ALTEREDZINE

What is a vector line integral and how to evaluate it?

2026-10-16T16:06:12.842-04:00

In a recent post, we defined a line integral and showed how to evaluate it. We considered a function over its curve domain. We divided the domain curve into small pieces, took a point P inside each piece, evaluated it by the scalar function and multiply f(P) by the length of a piece Δs. We added all the products f(P)Δs, let Δs shrink to zero and took the limit of the sum. The result is the scalar line integral. Now the function considered is a vector field, can we use the same method used for the scalar line integral? Let's make the following considerations. A scalar line integral add up quantities like density or temperature along a curve. A vector field evaluates how much a field pushes in the direction of travel along a curve. We will make this consideration in evaluating the vector Mathematics and Physics entangle. Let's consider a physical problem to calculate the vector line integral. We consider the travel of a bird along a definite path and we want to evaluate how much the wind helps it move along that path.

Let

be the vector field that represents the force of the wind that moves the bird along the chosen path. This path is represented by a parameterized curve r(t) such that a ≤ t ≤ b. The wind is moving in a positive direction which is the direction given by the curve. The opposite direction would be negative.

Let P₀, P₁, P₂, ......... Pₙ divide the curve into n pieces. The distance from Pi-1 to Pi is represented by Δsᵢ. Let P*ᵢ be a point between those the correspond force is F(P*ᵢ). If the distance Δsᵢ is small, when the bird is moving from Pi-1 to Pi it's moving in the direction of the tangent T(P*i). The work done by the force of the wind is given by: F(P*i).(Δsᵢ.T(P*i)). The total work done along C is given by:

Let the arc length of C be arbitrarily small as n→∞, the work done by the wind in moving the bird along its path is represented by:

Definition

The vector line integral of a vector field F along a smooth curve is given by:

The unit tangent vector is given by T = r'(t)/𐽕r'(t)𐽕 and ds = 𐽕r'(t)𐽕.dt. Let's calculate F.Tds by substituting T and ds:

Then the vector line integral becomes:

Example

Solution

Let's use the formula:

Let's calculate F(r(t)) and r'(t):

Let's substitute F(r(t)) and r'(t) in the formula:

Practice

Find the value of the integral ∫_C F · dr, where C is the semicircle of radius 2 parameterized by r(t) = ⟨2 cos t, 2 sin t⟩, 0 ≤ t ≤ π, and F = ⟨−y, x⟩.

Scalar line integral: Independence of parameterization

2026-10-09T17:34:03.722-04:00

Objective: Show that the value of a line integral doesn't change if the parameterization of the curve changes.

In the previous post we calculated the value of a line integral. What if the parameterization of the curve changes? Let's use the same example and change the parameterization of the curve?

Example

Solution

Let's apply the formula to evaluate a line integral:

Let's calculate f(r(t)) by substituting x, y and z in the function:

Let's calculate the radical expression:

Let's substitute f(x,y.z), f(r(t)) and the radical expression in the formula:

We obtain the same result as in the previous example. Scalar line integrals are independent of parameterization.

Practice

What is a scalar line integral and how to evaluate it?

2026-10-05T17:15:29.319-04:00

Objective: Evaluate a scalar line integral

Let's consider a simple integral of the form:

The domain of integration [a b] is a line segment that can be considered as a curve in the x-axis even though it doesn't have a curvature. Let's say that we want to integrate a function over a domain of integration that is not a straight line. The resulting integral is a line integral. A line integral is a type of integral with a domain of integration that can be a line or a curve. The line integral can be the integral of a multivariable function or the the integral of a vector field over a domain of integration that is a curve. In the case of a multivariable function the line integral is called a scalar line integral. If the function is a vector field the line integral is called a vector line integral.

How to evaluate a scalar line integral?

We divide the curve C into small pieces of curve C₁, C₂, C₃,......Cₙ by the points P₀, P₁, P₂....Pₙ. Each piece has a length Δs₁, Δs₂, Δs₃.....Δsₙ. Let choose a point P*ᵢ inside the piece Δsᵢ and evaluate f at P we have f(P). Multiplying f(P*i) by Δsᵢ we have the product f(P*i)Δs. Add all the products and let the arc length of the pieces shrink to zero by taking a limit we have:

Calculating the integral using this formula would be cumbersome. The following theorems provide better ways to evaluate a scalar line integral.

