Double integrals

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

Functions, Partials, and Gradients

• A function of two variables $f: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ assigns to each ordered pair in its domain a unique real number. Examples include geometric formulas like Area = $\frac{1}{2}bh$, or algebraic surfaces like $z = f(x, y) = 2x + 3y$, $f(x, y) = x^2 + y^2$, and $f(x,y) = e^{x+y}$.

• Partial derivatives differentiate a multivariable function $f(x_1, x_2, \dots, x_n)$ with respect to one target variable while treating all other variables as constants. They measure how the function changes in a single specific direction.
  If $f(x, y) = 2x^2y^3$, then $\frac{\partial f}{\partial x} = f_x = 4xy^3$ and $\frac{\partial f}{\partial y} = f_y = 6x^2y^2$.
  If $f(x, y) = x^3y + y^2$, then $f_x = 3x^2y$ and $f_y = x^3+2y$.

• A point $(x_1, x_2, \dots, x_n)$ is considered a critical point if the gradient of $f$ at that point is the zero vector, or if the gradient does not exist.

Gradients, Level Surfaces, and Directional Derivatives

• The gradient $\nabla f$ is a vector containing all first-order partial derivatives of $f$. It geometrically points in the direction of the function’s steepest ascent. $\nabla f = \left\langle \frac{\partial f}{\partial x_1}, \frac{\partial f}{\partial x_2}, \dots, \frac{\partial f}{\partial x_n} \right\rangle$.

• In 2D space ($n = 2$), finding a critical point $(x_0, y_0)$ requires solving the system: $\frac{\partial f}{\partial x}(x_0, y_0) = 0$ and $\frac{\partial f}{\partial y}(x_0, y_0) = 0$.

• Theorem: The gradient vector is perpendicular (orthogonal) to the level curves or level surfaces of the function at any given point. This means if you walk along a contour line where the function’s value is constant, the direction of steepest uphill is exactly 90 degrees to your path.

The directional derivative $\frac{dw}{ds}\bigg\vert{}_{\vec{u}}$ represents the rate of change of a function $w$ as you move in the direction of a unit vector $\vec{u}$. It can be expressed or compactly written via the dot product of the gradient and the direction vector: $\frac{dw}{ds}\bigg\vert{}_{\vec{u}} = \nabla w \cdot \vec{u} = \vert{}\nabla w\vert{}\vert{}\vec{u}\vert{}\cos(\theta) = \vert{}\nabla w\vert{}\cos(\theta)$ where $\theta$ is the angle between the gradient $\nabla w$ and the unit vector $\vec{u}$.

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]. “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 the total or net 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 (typically the xy-plane). Instead of just finding or calculating a 2D area, with a double integral, we’re looking for the volume under the surface (or below the graph) $z = f(x, y)$ and above the region $R$ (Figure 1).

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

• $R$ is the region (the footprint) 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 given point $(x, y)$ within $R$.
• $dA$ represents a tiny or, being more precise, an infinitesimal area element within the region $R$. In Cartesian coordinates, $dA = dxdy$ or $dydx$ (Figure 2).

Formal definition (Riemann Sums)

To understand this better, imagine breaking or dividing the region R into a grid of small squares or rectangles, each with an area $\Delta A_i = \Delta x_i \Delta y_i$.

At each little rectangle, we pick a sample point $(x_i, y_i)$ and evaluate the function $f(x_i, y_i)$ to find the height of the surface at that specific spot. If we multiply that height by the area of the base rectangle, we get the volume of a tiny rectangular pillar. Summing up all these tiny pillars gives us a strong approximation of the total volume under the surface.

Mathematically, we approximate the volume using a double Riemann sum: $V \approx \sum_{i} f(x_i, y_i)\Delta A_i$

When we take the limit as the grid becomes infinitely dense (i.e., as $\Delta A_i \to 0$), this sum becomes the exact value of the double integral: $\boxed{\iint_R f(x, y)dA = \lim_{\Delta A \to 0}\sum_{i} f(x_i, y_i)\Delta A_i}$ (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. For each fixed value of x, imagine cutting the solid with a plane parallel to the yz‑plane. This produces a vertical slice whose cross‑sectional area is $S(x) =\int_{y_{min(x)}}^{y_{max(x)}} f(x, y)dy$.
   This integral measures the area under the curve $z = f(x, y)$ along that slice.
2. Integrate these slices with respect to x. Once you have calculated the area for each slice, you “stack” all slices by integrating with respect to x from $x_{min}$ to $x_{max}$.

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. Geometrically, 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 is as follows. Let f(x, y) be a continuous function on the rectangle $R = [a, b] \times [c, d]$. Pick a specific value $x_0 \in [a, b]$. Slicing the surface with the vertical plane $x = x_0$ gives a curve in the yz-plane. The area under this curve is given by the integral $A(x_0) = \int_{c}^{d} f(x, y)dy$. This is the area of the slice at $x_0$.

