: $\oint_C \vec{F} \cdot \mathbf{\hat{n}} ds = \iint_R \left( \frac{\partial M}{\partial x} + \frac{\partial N}{\partial y} \right) dA$

Physical Intuition: The integrand $\left( \frac{\partial M}{\partial x} + \frac{\partial N}{\partial y} \right)$ is the divergence of the field, denoted as $\text{div} \vec{F}$ or $\nabla \cdot \vec{F}$. It measures whether a point acts as a “source” (generating fluid) or a “sink” (draining fluid). The theorem states that the total fluid flowing outward through the boundary $C$ equals the sum of all the sources and sinks inside the region $R$.

Expanding into 3D: Triple Integrals

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:

1. $f(x, y, z)$ is a function of three variables over a three-dimensional region, representing the density or value of some quantity at each point $(x, y, z)$ within the region $R$. For example, if $f(x, y, z)$ represents density, the integral will give you the total mass. If $f(x,y,z) = 1$, the integral yields the total volume of the region $R$.
2. $dV$ represents an infinitesimal volume element within the region $R$. In Cartesian coordinates, $dV = dx \cdot dy \cdot dz$, but this element can look different in other coordinate systems.
3. The integral is calculated over the region $R$, which can be described by specific bounds on $x, y$, and $z$. (Figure 1).

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.

• Cartesian Coordinates $(x, y, z)$. Cartesian coordinates are used for regions that align well with the coordinate planes, such as rectangular prisms or boxes. The coordinates $(x, y, z)$ correspond directly to distances along the $x, y$, and $z$ axes, respectively. The differential volumen element is: dV = dxdydz. Limits are defined directly by the geometric bounds on the $x, y$, and $z$ axes.
• Cylindrical Coordinates $(r, \theta, z)$: Cylindrical coordinates are useful for regions with cylindrical symmetry, such as cylinders, cones, and paraboloids. In this system, the coordinates are:

1. $r$: The radial distance from the z-axis to the point, $r = \sqrt{x^2+y^2}$.
2. $\theta$: The angle in the xy-plane, measured counterclockwise, from the positive $x$-axis.
3. $z$: The height above the xy-plane, the same as in Cartesian coordinates.

Conversion to Cartesian Coordinates: $z = z, x = r\cos(\theta), y = r\sin(\theta)$, and the differential volume element is: $dV = rdr d\theta dz$.

• Spherical Coordinates $(\rho, \theta, \phi)$.

They are ideal for regions with spherical symmetry such as spheres, hemispheres, or cones. In this system, the coordinates or parameters are:

1. $\rho$: The radial distance from the origin to a given point, it ranges from 0 to infinity.
2. $\phi$: The polar angle, measured from the positive z-axis to the radial line (the line connecting the origin to the point P). It describes how far the point is from the z-axis. It value increases from $0$ (positive z-axis), $\frac{\pi}{2}$ (xy-plane), and $\pi$ (negative z-axis), that is, $0 \le \phi \le \pi$.
3. $\theta$: The azimuthal angle, measured from the positive x-axis to the projection of the radial line on the xy-plane. It ranges from $0$ to $2\pi$, representing a full rotation around the z-axis. It describes the rotation around the z-axis.

The differential volume element $dV = \rho^2 \sin(\phi)d\rho d\theta d\phi$. Conversion to Cartesian Coordinates: $x = r \cos(\theta) = \rho \sin(\phi)\cos(\theta), y =r \sin(\theta) = \rho\sin(\phi)\sin(\theta)$, and $z = \rho\cos(\phi)$.

Applications of Triple Integrals

• When f(x, y, z) = 1, the triple integral $\int \int \int_{R} dV$ represents the volume of the region R in three-dimensional space.
• If f(x, y, z) represents a density function, then the triple integral $\int \int \int_{R} f(x, y, z)dV$ gives the total mass of the region R.
• Triple integrals can also be used to calculate quantities like center of mass, moments of inertia, and electric charge distribution, depending on the function f(x, y, z) and the physical context.

Solved exercises

• Calculate the volume of the solid region bounded below by the $xy$-plane and above by the plane $z = 6 - 3x - 2y$, restricted over the unit square region $R = \{(x,y) \mid 0 \le x \le 1, 0 \le y \le 1\}$.

1. 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 is the region in the xy-plane where both x and y range from 0 to 1.