Theorem I

Theorem II

Let f be a continuous function with a domain that includes a smooth curve C with parameterization:

Let's substitute:

in the previous formula. Then we have a definite expression to calculate the line integral:

Example

Solution

Let's use the formula in theorem II :

Let's calculate f(r,t) and the radical expression:

Let's substitute the given scalar line integral and these 2 expressions in the formula:

Evaluating the integral on the right we have:

Finally the value of the given integral is:

Practice

Vector field of a fluid

2026-07-04T13:40:31.118-04:00

Objective: Define the vector field of a fluid

Let's observe the movement of a fluid: air, water, oil. Each point of the fluid moves in a direction and at a certain speed at any given time. The speed and the direction of the movement of the fluid together represent the velocity of the fluid at that point. The concept of velocity and speed can be both understood mathematically. The velocity at a point is represented by a vector. The magnitude of the vector represents the speed of the fluid at that point. The set of all vectors drawn at different points represent the vector field of the fluid.

Example

Solution

Practice

Using vector field to compare two physical phenomenas : a storm and a whirlpool

2026-06-27T15:02:10.950-04:00

Objectives:

1) Compare two physical phenomenas: a storm and a whirlpool

2) Use the vector field to show the physical similarities and one distinction

When scientists study a storm, they can see the movement inside and the direction of the movement. They can also locate the center of the storm. As the movement goes further from the center, the intensity of the movement increases. In the previous post, we have used the vector field to model the shapes, the direction and the relative intensity of these movements.

A whirlpool is a body of rotating water in which the flow moves in a circular pattern around a central point. The defining physical characteristic of a whirlpool is that the water spins faster near the centre and slower farther away. This is quite different in the case of a storm. We are going to use the same vector field method to study the physical characteristics of the whirlpool. We will also show that the vector field of the storm and that of the whirlpool confirm the physical characteristics of of these two phenomena. Let's start by drawing the vector field of the whirlpool.

Let's consider a whirlpool that can be modeled by the following function:

Choosing the base points, we obtain the following table of values:

Let's draw the vector field:

This field confirms the physical characteristics of the whirlpool. It moves in a circular clockwise direction around a central point.

Let's calculate the norm of any vector F(a,b) of the field:

As the vectors approach to the origin, the magnitude goes to infinity or become very large. This confirms the physical characteristic of the whirlpool where the intensity of the movement of the water grows as we approach to the center.

If we compare the vector field of the whirlpool to that of the storm, we can see that they are both rotational. The only difference is that in the case of the vector field of the storm the magnitude of the vectors increase as we goes farther from the origin. Therefore the intensity of the movement of the storm increases as we move away from the center. In the case of the vector field of the whirlpool the magnitude of the vectors increase as we approach to the center. Therefore the intensity of the movement increases as we approach to the center.

Drawing a rotational field

2026-06-13T12:48:42.726-04:00

Objective: Draw a rotational field

Rotational fields are used to study major storm systems including cyclones and hurricanes. In the northern hemisphere, storms rotate counterclockwise. In the southern hemisphere they rotate clockwise. This is an effect of the earth's rotation around its axis called Coriolis effect.

Example

Draw the rotational field F<x,y> = (y, -x)

Solution

The vector field is now complete. We can show now that the vectors are tangent to the circles of radius r = 1, r -⎷ 2, r = 2.

We can also compare the rotational field to the radial field and show that the vectors of the rotational field are perpendicular to the ones in the radial field. The following figure shows that.

We can confirm the perpendicularity of the vectors of the radial field to the vectors of the rotational by showing that their dot product is equal to zero.

Drawing a radial vector field

2026-06-06T18:58:56.866-04:00

Objective: draw a radial field

Considerations

A radial field models certain types of gravitational fields and energy source fields. To draw a vector field you have to draw a vector at each point of a set of points. Let's say that you have to draw the vector F(2.1) = <1.1/2>, you draw first the point (2,1) then you move one unit to the right and half unit up. The tail of the vector is at the point (2,1).

