Double integrals

Give me six hours to chop down a tree and I will spend the first four sharpening the axe, Abraham Lincoln.

Recall

• A function of two variables f: ℝ x ℝ → ℝ assigns to each ordered pair in its domain a unique real number, e.g., Area = $\frac{1}{2}b·h$, z = f(x, y) = 2x + 3y, f(x, y) = x² + y², e^x+y, etc.

• Partial derivatives are derivatives of a function of multiple variables, say f(x₁, x₂, ···, x_n) with respect to one of those variables, with the other variables held constant. They measure how the function changes as one variable changes, while keeping the others constant. $\frac{\partial f}{\partial x}(x_0, y_0) = \lim_{\Delta x \to 0} \frac{f(x_0+\Delta x, y_0)-f(x_0, y_0)}{\Delta x}$

Geometrically, the partial derivative $\frac{\partial f}{\partial x}(x_0, y_0)$ at a point (x₀, y₀) can be interpreted as the slope of the tangent line to the curve formed by the intersection of the surface defined by f(x,y) and the plane y = y₀. Similarly, $\frac{\partial f}{\partial y}(x_0, y_0)$ represents the slope of the tangent line to the curve formed by the intersection of the surface defined by f(x,y) and the plane x = x₀

• $\frac{dw}{ds}\bigg|_{\vec{u}}$ is the directional derivative in the direction of ${\vec{u}}$, the rate of change of the function w as we move in the direction of $\vec{u}$.

$\frac{dw}{ds}\bigg|_{\vec{u}} = ∇w·\vec{u} = |∇w|·|\vec{u}|cos(θ) = |∇w|·cos(θ)$ where θ is the angle between the gradient and the given unit vector.

• The directional derivative is maximized when cos(θ) = 1 ↭ θ = 0, that means, $\vec{u}$ is in the same direction as the gradient ∇w, $\vec{u}=dir(∇w).$

The gradient is the direction in which the function increases fastest (the direction of the steepest ascent) at a given point, and |∇w| = $\frac{dw}{ds}\bigg|_{\vec{u}=dir(∇w)}$.

The directional derivative is minimized when cos(θ) = -1 ↭ θ = 180 ↭ $\vec{u}$ is in the opposite direction of the gradient ∇w. Futhermore, $\frac{dw}{ds}\bigg|_{\vec{u}} = 0$ ↭ cos(θ) = 0 ↭ θ = 90° ↭ $\vec{u}$ ⊥ ∇w.

• The Chain Rule. Let w = f(x, y, z) where x = x(t), y = y(t), and z = z(t) are functions of another variable t, then the derivate of f with respect to t is given by $\frac{dw}{dt} = f_x\frac{dx}{dt} +f_y\frac{dy}{dt}+f_z\frac{dz}{dt}, \frac{dw}{dt} = w_x\frac{dx}{dt} +w_y\frac{dy}{dt}+w_z\frac{dz}{dt} = ∇w·\frac{d\vec{r}}{dt}~$ where ∇w = ⟨w_x, w_y, w_z⟩ and $\frac{d\vec{r}}{dt}=⟨\frac{d\vec{x}}{dt}, \frac{d\vec{y}}{dt}, \frac{d\vec{z}}{dt}⟩$

• The gradient vector is defined as follows: ∇w = ⟨w_x, w_y, w_z⟩ = $⟨\frac{\partial w}{\partial x}, \frac{\partial w}{\partial y}, \frac{\partial w}{\partial z}⟩$. And $\frac{d\vec{r}}{dt}$ is the vector of derivatives of x, y, and z with respect to t: $\frac{d\vec{r}}{dt} = ⟨\frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt}⟩$.

The gradient vector is a vector that points in the direction of the steepest increase of the function at a given point. It is perpendicular (orthogonal) to the level surfaces of the function. This means that the gradient points in the direction of the steepest ascent or increase, and there is no change in the function’s value along the level surface, making the gradient perpendicular to it.