2. Set up the integral. In Cartesian coordinates, the volume is given by the triple integral $\iiint_R 1\,dV$
   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: $0 \le x \le 1, 0 \le y \le 1$. The region is bounded above by the plane $z = 6 −3x −2y$ and below by the xy-plane, where $z = 0$ (Figure C).
   So the integral in Cartesian coordinates is set up as: $= \int_{0}^{1} \int_{0}^{1} \int_{0}^{6-3x-2y} dzdydx$

3. Compute the inner integral: $\int_{0}^{6-3x-2y} dz = z\bigg|_{0}^{6-3x-2y} = 6-3x-2y$

4. Middle integral with respect to y: $\int_{0}^{1} (6-3x-2y)dy = 6y -3xy -y^2\bigg|_{0}^{1} = 6 -3x -1 -(6(0)−3x(0)−0^2)= 5 -3x.$

5. Outer integral with respect to x: $\int_{0}^{1} (5 -3x)dx = 5x -\frac{3x^2}{2}\bigg|_{0}^{1}$
   Evaluate at limits: $= 5 -\frac{3}{2} = \frac{7}{2}$.

6. Conclusion. So the volume of the region between the $xy$-plane and the plane $z = 6 −3x −2y$ over the unit square $R$ is: $V = \frac{7}{2}$.

• Find the volume of a prism in the first octant bounded by the planes $z = 4$ and $y = 6 -3x$.

1. Understanding the Problem (determine the bounds) We want to find the volume of the region bounded by:
   $\star$ $z$-bounds: The floor is $z=0$ and the ceiling is $z=4$.
   $\star$ $y$-bounds: The $y$-axis starts at $y=0$ and hits the slanting wall $y = 6 - 3x$.
   $\star$ $x$-bounds: The maximum $x$ occurs when $y=0$. Setting $0 = 6 - 3x$ gives $x=2$. Thus, $x$ ranges from $0$ to $2$.

2. 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. So the integral in Cartesian coordinates is set up as: $= \int_{0}^{2} \int_{0}^{6-3x} \int_{0}^{4} dzdydx$

3. Compute the inner integral with respect to z: $\int_{0}^{4} dz = z\bigg|_{0}^{4} = 4$

4. Middle integral with respect to y: $\int_{0}^{6-3x} 4dy = 4y\bigg|_{0}^{6-3x} = 24-12x$

5. Outer integral with respect to x: $\int_{0}^{2} (24-12x)dx = 24x -6x^2\bigg|_{0}^{2} = 48 -6·4 -(0)=24$.

6. Conclusion. So the volume of the prism bounded by the planes z = 4 and y = 6 −3x is: V = 24 cubic units.

Triple Integrals in Cylindrical Coordinates

• Calculate the volume of the region between the paraboloids $z = x^2 + y^2, z = 4-x^2 -y^2$ (Figure 2 and 3).

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 $dxdydz$. 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: $x^2 + y^2 = 4 - x^2 -y^2 \implies 2(x^2 + y^2) = 4 \implies x^2 + y^2 = 2$, this equation describes a circle of radius $\sqrt{2}$ in the xy-plane.

While Cartesian coordinates $(x, y, z)$ are perfect for rectangular solids, they quickly become an algebraic nightmare when dealing with circular, cylindrical, or spherical symmetries. Integrating over circles in Cartesian coordinates introduces complex square root bounds like $y = \pm\sqrt{R^2 - x^2}$. To bypass this complexity, we transition to Cylindrical Coordinates.

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 $(r, \theta, z)$ (see Figure 4 and 5 where $r = a$ is a cylindrical surface that lies at a fixed radial distance “a” from the z-axis), where

1. $r$ is the radial distance from the $z$-axis to the Point. In our particular example, $r$ ranges between $0$ to the radius of the intersection circle $x^2 + y^2 = 2$, that is, $\sqrt{2}.$
2. $\theta$, the azimuthal angle, is the angle measured counterclockwise from the positive x-axis to the projection of $P$ onto the xy plane. It ranges from $0$ to $2\pi$ since we want to cover the entire circle, $x^2 + y^2 = 2$.
3. $z$ is the standard vertical height of $P$ above (or below) the $xy$-plane. It ranges from $z_{\text{bottom}}= x^2+y^2 = r^2 ~\text{to}~z_{\text{top}} = 4-x^2-y^2 = 4-r^2$

When we evaluate a triple integral $\iiint dV$, $dV$ represents an infinitesimally small block of volume. In Cartesian coordinates, this block is a tiny box: $dV = dx \, dy \, dz$.