In a radial field all the vectors point directly toward or away from the origin. The vector located at (x,y) is perpendicular to the circle centered at the origin and passing through (x,y). All the vectors on that circle have the same magnitude.

Method to draw a radial field

1) Choose a sample of points from each quadrant in the system of coordinates plane. It's better to choose a set of points at the intersection of the grid lines.

2) Draw the corresponding vector at each point

Example

Solution

Figure (a) shows the vector field. Figure (b) shows circles overlain on the vector field.

Practice

What are vector fields in Calculus?

2026-06-02T17:05:54.520-04:00

Objective: Define vector field

Introduction

Vector fields are used to describe physical phenomena such as gravitation, electromagnetism, atmospheric storms and deep-sea ocean currents. Let's give some examples.

Example (a). Gravitational force exerted by two astronomical objects such as a planet and a star or a planet and its moon.

The figure (a) models the gravitational force exerted by these two objects:

At any point in the figure, a vector associated with that point gives the net gravitational force exerted by the two objects on a unit of mass of an object. The vectors of the greater magnitude are associated with the largest object. The greater the mass of an object, the stronger is its gravitational force. The gravitational force of the larger object is stronger than that of the smaller object. The vectors of smaller magnitude are associated with the smaller object. The mass of this object is smaller. Its gravitational force is less strong than the larger object.

Example (b). Velocity of a river at points on its surface.

The vector field velocity of water on a river shows the varied speeds of water. Vectors in red are of greater magnitude. Therefore, the water flows quickly. Vectors in blue are smaller in magnitude. They show a smaller flow of water.

As we can see in these two examples, a vector field is a map of vectors. In the following posts we will study vectors fields in R² and R³.

Definition

Example

Solution

Practice

Solving a triple integral using the change of variables

2026-05-23T20:11:35.686-04:00

In the previous posts we spent time establishing formulas to solve a triple integral using a change of variables. Let's now solve an example.

Example

Solution

Let's apply the formula:

Let's take the following steps to apply the formula:

1) Let's find x, y, z from the given values of u, v, w. This leads to solve the following system of equations:

u = 2x-y/2 (1)

v = y/2 (2)

w = z/3 (3)

Solving this system, we find x = (u + v)/2 y = 2v z = 3w

2) Let's find the new function H(u,v,w) to be integrated by substituting x, y, z in the function F(x,y,z) = x +z/3.

3) Let's find the limits of integration of the new function i.e u.v.w:

Let's isolate x, y, z as limits of integration of the given integral. We have:

x = y/2 x = y/2 + 1

y = 0 y = 4

z = 0 z = 3

These equations represent the planes that bound the surface G in the space xyz.

From the equation x=y/2 we have 2x = y 2x-y = 0 ⇒2x-y/2 = 0 ⇒ u = 0 since u = 2x-y/2 (1)

From the equation x = y/2 + 1 we have 2x = y+2 ⇒ 2x-y = 2 ⇒ 2x-y/2 = 1 ⇒ u = 1.

Let's use the equation v = y/2:

For y = 0 we have v = 0. For y = 4 v =2

Let's use the equation w = z/3

For z = 0 w = 0

For z =3 w = 1

Let's calculate the jacobian:

Change of variables for triple integrals

2026-05-16T17:33:31.358-04:00

Objective: Set the formula for the change of variables in triple integrals.

Planar transformation in tridimensional space

a.The change of variables in triple integrals works the same way as in double integrals. We consider a transformation from a tridimensional space (u,v.w) to a tridimensional space (x,y,z). We define the Jacobian in the (u,v,w) space. Then we establish the formula for the triple integral defined in (u,v,w).

Suppose that G i s a region in the uvww space transformed in another region D in the xyz space by a C¹ transformation by the transformation T so that T(u,v,w) = (x,y,z) where x = g(u,v,w) y = h(u,v.w) z = k(u,v,w).