• Lagrange multipliers are a strategy for finding the local maxima and minima of a function, say f(x, y, z), where the variables x, y, and z are not independent, but subject to an equality constraint, g(x, y, z) = c. This method uses the fact that any point where f(x, y, z) reaches a constraint extremum (maximum or minimum), the gradient of f is parallel to the gradient of g. Formally, this relationship is expressed as ∇f = λ∇g where λ is the Lagrange multiplier.

Double integrals

Single Variable Integration

To begin with, let’s talk about a concept you might already be familiar with: the definite integral of a function of one variable.

The definite integral of a function of one variable, denoted as $\int_{a}^{b} f(x)dx$, provides the signed area between the graph of the function and the x-axis over the interval [a, b]. This signed area means that areas above the x-axis contribute positively to the total, while areas below the x-axis contribute negatively. The result gives you a single number representing that total area.

Double Integrals

Now, let’s extend this idea to two dimensions, where things get a bit more interesting. A double integral is an extension of the previous concept to functions of two variables.

A double integral allows us to integrate a function of two variables, say f(x, y), over a region R in a two-dimensional space (e.g. in the xy-plane). Instead of just finding or calculating an area, with a double integral, we’re looking for the volume under the surface (or below the graph) z = f(x, y) (Figure 1).

It is denoted or written as $\iint_R f(x, y) dA$ where:

• R is the region or the piece of area in the xy-plane over which the integration is performed.
• f(x,y) is the integrand, the function we want to integrate. Think of it as describing the height of a surface at any point (x, y) in R.
• dA represents a tiny or, being more precise, infinitesimal area element within the region R (Figure 2).

Formal definition

To understand this better, imagine breaking or dividing the region R into small squares or rectangles, each with an area ΔA_i.

At each little square, we can evaluate the function f(x, y) to find out how high the surface is above that point. If we multiply that height by the area of the square, we get a tiny bit of volume above that square. Summing up all these tiny volumes gives us a relatively good approximation of the total volume under the surface.

Mathematically, we approximate the double integral by the sum $\sum_{i} f(x_i, y_i)ΔA_i$ where $ΔA_i = Δx_iΔy_i$.

When we make the squares infinitely small (i.e., taking the limits as ΔA_i→ 0), this sum becomes the exact value of the double integral: $\int \int_R f(x, y)dA = \lim_{\Delta x \to 0}\sum_{i} f(x_i, y_i)ΔA_i$ (Figure 3).

Calculating a Double Integral

To actually calculate a double integral, we usually break it down into two steps:

1. Set up the limits of integration. Fix x and integrate with respect to y by slicing the surface parallel to the yz-plane at each x. The area of each slice is S(x) = $\int_{y_{min(x)}}^{y_{max(x)}} f(x, y)dy$
2. Integrate these slices with respect to x. Once you have calculated the area for each slice, you can add them up by integrating with respect to x.

Thus, the volume can be found by integrating these slices with respect to x: $\iint_R f(x, y) dA\ = \int_{x_{min}}^{x_{max}} S(x)dx = \int_{x_{min}}^{x_{max}} [\int_{y_{min(x)}}^{y_{max(x)}} f(x, y)dy]dx$ This expression is the double integral written as an iterated integral. It represents the volume under the surface defined by z = f(x, y) and above the region R in the xy-plane (Figure 3 and 4).

Another Perspective

Another way of seeing this. Let f(x, y) be a continuous function on R = [a, b] x [c, d]. ∀x₀ ∈ [a, b], we will slice the surface z = f(x, y) with the plane x = x₀ where x₀ is fixed.

The area A under the curve (between the curve and the xy-plane) is given by the integral A(x₀) = $\int_{c}^{d} f(x, y)dy$. To find the total volume V of the solid under the surface z = f(x, y), but above the region R, you then integrate this area as x moves across the interval [a, b]: V = $\int_{a}^{b} A(x)dx = \int_{a}^{b} [\int_{c}^{d} f(x, y)dy]dx$ (Figure 0).