In cylindrical coordinates, $dV$ is a “curved box” (a polar wedge). As you move further from the origin (as $r$ increases), the outer edge of this wedge gets wider. To account for this geometric stretching, we must multiply by the radial distance $r$. This extra $r$ is the determinant of the Jacobian matrix.The Cylindrical Volume Element: $dV = r \, dr \, d\theta \, dz$

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 $\int \int \int 1dV = \int_{-\sqrt{2}}^{\sqrt{2}} \int_{-\sqrt{2-x^2}}^{+\sqrt{2-x^2}} \int_{x^2+y^2}^{4-x^2-y^2} dzdydx$

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 from $-\sqrt{2-x^2}$ 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: $\int_{x^2+y^2}^{4-x^2-y^2} dz = z\bigg|_{x^2+y^2}^{4-x^2-y^2} = 4-2x^2-2y^2$

$\int_{-\sqrt{2}}^{\sqrt{2}} \int_{-\sqrt{2-x^2}}^{+\sqrt{2-x^2}} \int_{x^2+y^2}^{4-x^2-y^2} dzdydx = \int_{-\sqrt{2}}^{\sqrt{2}} \int_{-\sqrt{2-x^2}}^{+\sqrt{2-x^2}} 4-2x^2-2y^2 dydx$

Evaluating triple integrals can be quite complex, so we are better off figuring out some techniques to simplify calculations.

Recall $z = z, x = r\cos(\theta), y = r\sin(\theta)$. Combining these contributions together, we get the volume element dV in cylindrical coordinates: $dV = rdr d\theta dz$ (see Figure 6).

Evaluate the Integral: $\int_{0}^{2\pi} \int_{0}^{\sqrt{2}} \int_{r^2}^{4-r^2} rdz dr d\theta$.

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 $\theta$: $\int_{0}^{2\pi} \int_{0}^{\sqrt{2}} \int_{r^2}^{4-r^2} dz r dr d\theta = \int_{0}^{2\pi} 2d\theta = 2\theta \bigg|_{0}^{2\pi} = 4\pi$

Therefore, the volume of the region between the two paraboloids is $4\pi$.

Identifying Common Surfaces in Cylindrical Coordinates

Before diving into integrals, it is vital to recognize basic 3D shapes solely from their cylindrical equations:

• $r = 2$ (A Cylinder): This dictates that the distance from the $z$-axis is always $2$, regardless of the angle $\theta$ or height $z$. This forms an infinite cylinder of radius $2$ centered on the $z$-axis.
• $\theta = \pi/3$ (A Half-Plane): This fixes the angle but allows radius $r$ and height $z$ to be anything. It creates a vertical plane extending outward from the $z$-axis at a $60^\circ$ angle.
• $z = r^2$ (A Paraboloid): Since $r^2 = x^2 + y^2$, this translates to $z = x^2 + y^2$. It is a standard circular paraboloid opening upwards from the origin.
• $r = z$ (A Cone): For $z \ge 0$, the radius grows linearly with height. This traces out the upper nappe of a right circular cone.
• $r = 4\cos(\theta)$ (A Shifted Cylinder): This one is tricky. Multiply both sides by $r$ to get $r^2 = 4r\cos(\theta)$.
  Converting to Cartesian gives $x^2 + y^2 = 4x$.
  By completing the square, $(x-2)^2 + y^2 = 4$. This is a cylinder of radius $2$, shifted so its central axis runs vertically through $(2, 0, 0)$.

• Sketch the solid whose volume is $\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{\sqrt{x^2+y^2}}^{1} dzdydx$. Find the volume.

Decoding the Cartesian Bounds (Understanding the Region). To sketch the solid, we must read the limits of integration from the outside in:

• The $xy$-shadow (Outer Integrals): The variable $x$ goes from $-1$ to $1$. For any fixed $x$, $y$ goes from the bottom half of a circle ($y = -\sqrt{1-x^2}$) to the top half of a circle ($y = \sqrt{1-x^2}$). Together, these bounds describe a solid disk of radius $1$ centered at the origin in the $xy$-plane.
• The $z$-height (Inner Integral): The variable $z$ ranges from a lower surface $z = \sqrt{x^2+y^2}$ to an upper surface $z = 1$. The lower boundary is the equation of a cone with its vertex at the origin, opening upward. The upper boundary is a flat, horizontal plane.

Convert to cylindrical coordinates. Evaluating this in Cartesian coordinates is unnecessarily difficult due to the square roots. Because the region has perfect circular symmetry around the $z$-axis, we convert to cylindrical coordinates $P = (r, \theta, z)$ where:

1. r is the radial distance from the z-axis to the point. In our particular example, $r$ ranges between $0$ to $1$ (the radius of the unit circle).
2. $\theta$ is the angle measured counterclockwise from the positive x-axis to the projection of $P$ onto the $xy$-plane. It ranges from $0$ to $2\pi$ since we want to cover the entire circle, $x^2 + y^2 = 1$.
3. z is the height of $P$ above (or below) the $xy$-plane. It ranges from ${\sqrt{x^2+y^2} = \sqrt{r^2}=r} ~\text{(the cone) to}~{1}~\text{(the plane)}$

Recall $z = z, x = r\cos(\theta), y = r\sin(\theta)$. Combining these contributions together, we get the volume element $dV$ in cylindrical coordinates: $dV = rdrd\theta dz$.