Any function F(x.y,z) defined on D can be thought of another function H(u,v. w) defined G such that:

F(x.y.z) = F(g(u,v,w), h(u,v,w), k(u,v,w)) = H(u,v,w), Now let's define the Jacobian in a tridimensional space (u,v,w

Jacobian for three variables. Definition

The jacobian J(u,v,w) for three variables is defined as follows:

This is the same as:

Theorem

Change of Variables for Triple Integrals

Let T(u, v, w) = (x, y, z) where x = g(u, v, w), y = h(u, v, w), and z = k(u, v, w), be a one-to-one C¹ transformation, with a nonzero Jacobian, that maps the region G in the uvw-space into the region R in the xyz-space. As in the two-dimensional case, if F is continuous on R, then

∫∫∫_R F(x,y,z) dV = ∫∫∫_G F(g(u,v,w), h(u,v,w), k(u,v,w)) | ∂(x,y,z) ∂(u,v,w) | du dv dw
= ∫∫∫_G H(u,v,w) |J(u,v,w)| du dv dw.

Example

Obtaining Formulas in Triple Integrals for Cylindrical and Spherical Coordinates

Derive the formula in triple integrals for

cylindrical and
spherical coordinates.

Solution

Let's apply the formula for change of variables in triple integrals:

∫∫∫_R F(x,y,z) dV = ∫∫∫_G H(u,v,w) |J(u,v,w)| du dv dw

In cylindrical coordinates it becomes:

∫∫∫_D f(x,y,z) dV = ∫∫∫_G f(r cosθ, r sinθ, z) |J(r,θ,z)| dr dθ dz

J(r, θ, z) = |

∂x/∂r	∂x/∂θ	∂x/∂z
∂y/∂r	∂y/∂θ	∂y/∂z
∂z/∂r	∂z/∂θ	∂z/∂z

| = |

cos θ	−r sin θ	0
sin θ	r cos θ	0
0	0	1

| = r cos²θ + r sin²θ = r (cos²θ + sin²θ) = r.

We know that r ≥ 0, so |J(r,θ,z)| = r. Then the triple integral becomes by substituting the value of the jacobian:
∫∫∫_D f(x,y,z) dV = ∫∫∫_G f(r cos θ, r sin θ, z) r dr dθ dz.

b. Let's find the formula for the triple integral in spherical coordinates by applying the same process.

For spherical coordinates, the transformation is T(ρ, θ, φ) = (x, y, z) from the Cartesian pθφ-plane to the Cartesian xyz-plane . Here x = ρ sin φ cos θ, y = ρ sin φ sin θ, and z = ρ cos φ. The expression of the triple integral can be written as:

∫∫∫_D f(x,y,z) dV = ∫∫∫_G f(ρ sinφ cosθ, ρ sinφ sinθ, ρ cosφ) |J(ρ,θ,φ)| dρ dφ dθ. Let's calculate the jacobian.

First write it as a quotient of partial derivatives:

J(ρ,θ,φ) = ∂(x,y,z) ∂(ρ,θ,φ)

Now write it as a determinant:

J(ρ,θ,φ) = |

∂x/∂ρ	∂x/∂θ	∂x/∂φ
∂y/∂ρ	∂y/∂θ	∂y/∂φ
∂z/∂ρ	∂z/∂θ	∂z/∂φ

Substitute the partial derivatives:

J(ρ,θ,φ) = |

sinφ cosθ	-ρ sinφ sinθ	ρ cosφ cosθ
sinφ sinθ	ρ sinφ cosθ	ρ cosφ sinθ
cosφ	0	-ρ sinφ

Expand along the third row:

J = cosφ |

-ρ sinφ sinθ	ρ cosφ cosθ
ρ sinφ cosθ	ρ cosφ sinθ

| - ρ sinφ |

sinφ cosθ	-ρ sinφ sinθ
sinφ sinθ	ρ sinφ cosθ

Calculate the first determinant:

(-ρ sinφ sinθ)(ρ cosφ sinθ) - (ρ cosφ cosθ)(ρ sinφ cosθ)

= -ρ² sinφ cosφ sin²θ - ρ² sinφ cosφ cos²θ

= -ρ² sinφ cosφ

Calculate the second determinant:

(sinφ cosθ)(ρ sinφ cosθ) - (-ρ sinφ sinθ)(sinφ sinθ)

= ρ sin²φ cos²θ + ρ sin²φ sin²θ

= ρ sin²φ

Therefore,

J = cosφ(-ρ² sinφ cosφ) - ρ sinφ(ρ sin²φ)

J = -ρ² sinφ cos²φ - ρ² sin³φ

J = -ρ² sinφ(cos²φ + sin²φ)

J(ρ,θ,φ) = -ρ² sinφ

So,

|J(ρ,θ,φ)| = ρ² sinφ

Let's substitute the jacobian in the expression of the integral. Then the triple integral becomes:

∫∫∫_D f(x,y,z) dV = ∫∫∫_G f(ρ sinφ cosθ, ρ sinφ sinθ, ρ cosφ) ρ² sinφ dρ dφ dθ.

Change of variables for double integrals

2026-05-09T18:28:00.001-04:00

We have seen that under the change of variables T(u,v) = (x,y) where x = g(u,v) and y = h(u,v), a small region ΔA in the xy plane is related to the area formed by the product 𝜟u 𝜟v in the uv plane by the approximation: 𝜟A ≃J(u,v)𝜟u𝜟

Remember that the double integral is defined as:

In the following figure we divide the region S in small rectangles Sᵢⱼ and the region R in small rectangles Rᵢⱼ. A small rectangle Rᵢⱼ is the image of a small rectangle Sᵢⱼ under the transformation T.

Let's substitute f(xᵢⱼ, yᵢⱼ) and ΔA in the definition of the integral:

∫∫_R f(x,y) dA = ∫∫_S f(g(u,v),h(u,v)) |J(u,v)| du dv

Let's substitute J(u,v) in in the expression on the right side of the equality sign, we have:

Theorem

Example

Solution

Let's do the change of variables in the expression √x² + y² . By substituting x = rcosθ and y = sinθ, the expression becomes equal to r.

The expression dydx becomes J(r,θ) drdθ. In a previous example J(r,θ) was equal to r. The integral becomes when we substitute everything:

Practice

What are Jacobians in multiple integrals?

2026-05-02T18:02:00.121-04:00

Let's consider a one-to-one C¹ transformation defined by:

We want to see how it transforms a small rectangle with sides Δu and Δv in the (u,v) plane.

Let's call ΔA the area of the curved region R of sides Δurᵤ and Δvrᵥ. This area is approximately equal to the norm of the product of the two sides. It can be written as:

ΔA≃ ॥Δurᵤ॥.॥Δvrᵥ॥ = ॥Δu॥॥rᵤ॥. ॥Δv॥rᵥ॥ = Δu॥rᵤ॥.Δv॥rᵥ॥ = ॥rᵤ॥॥rᵥ॥ ΔuΔv

Let's find rᵤ and rᵥ in order to calculate the norm of their product.

Let's calculate the product of the vectors rᵤ and rᵥ. This is the tricky part. The surface is sitting in a tridimensional space, which is R³. We can rewrite rᵤ and rᵥ as:

rᵤ = ∂x/∂ui + ∂x/∂uj + 0k

rᵥ = ∂x/∂vi + ∂x/∂vj + 0k

So, the coordinates of the tangent vectors rᵤ and rᵥ at (u₀, v₀) are:

𝐫ᵤ = ( ∂x/∂u, ∂y/∂u, 0 )

𝐫ᵥ(u₀,v₀) = ( ∂x/∂v, ∂y/∂v, 0 )

Let's calculate the product of the vectors now:

Let's take the norm of the product:

॥rᵤ 🗙 rᵥ॥ =∥ ∂x/∂u)(∂y/∂v) − (∂x/∂v)(∂y/∂u) ) 𝐤 ) ∥

= ॥ (∂x/∂u)(∂y/∂v) − (∂x/∂v)(∂y/∂u)॥ ∥𝐤∥

= (∂x/∂u)(∂y/∂v) − (∂x/∂v)(∂y/∂u) since ∥𝐤∥ = 1

Let's substitute in the expression of ΔA. We have:

The jacobian can finally be written as:

J(u,v) =

Example. Find the jacobian of the transformation given in the following example:

Solution

Finding the image of a triangle by a transformation

2026-04-21T15:40:00.000-04:00

The previous example consists in showing that a transformation is a one-to-one transformation. Now we are concerned about how to find the image of a triangle under a transformation. Let's use an example in this case.