Properties of Double integrals

Double integrals extend the concept of integration from single-variable functions to functions of two variables over a region R in the xy-plane. Understanding the properties of double integrals can help make calculations more intuitive and manageable.

1. Linearity (Sum of Functions).$\int \int_R [f(x, y) + g(x, y)]dA = \int \int_R f(x, y)dA + \int \int_R g(x, y)dA$. If you have two functions, f(x, y) and g(x, y), and you want to integrate their sum over the same region R, you can do this in two steps: A. Integrate f(x, y) and g(x, y) over R. B. Then, just add the two results together.
2. Scalar Multiplication. $\int \int_R cf(x, y)dA = c\int \int_R f(x, y)dA$. If you multiply your function f(x, y) by a constant c, the double integral of this new function over the region R is just the original double integral multiplied by c.
3. Comparison (Order of Functions). If f(x, y) ≥ g(x, y) ∀x, y ∈ R (for all x, y in R), then $\int \int_R f(x, y)dA ≥ \int \int_R g(x, y)dA$. If f(x, y) is always greater than or equal to g(x,y) across the entire region R, then the double integral of f(x,y) over R will also be greater than or equal to the double integral of g(x, y) over the same region.

Fubini’s Theorem

Fubini’s Theorem is a powerful and fundamental result in calculus, particularly in the context of double integrals. It provides a method to evaluate double integrals by breaking them down into two simpler, single integrals.

Suppose f(x, y) is a continuous function defined on a rectangular region R = [a, b] x [c, d] in the xy-plane. Fubini’s Theorem states that the double integral of f(x, y) over the region R can be computed as an iterated integral in either of two orders:

$\int \int_R f(x, y)dA = \int_{a}^{b}\int_{c}^{d} f(x, y)dydx = \int_{c}^{d}\int_{b}^{a} f(x, y)dxdy$

The expression $\int \int_R f(x, y)dA$ is the double integral of f(x, y) over the region R and gives us the “volume” under the surface defined by z = f(x, y) over the area R in the xy-plane.

This means we can first integrate f(x, y) with respect to one variable (say, y) while treating the other variable (x) as a constant, and then integrate the resulting expression with respect to x.

Alternatively, we can reverse the order: first integrate with respect to x while treating the other variable (y) as a constant, and then integrate the result with respect to y.

Solved examples

• $\int_{0}^{5}(\int_{0}^{3} (5-x)dy)dx$

Evaluate the Inner Integral: $\int_{0}^{3} (5-x)dy = (5-x)\int_{0}^{3} 1dy = (5-x)y\bigg|_{0}^{3} = 3(5-x).$

Substitute into and calculate the Outer Integral. $ = 3\int_{0}^{5} (5-x)dx = 3(5x -\frac{x^2}{2})\bigg|_{0}^{5}$

$3(5·5-\frac{25}{2}) = 3·(\frac{50}{2}-\frac{25}{2})=\frac{3·25}{2} = \frac{75}{2}$

The integral we just calculated corresponds to the volume of a right triangular prism. The base of the prism is a triangle with a base length of 5 and height 3, and the height of the prism (along the z-axis) is determined by the function 5−x (Figure A), V = $\frac{1}{2}·baseLength·weightBase·HeightPrisma = \frac{1}{2}·5·3·5 = \frac{75}{2}.$ This confirms our integral calculation is correct.

• $\int_{0}^{2}(\int_{1}^{3} xy^2dy)dx$

Compute the Inner Integral. Integrate with respect to y, then evaluate the inner integral.

$\int_{1}^{3} xy^2dy = \frac{xy^3}{3}\bigg|_{1}^{3}=\frac{x}{3}[3^3-1^3]=\frac{x}{3}(27-1) = \frac{26}{3}x$

Compute the Outer Integral. Integrate with respect to x, then evaluate the outer integral.