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 slice‑areas 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).

This viewpoint emphasizes that a double integral is simply accumulating areas of vertical slices as you sweep across the region.

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).$\int \int_R [f(x, y) \pm g(x, y)]dA = \int \int_R f(x, y)dA \pm \int \int_R g(x, y)dA$. If you have two functions, f(x, y) and g(x, y), and you want to integrate their sum (difference) 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 (subtract) 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 $f(x, y) \ge g(x, y)$ for all $(x, y)$ in $R$, then $\int \int_R f(x, y)dA \ge \int \int_R g(x, y)dA$. 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.
4. Additivity of Regions. If region $R$ is split into two non-overlapping sub-regions $R_1$ and $R_2$, then: $\iint_R f(x, y)dA = \iint_{R_1} f(x, y)dA + \iint_{R_2} f(x, y)dA$

Fubini’s Theorem

Fubini’s Theorem is a powerful and fundamental result in calculus, particularly in the context of double integrals. It provides a practical method to evaluate double integrals by breaking them down into two standard, single-variable integrals (known as an iterated integral).

Suppose $f(x, y)$ is a continuous function defined on a rectangular region $R = [a, b] \times [c, d]$ 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 and yet get the exact same result: $\int \int_R f(x, y)dA = \int_{a}^{b}\int_{c}^{d} f(x, y)dydx = \int_{c}^{d}\int_{a}^{b} f(x, y)dxdy$

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, effectively taking 2D “slices” of the volume, finding the area of each slice, and then integrate those areas across the remaining dimension (x) to find the total volume.

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.

Integrating Over General Regions

Most regions $R$ are not perfect rectangles. When dealing with general regions, the limits of the inner integral will be functions rather than constants. We categorize these into two types:

Type I Regions (Vertically Simple): The region is bounded on the left and right by constants $x = a$ and $x = b$, and bounded above and below by continuous curves $y = g_2(x)$ and $y = g_1(x)$.
$\iint_R f(x, y) dA = \int_{a}^{b} \int_{g_1(x)}^{g_2(x)} f(x, y) dy dx$

Type II Regions (Horizontally Simple): The region is bounded below and above by constants $y = c$ and $y = d$, and bounded on the left and right by continuous curves $x = h_1(y)$ and $x = h_2(y)$.
$\iint_R f(x, y) dA = \int_{c}^{d} \int_{h_1(y)}^{h_2(y)} f(x, y) \,dx \,dy$

Solved examples

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

1. Evaluate the Inner Integral (with respect to y): $\int_{0}^{3} (5-x)dy = (5-x)\int_{0}^{3} 1dy = (5-x)y\bigg|_{0}^{3} = 3(5-x).$
2. Substitute into and calculate the Outer Integral (with respect to x): $ = 3\int_{0}^{5} (5-x)dx = 3(5x -\frac{x^2}{2})\bigg|_{0}^{5}$
3. Simplify, $3(5\cdot 5-\frac{25}{2}) = 3·(\frac{50}{2}-\frac{25}{2})=\frac{3\cdot 25}{2} = \frac{75}{2}$

The integral we just calculated corresponds to the volume of a right triangular prism. The base of the prism is the rectangle $0 \le x \le 5, 0 \le y \le 3$ with base length 5 (in the 𝑥-direction), and width 3 (in the 𝑦-direction), and the height of the prism (along the z-axis) is determined by the function 5 − x, namely 5 (the maximum value of $5-x$, Figure A), V = $\frac{1}{2}\cdot \text{baseLength}\cdot \text{widthBase} \cdot \text{heightPrism} = \frac{1}{2}\cdot 5 \cdot 3 \cdot 5 = \frac{75}{2}.$ This confirms our integral calculation is correct.

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

1. Compute the Inner Integral. Integrate with respect to y, then evaluate the inner integral.
   $\int_{1}^{3} xy^2dy = \frac{xy^3}{3}\bigg|_{1}^{3}=\frac{x}{3}[3^3-1^3]=\frac{x}{3}(27-1) = \frac{26}{3}x$
2. 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 $R = [0, 1] \times [0, 1]$ (a square on the xy-plane from 0 to 1 in both x and y directions), we want to demonstrate the inequality: $0 \le \iint_R \sin(x+y)dA \le 1$