Example

Solution

The triangle and its image are shown in the figure below. To understand how the sides of the triangle transform, call the side that joins (0,0) and (0,1) side A, the side that joins (0,0) and (1,1) side B and the side that joins (0,1) and (1,1) side C.

Practice

Change of variables in multiple integrals. Planar Transformations

2026-04-17T16:01:00.004-04:00

Objective: Determine the image of a region under a given transformation of variables

Planar Transformation

A planar transformation is a function that transforms a region G in one plane into another region R into another plane by a change of variables. Both G and R are subsets of R². The figure bellow shows a region G in the (u,v) plane transformed into another region R in the (x,y) plane by the change of variables x = g(u,v) and y = h(u,v).

Definition

A transformation T: G → R defined as T(u,v) = (x,y) is said to be one- to- one transformation if no two points map the same image point.

Example

Solution

first quadrant of the xy plane. Hence R is a quarter circle bounded by x² + y² = 1 in the first quadrant.

Finding the moments of inertia of a solid in three dimensions

2026-04-06T11:10:00.003-04:00

The formulas to calculate the moments of inertia being known, let's solve a problem to apply them.

Example

Suppose the region Q is bounded by the plane x+2y+3z = 0 and the coordinates planes with density ⍴ = x²yz (see figure in this example). Find the moments of inertia about the yz plane, the xz plane, the xy plane.

Solution

Let's use the formulas already established

Practice

Consider the same region Q with density function ⍴(x,y,z) = xy²z. Find the moments of inertia about the three coordinate planes.

Finding the center of mass of a solid in 3 dimensions

2026-04-02T20:22:00.000-04:00

Goal: Find the center of mass of a solid in 3 dimensions

We already stated the formulas to calculate the center of mass of a solid in 3 dimensions. Let's solve an example.

Example

Suppose Q is a solid region bounded by the plane x + 2y + 3z = 0, the coordinates planes with density ϼ(x,y,z) = x²yz (see figure in the example in the previous post). Find the center of mass using decimal approximation. Use the mass found in the previous example.

Solution

Practice

Consider the same region Q and the density function ρ(x,y,z) = xy²z. Find the center of mass using the following figure used in this example.

Center of mass and moments

2026-03-21T08:56:00.000-04:00

The expressions of mass, center of mass, moments of inertia expressed in double integrals can be modified by replacing the double integrals with triple integrals.

Example

Solution

The region Q is a tetrahedron meeting the axes at the point (6,0,0), (0,3,0) and (0,0,2) (see figure below). To find the limits of integration, let z = 0 in the slanted plane z = 1/3(6-x-2y). Then for x and y find the projection of Q on the the xy plane which is bounded by the axes and the line x_+ 2y = 6. The mass is calculated as follow:

Finding the moments of inertia of a solid in two dimensions.

2026-03-14T17:21:00.022-04:00

Goal: Find the moment of inertia of a solid in two dimensions

Let's go back to the lamina considered in earlier post where we calculated its mass. In order to do that we considered the region R occupied by the lamina. We divided the lamina into tiny subrectangles. Our goal now is to find the different moments of inertia of the solid.

Let's find the moment of inertia about the x-axis.

The moment of inertia of a solid about an axis is equal to its mass by the square of its distance from this axis.

The moment of inertia of the lamina about the x-axis is equal to the sum of the moments of the tiny subrectangles of the lamina.

The moment of inertia of a subrectangle about the x-axis is equal to the product of its mass by the square of its distance from the x-axis. This moment is calculated as:

Let's add all the moments of the subrectangles. This comes to take the Rieman sum of the product and to determine its limit.