$\int_{0}^{2}(\int_{1}^{3} xy^2dy)dx = \int_{0}^{2} \frac{26}{3}xdx = \frac{26}{3}\frac{x^2}{2} = \frac{13}{3}x^2 \bigg|_{0}^{2} = \frac{13}{3}(4-0)=\frac{52}{3}$

• Given the region R = [0, 1] x [0, 1] (a square on the xy-plane from 0 to 1 in both x and y directions), we want to demonstrate the inequality: $0 ≤ \int \int_R sin(x, y)dA ≤ 1.$

The region R is [0, 1] x [0, 1] ⇒ 0 ≤ x + y ≤ 2 < π ⇒[We know that sin(θ) for any angle θ always lies between -1 and 1. But here, x + y varies between 0 and π, and within this interval sin(x+y) ranges between 0 and 1] 0 ≤ sin(x + y) ≤ 1.

By Property 3 of double integrals. If f(x, y)≥ g(x, y) ∀x, y ∈ R (for all x, y in R), then $\int \int_R f(x, y)dA ≥ \int \int_R g(x, y)dA$. Here, we can consider f(x,y) = sin(x + y) and g(x, y)=0 because 0 ≤ sin(x + y) everywhere in R. Also, since sin(x + y) ≤ 1, we can compare sin(x+y) to 1. Therefore, $\int \int_R 0dA ≤ \int \int_R sin(x, y)dA ≤ \int \int_R sin(x, y)dA$.

Futhermore, $\int \int_R 0dA = 0$, since the integral of zero over any region is zero. $\int \int_R 1dA$ is just the area of the region R. Since R is a square of side length 1, its area is 1 × 1 = 1.

Therefore, $0 ≤ \int \int_R sin(x, y)dA ≤ 1$∎

• Find the volume of the solid that lies below the surface z = xy and above a triangle region with vertices (1, 1), (4, 1), and (1, 2) (Figure iv).

The problem essentially requires you to compute a double integral over the triangular region R, where the integrand is f(x,y) = xy, which represents the height of the solid above the xy-plane at any point (x, y) within the region R, V = $\int \int_R xydA$

Determining the Limits of Integration: The region R is a triangle in the xy-plane with vertices at (1, 1), (4, 1), and (1, 2). To integrate over this region, we need to express the limits of integration for x and y. y ranges from 1 (the bottom edge of the triangle) to 2 (the top edge of the triangle). For each fixed value of y, x ranges from 1 (on the left) to the line x = 7−3y (the right side of the triangle).

The equation of the line joining the points (4,1) and (1,2) can be found by calculating the slope and intercept. The slope m = $\frac{2-1}{1-4} = \frac{-1}{3}$. Point-slope: y -y₀ = m(x -x₀) ⇒ $y-1 = \frac{-1}{3}(x-4) ↭ 3y -3 = -x + 4 ↭ x = 7 -3y.$

Calculation of the Double Integral. Therefore, the volume integral is set up as: $\iint_R f(x, y) dA\ = \int_{1}^{2}(\int_{1}^{7-3y} xydx)dy =$

Compute the Inner Integral: $\int_{1}^{7-3y} xydx = \frac{x^2y}{2}\bigg|_{1}^{7-3y} = \frac{(7-3y)^2y}{2}-\frac{y}{2}=\frac{(49-42y+9y^2)y-y}{2} = \frac{1}{2}(49y-42y^2+9y^3-y)=\frac{48y-42y^2+9y^3}{2}$

Compute the Outer Integral: V = $\frac{1}{2}\int_{1}^{2} (48y-42y^2+9y^3)dy = \frac{1}{2}(24y^2-14y^3+\frac{9}{4}y^4)\bigg|_{1}^{2} = \frac{1}{2}[96-112+36-(24-14+\frac{9}{4})] = \frac{31}{8}$. Thus, the volume V of the solid is $\frac{31}{8}$ cubic units.

• Compute $\int_{0}^{1} \int_{1}^{2} (x^4-y^2)dxdy$.

Compute the inner integral: $\int_{1}^{2} (x^4-y^2)dx = \frac{x^5}{5}-xy^2\bigg|_{1}^{2} = \frac{32}{5}-2y^2-\frac{1}{5}+y^2 = \frac{31}{5}-y^2$

Compute the outer integral: $\int_{0}^{1} (\frac{31}{5}-y^2)dy = \frac{31}{5}y-\frac{y^3}{3}\bigg|_{0}^{1} = \frac{31}{5} - \frac{1}{3} = \frac{93}{15}-\frac{5}{15} = \frac{88}{15}$

• Compute the double integral of the function z = f(x, y) = 1 -x² -y² (it is a paraboloid opening downwards) over the region R in the xy-plane, where R is the square defined by 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 (Figure A, B, and C).

Please notice that f(0.9, 0.9) = -0.62, so the region is a square in the xy-plane, and f(x, y) can be negative within the region, but it does not affect the double integral calculation directly.

Calculation of the Double Integral: $\iint_R f(x, y) dA\ = \int_{0}^{1}(\int_{0}^{1} 1-x^2-y^2dy)dx =$[🚀]

Compute the Inner Integral. Integrate with respect to y, then evaluate the inner integral)

$\int_{0}^{1} (1-x^2-y^2)dy = y -x^2y -\frac{y^3}{3}\bigg|_{0}^{1} = 1 -x^2 -\frac{1}{3} = \frac{2}{3} -x^2$

Compute the Outer Integral. Integrate with respect to x, then evaluate the outer integral.

=[🚀] $\int_{0}^{1} (\frac{2}{3} -x^2)dx = \frac{2}{3}x - \frac{1}{3}x^3\bigg|_{0}^{1} = \frac{2}{3} - \frac{1}{3} - 0 + 0 = \frac{1}{3}$

Conclusion: $\iint_R f(x, y) dA\ = \frac{1}{3}$. The volume of the solid lying below the paraboloid z = 1 -x² -y² and above the square region R in the xy-plane, is $\frac{1}{3}$ cubic units.

• $\int_{1}^{2} \int_{0}^{1}(x^4-y^2)dydx$

Compute the inner integral. $\int_{0}^{1}(x^4-y^2)dy = x^4y-\frac{y^3}{3}\bigg|_{0}^{1} = x^4-\frac{1}{3}$.

Compute the outer space. $\int_{1}^{2} (x^4-\frac{1}{3})dx = \frac{x^5}{5}-\frac{1}{3}x\bigg|_{1}^{2} = (\frac{32}{5}-\frac{2}{3})-(\frac{1}{5}-\frac{1}{3}) = \frac{88}{15}.$

• Compute the double integral: $\int_{0}^{1} \int_{πy}^{π}\frac{sin(x)}{x}dxdy$

Changed the order of integration to simplify the computation. Sometimes, changing the order of integration makes the computation easier. The original double integral is set up with y ranging from 0 to 1 and x ranging from πy to π. To change the order of integration, consider the region of integration: x ranges from 0 to π. For each fixed x, y ranges from 0 to x/π (Figure B).

Thus, the double integral can be rewritten by swapping the order of integration:

$\int_{0}^{1} \int_{πy}^{π}\frac{sin(x)}{x}dxdy = \int_{0}^{π} \int_{0}^{\frac{x}{π}}\frac{sin(x)}{x}dydx$

Compute the inner integral: $\int_{0}^{\frac{x}{π}}\frac{sin(x)}{x}dy = \frac{sin(x)}{x}y\bigg|_{0}^{\frac{x}{π}} =\frac{sin(x)}{x}\frac{x}{π} = \frac{sin(x)}{π}$.

Compute the outer integral: $\int_{0}^{π} \frac{sin(x)}{π} dx = \frac{1}{π}\int_{0}^{π} sin(x)dx = \frac{1}{π}·(-cos(x))\bigg|_{0}^{π} = \frac{1}{π}·(-cos(π)-(cos(0))) = \frac{2}{π}$

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