Triple Integrals 3. Spherical coordinates

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.

Recall

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 P₀ and P₁ are the initial and final points of the curve C, respectively.

Equivalent Properties of Conservative Vector Fields

1. 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 $\int_{C} \vec{F}·d\vec{r} = 0$.

2. 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.

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 $\vec{F} = ∇f$, 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: $\int_{C} \vec{F}·d\vec{r} = f(P_1)-f(P_0).$

4. 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.

Criterion for a Conservative 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

• $\oint_C \vec{F} \cdot{} d\vec{r}$ represents the work or line integral of the vector field $\vec{F}$ around the curve C.
• $\iint_R (\frac{∂N}{∂x}-\frac{∂M}{∂y})dA$ represents the double integral over the region R, with the integrand being the difference between the partial derivatives of N and M.

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

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.
2. dV represents an infinitesimal volume element element within the region R. In Cartesian coordinates, dV = dx·dy·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 volument element is: dV = dxdydz.
• Cylindrical Coordinates (r, θ, z): Cylindrical coordinates are useful for regions with cylindrical symmetry, such as cylinders, cones, and circular regions. 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. θ: 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 = rcos(θ), y = rsin(θ), and the differential volume element is: dV = rdrdθdz.

• Spherical Coordinates (ρ, θ, ϕ).

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

1. ρ: The radial distance from the origin to a given point, it ranges from 0 to infinity.
2. φ: The polar angle, measured from the positive z-axis to the radial line r (the line connecting the origin to the point P). It describes how far the point is from the z-axis. It values increases from 0 (positive z-axis), π/2 (xy-plane), and π (negative z-axis), that is, 0 ≤ φ ≤ π.
3. θ: 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 2π, representing a full rotation around the z-axis. It describes the rotation around the z-axis.

The differential volume element dV = ρ²sin(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

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.

Parameters in Spherical Coordinates

• r or ρ is the radial distance. This is the distance from the origin to a given point P in space. It values range from 0 to infinity.
• φ is the polar angle, measured from the positive z-axis to the radial line r (the line connecting the origin to the point P). It values increases from 0 (positive z-axis), π/2 (xy-plane), and π (negative z-axis), that is, 0 ≤ φ ≤ π. It describes how far the point is from the z-axis.
• θ is 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 2π, representing a full rotation around the z-axis. It describes the rotation around the z-axis.

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).

Conversion to Cartesian Coordinates

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.

Calculating Volume Element dV

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 = ρ²sin(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.

Solved examples

• Sketch the surfaces in spheric coordinates:

1. ρ = 3.The equation ρ = 3 represents a sphere with a radius of 3 centered at the origin. In spherical coordinates, ρ is the radial distance from the origin to a point in space. Since ρ is constant and equals 3, this describes all points that are exactly 3 units away from the origin. (Figure a).
2. θ = ^π⁄₄. It describes a vertical plane in 3D space that extends infinitely in the z-direction. θ is the azimuthal angle measured in the xy-plane from the positive x-axis. A constant θ value indicates all points that lie on the same half-plane emanating from the z-axis, hence the surface is a half-plane that makes a 45-degree angle with the positive x-axis in the xy-plane. (Figure b).
3. φ = ^π⁄₄ is an upper half cone with an opening angle of 45° (Figure c). φ is the polar angle, which is measured from the positive z-axis. When φ is constant at ^π⁄₄, it traces out a conical surface. φ = ^3π⁄₄ is the lower half-cone, which is a cone opening downward at an angle of 135 degrees with respect to the z-axis. (Figure d).
4. ρ = 2cos(φ). The equation ρ = 2cos(φ) can be transformed into Cartesian coordinates for a clearer interpretation. $x^2+y^2+z^2 = 2z↭ x^2+y^2+(z-1)^2=1$. It is a sphere of center at (0, 0, 1) with a radius of 1 (Figure e).