1. The region R is $R = [0, 1] \times [0, 1] \implies 0 \le x + y \le 2 < \pi$. Since we know that $\sin(\theta)$ for any angle $\theta$ always lies between -1 and 1. But here, x + y varies between 0 and $\pi$, and within this interval $\sin(x+y)$ ranges between 0 and 1: $0 \le \sin(x + y) \le 1$.
2. By Property 3 of double integrals. If $f(x, y) \ge g(x, y)$ for all $(x, y)$ in $R$, then $\int \int_R f(x, y)dA \ge \int \int_R g(x, y)dA$.
   Here, we can consider $f(x,y) = \sin(x + y)$ and $g(x, y)=0$ because $0 \le \sin(x + y)$ everywhere in R. Also, since $\sin(x + y) \le 1$, we can compare $\sin(x+y)$ to 1. Therefore, $\int \int_R 0dA \le \int \int_R sin(x, y)dA \le \int \int_R 1dA$.
3. Furthermore, $\int \int_R 0dA = 0$, 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 $1 \times 1 = 1$.
4. Conclusion. Therefore, $0 ≤ \int \int_R sin(x, y)dA \le 1$

• Find the volume of the solid that lies below the surface $z = xy$ and above a triangle region $R$ 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 = $\int \int_R xydA$

1. 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). This is a Type II Region.
   

   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: $y -y_0 = m(x -x_0) \implies y -1 = \frac{-1}{3}(x-4) \implies 3y -3 = -x + 4 \implies x = 7 -3y.$

2. Calculation of the Double Integral. Therefore, the volume integral is set up as: $\iint_R f(x, y) dA = \int_{1}^{2}(\int_{1}^{7-3y} xydx)dy$
3. Compute the Inner Integral: $\int_{1}^{7-3y} xydx = \frac{x^2y}{2}\bigg|_{1}^{7-3y} = \frac{(7-3y)^2y}{2}-\frac{y}{2}=\frac{(49-42y+9y^2)y-y}{2} = \frac{1}{2}(49y-42y^2+9y^3-y)=\frac{48y-42y^2+9y^3}{2}$
4. Compute the Outer Integral: V = $\frac{1}{2}\int_{1}^{2} (48y-42y^2+9y^3)dy = \frac{1}{2}(24y^2-14y^3+\frac{9}{4}y^4)\bigg|_{1}^{2} = \frac{1}{2}[96-112+36-(24-14+\frac{9}{4})] = \frac{31}{8}$.
5. Conclusion. Thus, the volume V of the solid is $\frac{31}{8}$ cubic units.

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

1. Compute the inner integral: $\int_{1}^{2} (x^4-y^2)dx = \frac{x^5}{5}-xy^2\bigg|_{1}^{2} = \frac{32}{5}-2y^2-\frac{1}{5}+y^2 = \frac{31}{5}-y^2$
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 downward-opening paraboloid $z = 1 - x^2 - y^2$ over the region R in the xy-plane, where R is the square defined by $R = [0, 1] \times [0, 1]$ (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 the function f(x, y) dips below the xy-plane within the region. Because the surface crosses the xy-plane, our final result will represent the net signed volume (the volume above the xy-plane minus the volume below it).

1. Calculation of the Double Integral: $\iint_R f(x, y) dA = \int_{0}^{1}(\int_{0}^{1} 1-x^2-y^2dy)dx$.
2. Compute the Inner Integral. Integrate with respect to y, then evaluate the inner integral.
   $\int_{0}^{1} (1-x^2-y^2)dy = y -x^2y -\frac{y^3}{3}\bigg|_{0}^{1} = 1 -x^2 -\frac{1}{3} = \frac{2}{3} -x^2$
3. Compute the Outer Integral. Integrate with respect to x, then evaluate the outer integral.
   $\iint_R f(x, y) dA = \int_{0}^{1} (\frac{2}{3} -x^2)dx = \frac{2}{3}x - \frac{1}{3}x^3\bigg|_{0}^{1} = \frac{2}{3} - \frac{1}{3} - 0 + 0 = \frac{1}{3}$
4. Conclusion: $\iint_R f(x, y) dA = \frac{1}{3}$. The volume of the solid lying below the paraboloid $z = 1 - x^2 - y^2$ 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 integral. $\int_{1}^{2} (x^4-\frac{1}{3})dx = \frac{x^5}{5}-\frac{1}{3}x\bigg|_{1}^{2} = (\frac{32}{5}-\frac{2}{3})-(\frac{1}{5}-\frac{1}{3}) = \frac{88}{15}.$

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

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 $\pi y$ to $\pi$. To change the order of integration, consider the region of integration: x ranges from 0 to $\pi$. For each fixed x, y ranges from 0 (it starts at the x-axis $y = 0$) to $\frac{x}{\pi}$ (Figure B).

Thus, the double integral can be rewritten by swapping the order of integration: $\int_{0}^{1} \int_{\pi y}^{\pi}\frac{sin(x)}{x}dxdy = \int_{0}^{\pi} \int_{0}^{\frac{x}{\pi}}\frac{sin(x)}{x}dydx$

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

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

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