In simple terms, this solid is the “inside” of a right circular cone, capped off at a height of $z = 1$.

Set up the integral. $V = \int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{\sqrt{x^2+y^2}}^{1} dzdydx = \int_{0}^{2\pi}\int_{0}^{1}\int_{r}^{1} rdzdrd\theta$

Evaluate the integral.
Compute the inner integral (with respect to $z$): $\int_{r}^{1} rdz = rz\bigg|_{r}^{1} = r(1-r)$
Compute the middle integral (with respect to $r$): $\int_{0}^{1} r(1-r)dr = \frac{r^2}{2}-\frac{r^3}{3}\bigg|_{0}^{1} = \frac{1}{2}-\frac{1}{3} = \frac{3}{6}-\frac{2}{6} =\frac{1}{6}$
Compute the outer integral (with respect to $\theta$): $V = \int_0^{2\pi} \frac{1}{6} \, d\theta = \frac{1}{6} \left[ \theta \right]_0^{2\pi} = \frac{2\pi}{6} = \frac{\pi}{3}$

Sanity Check. The classical geometric formula for the volume of a cone is $V = \frac{1}{3}\pi r^2 h$. With a radius $r = 1$ and a height $h = 1$, the formula yields $\frac{\pi}{3}$. Our calculus exactly matches the geometry.

• Set up the integral used to evaluate $\iiint_E \sqrt{x^2+y^2} \, dV$ if $E$ is the solid bounded by the paraboloid $z = 9 - x^2 - y^2$ and the $xy$-plane.

Understanding the problem. First, take note of the integrand. Because we are integrating the function $f(x,y,z) = \sqrt{x^2+y^2}$ and not just $1$, we are not calculating volume. We are calculating the total accumulation of a scalar field over a volume.

Physically, if the density of this solid is proportional to its distance from the $z$-axis ($\rho = \sqrt{x^2+y^2}$), this integral calculates the total mass of the solid. The material gets denser the further you move toward the outer edges.

Determine the Bounds of Region $E$. The solid is bounded above by the paraboloid (which opens downward from the vertex at $z=9$) and below by the floor ($z=0$). To find the shadow on the $xy$-plane, we find the intersection where the paraboloid hits the floor: $0 = 9 - x^2 - y^2 \implies x^2 + y^2 = 9$

This is a circle centered at the origin with a radius of $3$. Refer to Figure A for a visual representation and aid in understanding it.

Convert to Cylindrical Coordinates. Because of the circular base, we translate to Cylindrical Coordinates, $(r, \theta, z)$:

• $r$-bounds: The circular base has radius $3$, so $0 \le r \le 3$.
• $\theta$-bounds: A full rotation, so $0 \le \theta \le 2\pi$.
• $z$-bounds: From the floor $z = 0$ to the paraboloid ceiling $z = 9 - r^2$.

Recall $z = z, x = r\cos(\theta), y = r\sin(\theta)$. Combining these contributions together, we get the volume element dV in cylindrical coordinates: $dV = rdrd\theta dz$. Therefore, the full integrand becomes $(r)(r \, dz \, dr \, d\theta) = r^2 \, dz \, dr \, d\theta$.

Set up the Integral: $I =\int_{0}^{2\pi}\int_{0}^{3}\int_{0}^{9-r^2} r·rdzdrdθ$

Let $I$ represent the value of the integral: $I = \int_0^{2\pi}\int_0^3\int_0^{9-r^2} r^2 \, dz \, dr \, d\theta$

Compute the inner integral (with respect to $z$): $\int_0^{9-r^2} r^2 \, dz = r^2 \left[ z \right]_0^{9-r^2} = r^2(9 - r^2) = 9r^2 - r^4$

Compute the middle integral (with respect to $r$): $\int_0^3 (9r^2 - r^4) \, dr = \left[ 3r^3 - \frac{r^5}{5} \right]_0^3$

$= \left( 3(27) - \frac{243}{5} \right) - 0 = 81 - \frac{243}{5} = \frac{405}{5} - \frac{243}{5} = \frac{162}{5}$

Compute the outer integral (with respect to $\theta$): $I = \int_0^{2\pi} \frac{162}{5} \, d\theta = \frac{162}{5} \left[ \theta \right]_0^{2\pi} = \frac{162(2\pi)}{5} = \frac{324\pi}{5}$

The value of the scalar integral over the solid $E$ is $\frac{324\pi}{5}$.

Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn, Andrew Misseldine, and MathMajor, YouTube’s channels.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. MIT OpenCourseWare, 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007, YouTube.
8. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences. c{1}{6}\theta \bigg|_{0}^{2\pi} = \frac{2\pi}{6} = \frac{\pi}{3}$.