• Find the mass of a sphere of radius 4 centered at the origin, with density function f(ρ, φ, θ) = 5 -ρ.

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 = ρ²sin(φ)dρdφdθ.

Boundaries in Spherical Coordinates:

1. Radial Distance (ρ): This is the distance from the origin to a given point. For a sphere of radius 4, ρ ranges from 0 (at the center of the sphere) to 4 (at the surface of the sphere).
2. Polar Angle (φ): This angle is measured between the positive z-axis and the line connecting the origin to the point. For a full sphere, ϕ ranges from 0 (pointing directly upwards along the positive z-axis) to π (pointing directly downwards along the negative z-axis).
3. Azimuthal Angle (θ): The angle is measured in the xy-plane from the positive x-axis and the projection of the line connecting the origin to the point onto the xy-plane. For a full rotation around the z-axis, θ ranges from 0 to 2π.

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

1. Calculating the inner integral with respect to ρ: $\int_{0}^{4} (5ρ^2-ρ^3)dρ = (\frac{5}{3}ρ^3-\frac{1}{4}ρ^4)\bigg|_{0}^{4} = (\frac{5}{3}64-\frac{1}{4}256) = \frac{320}{3}-64=\frac{128}{3}$

2. Calculating the middle integral with respect to φ: $\int_{0}^{π} \frac{128}{3}sin(φ)dφ = -\frac{128}{3}cos(φ)\bigg|_{0}^{π} = -\frac{128}{3}(-1-1)=\frac{256}{3}$

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.

• We aim to find the volume of the unit sphere ρ = 1 that lies above the plane z = $\frac{1}{\sqrt{2}}$ (Figure A)

Understanding the problem: The unit sphere is defined by the equation: x² + y² + z² = 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 = ρ² sin(ϕ)dρdϕdθ.

Determining the Limits of Integration

1. Radial distance. ρ ranges from to the intersection with the plane $\frac{sec(ϕ)}{\sqrt{2}}$ (solved from ρ= $\frac{1}{\sqrt{2}}$) to the surface of the sphere at ρ = 1
2. Polar Angle. ϕ ranges from 0 (we start at the North Pole of the sphere) to $\frac{π}{4}$ (where the plane intersects the sphere, Figure B).

   The lower bound for ϕ (Figure B) is the point of the intersection between the sphere ρ = 1 and the plane $z = \frac{1}{\sqrt{2}} = ρcos(φ)⇒cos(φ) = \frac{1}{\sqrt{2}} ⇒ φ = \frac{π}{4}$

3. Azimuthal Angle. As we move all move all around the z axis, θ ∈ [0, 2π].

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)

$\int_{0}^{\frac{\pi}{4}} tan(φ)sec^2(φ) = \int_{0}^{\frac{\pi}{4}} udu = \frac{u^2}{2} = \frac{sec^2(φ)}{2}\bigg|_{0}^{\frac{\pi}{4}} = 1-\frac{1}{2} = \frac{1}{2}$.

• At $φ = \frac{\pi}{4}, sec(\frac{\pi}{4})=$1/(√2/2)=2/√2=(2/√2)·(√2/√2) = 2√2/2 = √2, $sec^2(\frac{\pi}{4})=2$
• At $φ = 0, sec(0)=1, sec^2(0)=1$

Finally, the outer integral is:

$\int_{0}^{2π} \frac{1}{3}(\frac{-\sqrt{2}}{2}+1)- \frac{1}{3·2^2·\sqrt{2}} dθ = 2π[ \frac{1}{3}(\frac{-\sqrt{2}}{2}+1)- \frac{1}{12·\sqrt{2}} ] = 2π[\frac{-\sqrt{2}+2}{3}- \frac{1}{12·\sqrt{2}}] = 2π[\frac{-2\sqrt{2}+4-1}{12\sqrt{2}}] = 2π[\frac{-2\sqrt{2}+3}{12\sqrt{2}}] = \frac{-4π\sqrt{2}+6π}{12\sqrt{2}} = \frac{-π}{3} + \frac{π}{2\sqrt{2}}$

• Find the volume inside the cone φ = ^π/₃ and inside the sphere ρ = 6 (Figure 1).

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 = ρ²sin(φ)dρdφdθ.

Boundaries in Spherical Coordinates:

1. ρ: Radial distance from the origin to the point. ρ varies from 0 to 6, which is the radius of the sphere.
2. φ: Polar angle, the angle between the positive z-axis and the line connecting the origin to the point. It varies from 0 to ^π⁄₃, which is the angle defining the cone.
3. θ: Azimuthal angle, the angle between the projection of the line onto the xy-plane and the positive x-axis. From 0 to 2π, corresponding to a full rotation around the z-axis.

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.

• $\int_{0}^{2} \int_{0}^{\sqrt{4-y^2}} \int_{\sqrt{3(x^2+y^2)}}^{\sqrt{16-x^2-y^2}} (x^2+y^2+z^2)dzdxdy$

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 (x²+y²+z² = 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

1. Azimuthal Angle 0 ≤ θ ≤ ^π⁄₂, around the z-axis for the first quadrant (x ≥ 0, y ≥ 0).
2. Radial Distance, 0 ≤ ρ ≤ 4, ρ ranges from the origin to the sphere’s surface (radius 4).
3. Polar Angle φ From the z-axis down to the cone’s surface, $z=\sqrt{3(x^2+y^2)}↭ ρcos(φ)=\sqrt{3(x^2+y^2)}=\sqrt{3(ρ^2sin^2(φ)(cos^2(θ)+sin^2(θ)))}↭ρcos(φ)=\sqrt{3ρ^2sin^2(φ)}=\sqrt{3}ρsin(φ)↭tan(φ) = \frac{1}{\sqrt{3}}↭ φ = \frac{π}{6}$, so 0 ≤ φ ≤ ^π⁄₆.

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 ρ² 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

1. Inner integral with respect to ρ: $\int_{0}^{4} (ρ^4)dρ = \frac{ρ^5}{5}\bigg|_{0}^{4} = \frac{4^5}{5} = \frac{1024}{5}$

2. Middle integral with respect to θ: $\int_{0}^{\frac{π}{2}} \frac{1024}{5}θ\bigg|_{0}^{\frac{π}{2}} = \frac{1024}{5}·\frac{π}{2} = \frac{1024π}{10} = \frac{512π}{5}$

3. 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: $\frac{512π}{5}(1-\frac{\sqrt{3}}{2})$

• Evaluate $\int\int\int_{E} (x^2+y^2)dV$ if E is the solid region below the sphere x² + y² + z² = 1 and above the xy-plane.

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 x² + y² + z² = 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 = ρ²sin(φ)dρdθdφ

Spherical Coordinates Boundaries

1. Azimuthal Angle 0 ≤ θ ≤ 2π, i.e., θ varies from 0 to 2π, covering the full rotation around the z-axis.
2. Radial Distance, 0 ≤ ρ ≤ 1, ρ ranges from the origin to the sphere’s surface (radius 1).
3. Polar Angle φ, from 0 to ^π⁄₂, covering the region from the positive z-axis down to the xy-plane.

$\int_{0}^{2π} \int_{0}^{\frac{π}{2}} \int_{0}^{1} (ρsin(φ))^2 ρ^2sin(φ)dρdφdθ$

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: $\int_{0}^{2π} \frac{1}{5}\frac{2}{3}dθ = \frac{2}{15}θ\bigg|_{0}^{2π} = \frac{2}{15}·2π = \frac{4π}{15}$.

• Set up integrals in all three systems that represent the volume of a cylinder of radius 2 and height 3 (Refer to Figure ii for a visual representation and aid in understanding it)

In rectangular (Cartesian) coordinates, the cylinder is aligned with the z-axis, and its equation is defined by the following constraints:

1. The cylinder has a radius of 2, which means that x² + y² ≤ 4 (this represents the circular cross-section of the cylinder).
2. The height of the cylinder is 3, which means z ranges from 0 to 3.

Explanation of the limits:

1. x ranges from −2 to 2, which represents the width of the cylinder’s base.
2. For each fixed x, y ranges from $-\sqrt{4-x^2}$ to $\sqrt{4-x^2}$, which is the projection of the circular cross-section of the cylinder onto the xy-plane.
3. z ranges from 0 to 3, representing the height of the cylinder.

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 radius of the cylinder is 2, so r ranges from 0 to 2.
• The height of the cylinder is 3, so z ranges from 0 to 3 (height of the cylinder).
• The angle θ covers the entire circle, so it ranges from 0 to 0 to 2π.

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 = ρ²sin(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 top of the cylinder corresponds to z = 3, which in spherical coordinates is ρcos(φ) = 3, hence ρ = 3sec(φ);

1. ρ for the top of the cylinder is bounded by 0 and 3sec(φ).
2. φ ranges from 0 to $tan^{-1}(\frac{2}{3}) = arctan(\frac{2}{3})$, which corresponds to the angle at the edge where the top intersects with the side of the cylinder. ⁡3. θ ranges from 0 to 2π, covering the full circle around the z-axis.

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

• The side of the cylinder is defined by r = 2, which in spherical coordinates becomes $\sqrt{x^2+y^2} = 2 ↭ \sqrt{ρ^2sin^2(φ)cos^2(θ) + ρ^2sin^2(φ)sin^2(θ)} = 2 ↭ \sqrt{ρ^2sin^2(φ)} = 2↭ ρsin(φ) = 2 ⇒ρ = 2csc(φ)$ The cosecant (csc) of an angle is the reciprocal of the sine of that angle: csc(φ) = 1 / sin(φ).

1. At the side, φ ranges from $tan^{-1}(\frac{2}{3})$, which corresponds to the angle at the edge of the cylinder to π/2, which corresponds to the xy-plane.
2. ρ is bounded by 0 and 2csc(φ).
3. θ again ranges from 0 to 2π, covering the full circle around the z-axis.

$\int_{0}^{2π} \int_{tan^{-1}(\frac{2}{3})}^{\frac{π}{2}} \int_{0}^{2csc(φ)} ρ^2sin(φ)dρdφdθ$

The total volume of the cylinder is the sum of these two integrals:

$\int_{0}^{2π} \int_{0}^{tan^{-1}(\frac{2}{3})} \int_{0}^{3sec(φ)} ρ^2sin(φ)dρdφdθ + \int_{0}^{2π} \int_{tan^{-1}(\frac{2}{3})}^{\frac{π}{2}} \int_{0}^{2csc(φ)} ρ^2sin(φ)dρdφdθ$

Calculation of Gravitational Force Exerted by an object

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:

1. F is the gravitational force acting between the two objects (masses).
2. m₁ and m₂ are the masses of the two objects.
3. r is the distance between the centers of their masses.
4. G is the gravitational constant.

Objective

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:

1. $|\vec{F}|$ is the magnitude of the gravitational force.
2. ρ is the distance from the differential mass to the origin: $ρ = \sqrt{x^2+y^2+z^2}$
3. ⟨x, y, z⟩ is the position vector of the differential mass.

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 = ρ²sin(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: $F_z = \int \int \int \frac{G·m·z}{ρ^3}δdV = Gm\int \int \int \frac{ρcos(φ)}{ρ^3}δ·ρ^2sin(φ)dρdφdθ = Gm \int \int \int δ·cos(φ)sin(φ)dρdφdθ.$

Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Calculus and Calculus 3e (Apex). Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn, and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
8. MIT OpenCourseWare 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007.
9. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.