The moment of inertia of a subrectangle about the y-axis is equal to the product of its mass by the square of its distance from the t-axis. Hence, the moment is given by:

Adding all the moments together allows to find the moment of the lamina about the y-axis. This leads to use the Rieman sum of the product and take its limit.

Example

Solution

Practice

Center of mass in two dimensions

2026-03-07T12:03:00.000-05:00

The center of mass of an object is called the center of gravity if the object is located in a gravitational field. If the object has a uniform density, the center of mass is the geometric center of the object, which is called the centroid. The following figure shows the point P as the center of mass of a lamina (flat plate).

Restating the center of mass in terms of integrals, we have:

If the object has uniform density, the density function ⍴(x,y) is constant and the formulas become:

Example

Solution

Practice

Application of double integrals: finding moments about x and y axes

2026-02-28T19:43:00.001-05:00

Objective: Finding moments about x and y axis

Let's find the moments with respect to x and y of a region R with density function ⍴(x,y). We divide the region in small rectangles for which the density is constant. We add the moments of each of these rectangles and take the limit of the sum as as the rectangles approach zero.

If we name Rᵢⱼ the small rectangle, the moment with respect to the x-axis is:

Similarly, the moment with respect to the y-axis is defined by:

Example

We use the same example in a previous blog post

Solution

Practice

Applications of double Integrals: Mass of an object in a two-dimensional space.

2026-02-21T21:34:00.005-05:00

Let's consider a lamina (a thin plate) that occupies the region R of a two-dimensional space. Let (x,y) be a point in the region R surrounded by a small rectangle. The density of the lamina at that point is a function ⍴(x,y). This function is determined by:

where 𝚫m and 𝚫A represent the mass and the area of the small rectangle.

Let's divide the region R into tiny rectangles of area ΔA. The mass of a tiny rectangle is given by:

Let's add the tiny rectangles together and take the limit of the sum when delta x and delta y approach zero. Since we have two variables, we know by experience that the limit is a double sum of Rieman. This limit represents a double integral that allows us to find the mass off the lamina.

Example

Solution

Let's sketch the region R:

Practice

Find the volume inside of an ellipsoid and outside of a sphere

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We are going to use an example to do that:

Example:

Solution

First let's find the volume using a = 75ft b = 80 ft c = 90ft using the result from the example treated previously about the volume of the ellipsoid. Hence the volume of the ellipsoid is:

From the result from the example about the volume of a sphere, we get:

Volume of an ellipsoid using spherical coordinates

2026-02-09T20:50:00.002-05:00

Let's use an example to calculate the volume of an ellipsoid using spherical coordinates.

Example

Solution

Let's use the change of variables that corresponds to an ellipsoid. We still change the variables from rectangular coordinates to spherical coordinates.

The volume V of the ellipsoid is given by:

Let's change dV = dxdydz in spherical coordinates:

Let's apply the change-of-variables formula

$dx\,dy\,dz=\left|\frac{\partial(x,y,z)}{\partial(\rho,\theta,\varphi)}\right|\;d\rho\,d\theta\,d\varphi.$

$quotient of the partial derivatives.$

So, the task is to compute the Jacobian:

$J(\rho,\theta,\varphi)=\frac{\partial(x,y,z)}{\partial(\rho,\theta,\varphi)}.$

Compute the partial derivatives

Differentiate $x, y, z with respect to$ $\rho,\theta,\varphi$ .

With respect to $\rho$ :

$\frac{\partial(x,y,z)}{\partial\rho} =\Big(a\cos\varphi\sin\theta,\; b\sin\varphi\sin\theta,\; c\cos\theta\Big).$

With respect to $\theta$ :

$\frac{\partial(x,y,z)}{\partial\theta} =\Big(a\rho\cos\varphi\cos\theta,\; b\rho\sin\varphi\cos\theta,\; -c\rho\sin\theta\Big).$

With respect to $\varphi$ :

$\frac{\partial(x,y,z)}{\partial\varphi} =\Big(-a\rho\sin\varphi\sin\theta,\; b\rho\cos\varphi\sin\theta,\; 0\Big).$

Form the Jacobian determinant

Put those three vectors as columns (or rows—just be consistent). Using columns:

$\left|\frac{\partial(x,y,z)}{\partial(\rho,\theta,\varphi)}\right| = \left| \begin{matrix} a\cos\varphi\sin\theta & a\rho\cos\varphi\cos\theta & -a\rho\sin\varphi\sin\theta\\ b\sin\varphi\sin\theta & b\rho\sin\varphi\cos\theta & \ \ b\rho\cos\varphi\sin\theta\\ c\cos\theta & -c\rho\sin\theta & 0 \end{matrix} \right|.$

Factor constants (the key simplification)

Factor $a$ from row 1, $b$ from row 2, $c$ from row 3
Factor $\rho$ from column 2 and $\rho$ from column 3

$\left|\frac{\partial(x,y,z)}{\partial(\rho,\theta,\varphi)}\right| = abc\,\rho^2 \left| \begin{matrix} \cos\varphi\sin\theta & \cos\varphi\cos\theta & -\sin\varphi\sin\theta\\ \sin\varphi\sin\theta & \sin\varphi\cos\theta & \ \cos\varphi\sin\theta\\ \cos\theta & -\sin\theta & 0 \end{matrix} \right|.$

$Evaluating the determinant we obtain:$

$\left|\frac{\partial(x,y,z)}{\partial(\rho,\theta,\varphi)}\right| =abc\,\rho^2\sin\theta.$

Final conversion of dV

$\boxed{dx\,dy\,dz=abc\,\rho^2\sin\theta\;d\rho\,d\theta\,d\varphi.}$

The volume of the ellipsoid is calculated as follows:

Volume of a sphere using spherical coordinates

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Let's use an example to calculate the volume of a sphere in spherical coordinates.

Example

Solution

Converting a triple integral in rectangular coordinates to spherical coordinates

2026-01-24T18:14:00.003-05:00

Converting a triple integral from rectangular coordinates to spherical coordinates. Let's do that through an example.

Example

Solution

Let's start by finding the ranges for θ, ⍴, 𝛗.

1) Look only at the outer two integrals (the x,y bounds):

0 ≤ y ≤ 3, 0 ≤ x ≤ √(9 − y²)

Rewrite the x-bound as an inequality:

0 ≤ x ≤ √(9 − y²)
⇔ x² ≤ 9 − y²
⇔ x² + y² ≤ 9

Combine this with x ≥ 0 and y ≥ 0.
This describes the first-quadrant portion of the disk x² + y² ≤ 9 in the xy-plane.

In polar (or spherical) coordinates, θ is the angle in the xy-plane measured from the positive x-axis.
The first quadrant therefore gives:

0 ≤ θ ≤ π⁄2

2) Let's find the ranges for ⍴:

The top z-surface is given by:

z = √(18 − x² − y²)

Square the equation (this is valid here since z ≥ 0):

z² = 18 − x² − y²
⇔ x² + y² + z² = 18

In spherical coordinates, the relation between rectangular and spherical variables is:

x² + y² + z² = ρ²

Therefore, this surface becomes:

ρ² = 18
ρ = 3√2

The range for ⍴ is then: 0 ≤ ⍴ ≤ 3⎷2

3) The bottom z-surface is given by:

z = √(x² + y²)

Let r = √(x² + y²).
Then the surface can be written as:

z = r

This represents a cone opening upward with vertex at the origin.

In spherical coordinates, the relationships are:

z = ρ cos φ
r = ρ sin φ

Substitute these into z = r:

ρ cos φ = ρ sin φ

For ρ > 0, divide both sides by ρ:

cos φ = sin φ

This implies:

tan φ = 1
φ = π/4

The original bounds satisfy z ≥ √(x² + y²), which means the region lies above the cone.
In spherical coordinates, this corresponds to angles smaller than π/4.

Therefore, the φ-range is:

0 ≤ φ ≤ π/4

From the coordinate transformation, we have:

x² + y² + z² = ρ²
dV = ρ² sin φ dρ dφ dθ

The integrand becomes:

x² + y² + z² = ρ²

Finally, the triple integral becomes: