Applications of Double Integrals

The best place to find a helping hand is at the end of your own arm, Swedish Proverb.

Core Concepts

A double integral allows us to integrate a function of two variables, $f(x, y)$, over a two-dimensional region $R$ in the xy-plane. While a single integral calculates the area under a curve, a double integral calculates the volume under a surface $z = f(x, y)$. It is denoted or written as $\iint_R f(x, y) dA$.

Fubini’s Theorem. If $f(x, y)$ is continuous on a rectangular region $R = [a, b] \times [c, d]$, the double integral can be computed as an iterated integral in either order without changing the result: $\int \int_R f(x, y)dA = \int_{a}^{b}\int_{c}^{d} f(x, y)dydx = \int_{c}^{d}\int_{a}^{b} f(x, y)dxdy$

Applications of Double Integrals

Double integrals are crucial in various applications, such as:

• Calculating Areas. The area of a 2D region $R$ can be found by integrating the constant function $f(x,y) = 1$ over $R$: $\text{Area} = \iint_R 1 dA$.
• Calculating Volumes. The volume of a region $R$ under a surface $z = f(x, y)$ is given by the double integral $\iint_R f(x, y) dA$.
• Finding the Average Value of a Function. The average value of a function $z = f(x, y)$ over a region $R$ is the double integral divided by the area of the region: $F_{\text{average}} = \frac{1}{\text{Area of } R} \iint_R f(x,y)dA$.

Double integrals are powerful mathematical tools that allow us to calculate a variety of important quantities, such as areas of regions, volumes under surfaces, average values of functions over regions, and physical quantities like mass, center of mass, and moment of inertia.

By understanding the techniques for setting up and evaluating double integrals, including the ability to convert to alternative coordinate systems such as polar or spherical coordinates when appropriate, we unlock the ability to solve a wide range of complex problems in multivariable calculus.

Solved exercises

Calculating Area

The integral of the constant function $f(x,y) = 1$ over the region $R$ gives the area of $R$. Mathematically, this is expressed as: $\text{Area} = \iint_R 1 dA$

1. Finding the Area of a Region. We need to find the area of the region bounded below by the curve $y = x^2$ and above by the line $y = 2x$ in the first quadrant (Refer to Figure iii for a visual representation and aid in understanding it).
2. Understanding the Region: $y = x^2$ represents a parabola that opens upwards, $y = 2x$ is a straight line passing through the origin with a slope of 2. The region of interest is where these two curves intersect in the first quadrant.
3. Finding the intersection points: Set the equations equal to each other: $x^2 = 2x \implies x(x-2) = 0$. The curves intersect at $x = 0$ and $x = 2$.
   The solutions are $x = 0$ (which corresponds to the origin, where y = 0) and $x = 2$ (where the curves intersect again). These are the points of intersection in the first quadrant.
4. Setting up the double integral. We just have to integrate the constant function f(x,y) = 1 over the region R to find the area. The limits for x are from 0 to 2, corresponding to the points of intersection we found. For each fixed x, y ranges from the lower curve $x^2$ to the upper line $2x$. Thus, the area A of the bounded region is given by $\int_{0}^{2} (\int_{x^2}^{2x}1dy)dx$
5. Evaluating the inner integral. We first evaluate the inner integral with respect to y: $\int_{x^2}^{2x}1dy = y\bigg|_{x^2}^{2x} = 2x -x^2$. This represents the length of a vertical slice of the region at a fixed value of x (denoted as x₀).
6. Substituting and evaluating the outer integral with respect to x to find the total area: Area = $\int_{0}^{2} (\int_{x^2}^{2x}1dy)dx = \int_{0}^{2} (2x -x^2)dx = x^2-\frac{x^3}{3}\bigg|_{0}^{2}= 4 -\frac{8}{3} = \frac{12}{3}-\frac{8}{3} = \frac{4}{3}$.
7. Conclusion: Thus, the area of the region bounded by $y = x^2$ and $y = 2x$ in the first quadrant is $\frac{4}{3}$. This approach demonstrates how double integrals can be used effectively to compute the area of a region in the plane.

• Calculating the area of a region R bounded by $y = -x^2$ and $x + y = -6$.

1. Finding the Area of a Region. We are tasked with finding the area of the region bounded below by the curve $y = -x^2$ (a downward-opening parabola) and above by the line $x + y = -6$. Refer to Figure 1 for a visual representation and aid in understanding it.
2. Understanding the Region: $y = -x^2$ represents a parabola that opens downwards, $x + y = -6$ is a straight line. The region of interest is where these two curves intersect.
3. Finding the intersection points. To find the points of intersection, we set the equations of the two curves equal to each other: $-x^2 = -x-6 \implies x^2 - x - 6 = 0 \implies[\text{This is a quadratic equation. We solve for x using factoring:}] (x+2)(x-3) = 0$. These correspond to the points of intersections (3, -9) and (-2, -4).
4. Setting up the double integral. We just have to integrate the constant function f(x,y) = 1 over the region R to find the area. The limits for x are from -2 to 3, corresponding to the points of intersection we found. For each fixed x, y ranges from the lower line $x + y = -6$ to the upper curve $y = -x^2$. Thus, the area A of the bounded region is given by $\int_{-2}^{3} (\int_{-x-6}^{-x^2}1dy)dx$
5. Evaluating the inner integral. We first evaluate the integral with respect to y: $\int_{-x-6}^{-x^2}1dy = y\bigg|_{-x-6}^{-x^2} = -x^2 -(-x-6)= -x^2 +x + 6.$ This represents the length of a vertical slice of the region at a fixed value of x (denoted as x₀).
6. Substituting and evaluating the outer integral with respect to x to find the total area: Area = $\int_{-2}^{3} (\int_{-x-6}^{-x^2}1dy)dx = \int_{-2}^{3} (-x^2 +x + 6)dx = \frac{-x^3}{3}+\frac{x^2}{2}+6x\bigg|_{-2}^{3}$.
   $(\frac{-27}{3}+\frac{9}{2}+18)-(\frac{8}{3}+2-12) = \frac{-54+27+108-16+60}{6} = \frac{125}{6}$.
7. Conclusion. Thus, the area of the region bounded by $y = -x^2$ and $x + y = -6$ is $\frac{125}{6}$.

• Calculating the area of a region R bounded by $y = x^2$, $y = 2 -x$ in the first quadrant, meaning x ≥ 0.

In general, the integral of the constant function f(x,y) = 1 over the region R gives the area of R. Mathematically, this is expressed as: $\text{Area} = \iint_R 1 dA$. Sometimes, it is a little more complicated than that.

1. Understanding the Region. We are tasked with finding the area of the region bounded by the curve $y = x^2$ (an upward-opening parabola) and $y = 2 -x$, a straight line with a negative slope. Refer to Figure 2 for a visual representation and aid in understanding it. The region of interest is where these two curves intersect in the first quadrant.
2. Finding the intersection points. To find the points of intersection, we set the equations of the two curves equal to each other: $x^2 = 2 - x \implies x^2 + x - 2 = 0 \implies[\text{This quadratic equation can be factored as:}] (x-1)(x+2) = 0$. Thus, the solutions are x = 1 and x = -2. Since we are only interested in the first quadrant where $x \ge 0$, we will only consider the point $x = 1$, and the intersection point is $(1, 1)$.
3. Setting up the double integral. Typically, we just have to integrate the constant function $f(x,y) = 1$ over the region R to find the area. We can calculate the area of the region by setting up two double integrals. The first one is bounded above by the upward parabola $y = x^2$ and below by the x-axis. The limits for x are 0 to 1. The second one is bounded above by the line $y = 2 -x$ and below by the x-axis. The limits for x are 1 to 2.
4. Calculating the integrals:
   $A_1 = \int_{0}^{1} x^2dx = \frac{x^3}{3}\bigg|_{0}^{1} = \frac{1}{3}$
   $A_2 = \int_{1}^{2} (2-x)dx = (2x-\frac{x^2}{2})\bigg|_{1}^{2} = 4-2-(2-\frac{1}{2}) = 4 -2 - 2 + \frac{1}{2} = \frac{1}{2}$
   Adding these areas together, we obtain the area of the region: $A = A_1 + A_2 = \frac{1}{3} + \frac{1}{2} = \frac{2}{6} + \frac{3}{6} = \frac{5}{6}$

• Calculating the Volume of a Quarter of a Unit Disk

Let’s consider an example where we compute the volume under the surface $z = 1 - x^2 - y^2$ over a quarter of a unit disk centered at the origin. The region of integration is defined by $x^2 + y^2 \le 1$, and $x, y \ge 0$, which corresponds to the first quadrant of the disk.

1. Setting Up the Integral in Cartesian Coordinates. In Cartesian coordinates the double integral would be: $\int \int_{x^2+y^2 \le 1, x, y \ge 0} (1-x^2-y^2)dA$
   However, due to the circular nature of the region R, converting the integral to polar coordinates simplifies the computation and seems to be a sensible thing to do.
2. Convert to polar coordinates. In polar coordinates, the relationship between Cartesian and polar coordinates is: $x = r\cos(\theta)$, $y = r\sin(\theta)$ and $x^2 + y^2 = r^2$.
   You can visualize the area element $\Delta A$ as a small “polar rectangle” or sector wedge (Figure ii).
   When you change the radius by $\Delta r$ and the angle by $\Delta \theta$, the resulting shape has: (i) A width of $\Delta r$; (ii) A curved length of $r \Delta \theta$ (using the arc length formula). For an infinitesimally small region, this sector wedge behaves just like a flat rectangle with sides $dr$ and $r d\theta$.
   $dA = (\text{width}) \times (\text{length}) = dr \cdot (r d\theta) = r \, dr \, d\theta$
   If you calculate the area exactly using geometry, you subtract the inner circular sector from the outer circular sector: $\Delta A = \frac{1}{2}(r + \Delta r)^2 \Delta \theta - \frac{1}{2}r^2 \Delta \theta$
   Expanding this expression gives: $\Delta A = \frac{1}{2}(r^2 + 2r\Delta r + (\Delta r)^2)\Delta \theta - \frac{1}{2}r^2 \Delta \theta = r\Delta r\Delta \theta + \frac{1}{2}(\Delta r)^2\Delta \theta$
   As $\Delta r$ and $\Delta \theta$ shrink toward zero ($dr$ and $d\theta$), the second-order term $\frac{1}{2}(dr)^2 d\theta$ becomes negligibly small and drops out, leaving exactly $dA = r \, dr \, d\theta$.
3. Setting up the double integral. The surface becomes $z = 1 - r^2$. The bounds for the quarter disk are $r$ from $0$ to $1$ and $\theta$ from $0$ to $\frac{\pi}{2}$.
   The integral now becomes: $V = \int_{0}^{\frac{\pi}{2}} \int_{0}^{1} (1-r^2)r dr d\theta$
4. Compute the Inner Integral (with respect to r): $\int_{0}^{1} (1-r^2)rdr = \frac{r^2}{2}-\frac{r^4}{4}\bigg|_{0}^{1} =\frac{1}{2}-\frac{1}{4} = \frac{1}{4}$
5. Compute the Outer Integral (with respect to $\theta$): $\int_{0}^{\frac{\pi}{2}} \frac{1}{4}d\theta = \frac{1}{4}\theta \bigg|_{0}^{\frac{\pi}{2}} = \frac{1}{4} \cdot \frac{\pi}{2} = \frac{\pi}{8}$

Basically, to calculate the volume we divide the region into rectangles boxes $R_{ij}$, each with area $\Delta A$; the volume of a very thin rectangular box above $R_{ij}$ is given by $f(x_{i}, y_{j}) \Delta A$, where $(x_{i}, y_{j})$ is an arbitrary sample point in each $R_{ij}$. Summing up the volumes of all these thin boxes gives an approximation of the total volume: $V \approx \sum_{i, j}f(x_i, y_j) \Delta A$. In the limit as the size of the rectangles approaches zero, this sum becomes the double integral: $\iint_R f(x, y) dA$.

• Volume Under a Surface. Find the volume V under the plane $z = 8x + 6y$ over the region $R = \{(x, y): 0 \le x \le 1, 0 \le y \le 2x^2 \}$

The volume under the surface $z = f(x, y)$ over a region R in the xy-plane is given by the double integral: $V = \iint_R f(x, y) dA$

Method A: Vertical Slices (Integrate $y$ first)

1. Setup: $x$ ranges from $0$ to $1$, $y$ ranges from $0$ to $2x^2$ (Refer to Figure i for a visual representation).
   Therefore, the volume V can be computed using vertical slices as follows: $V = \int_{0}^{1}\int_{0}^{2x^2} (8x + 6ydy)dx$.
2. Compute the Inner Integral (with respect to y): $\int_{0}^{2x^2} 8x + 6ydy = 8xy+3y^2\bigg|_{0}^{2x^2} = 8x(2x^2)+3(2x^2)^2−(8x⋅0+3⋅0^2) = 16x^3+12x^4.$
3. Compute the Outer Integral (with respect to x): $\int_{0}^{1} (16x^3+12x^4)dx = 4x^4+\frac{12}{5}x^5\bigg|_{0}^{1}=4+\frac{12}{5}=\frac{32}{5}$. This gives us the volume V = $\frac{32}{5}$.

Method B: Horizontal Slices (Integrate $x$ first)

1. Setup: $y$ ranges from $0$ to $2$. For a fixed $y$, $x$ ranges from the parabola $x = \sqrt{\frac{y}{2}}$ to the right boundary $x = 1$ (Refer to Figure ii for a visual representation and aid).
   Using horizontal slices, the volume V is given by $V = \int_{0}^{2}\int_{\sqrt{\frac{y}{2}}}^{1} (8x + 6ydx)dy$.

2. Compute the Inner Integral (with respect to x): $\int_{\sqrt{\frac{y}{2}}}^{1} (8x + 6y)dx = 4x^2+6xy\bigg|_{\sqrt{\frac{y}{2}}}^{1} = 4 + 6y -(2y+\frac{6}{\sqrt{2}}y\sqrt{y}) = 4 + 4y -3\sqrt{2}y^{\frac{3}{2}}$

3. Compute the Outer Integral (with respect to y): $\int_{0}^{2} (4 + 4y -3\sqrt{2}y^{\frac{3}{2}})dy = 4y +2y^2-\frac{6\sqrt{2}}{5}y^{\frac{5}{2}}\bigg|_{0}^{2} = 8 +8 -\frac{6\sqrt{2}\sqrt{32}}{5} = 16 -\frac{6\sqrt{64}}{5} = 16 -\frac{6 \cdot 8}{5}= 16 -\frac{48}{5} = \frac{80}{5}-\frac{48}{5} = \frac{32}{5}$.

4. Conclusion. Both methods, using vertical and horizontal slices, yield the same result, $V = \frac{32}{5}$. This consistency confirms the accuracy of our calculation.

• Volume Under a Surface. Find the volume V under the plane $3x + 2y -z = 0$ and above the region enclosed by the parabolas $y = x^2$ and $x = y^2$.

Double Integral Setup

The volume under the surface $z = f(x, y)$ over a region R in the xy-plane is given by the double integral: $V = \iint_R f(x, y) dA$.

The function $f(x, y) = 3x + 2y$ represents the height of the surface above each point (x, y) in the xy-plane. Our goal is to find the volume under this surface and above the region R in the xy-plane.

1. Finding the Region $R$. First, we find where the two boundary curves intersect. Substituting $y = x^2$ into $x = y^2$: $x = (x^2)^2 = x^4 \implies x^4 - x = 0 \implies x(x^3 - 1) = 0$.
   The solutions are $x = 0$ and $x = 1$. The corresponding intersection points are $(0, 0)$ and $(1, 1)$.
2. Method A: Vertical Slices (Integrate with respect to $y$ first).
   For vertical slices, $x$ ranges from 0 to 1. For any fixed $x$, $y$ ranges from the lower parabola $y = x^2$ to the upper curve $y = \sqrt{x}$ (since $x = y^2$ implies $y = \sqrt{x}$ in the first quadrant). Refer to Figure 3 for a visual representation.
   Therefore, the volume V can be computed using vertical slices as follows: $V = \int_{0}^{1}\int_{x^2}^{\sqrt{x}} (3x + 2y)dydx$.
3. Compute the Inner Integral (with respect to y): $\int_{x^2}^{\sqrt{x}} (3x + 2y)dy = 3xy + y^2\bigg|_{x^2}^{\sqrt{x}} = 3x\sqrt{x}+x-(3x^3+x^4)=3x^{\frac{3}{2}}+x-3x^3-x^4$
4. Compute the Outer Integral (with respect to x): $V = \int_{0}^{1}\int_{x^2}^{\sqrt{x}} (3x + 2y)dydx = \int_{0}^{1} (3x^{\frac{3}{2}}+x-3x^3-x^4)dx = \frac{6}{5}x^{\frac{5}{2}}+\frac{1}{2}x^2-\frac{3}{4}x^4-\frac{1}{5}x^5\bigg|_{0}^{1}$
   $\frac{6}{5}+\frac{1}{2}-\frac{3}{4}-\frac{1}{5} = \frac{24}{20} + \frac{10}{20}-\frac{15}{20}-\frac{4}{20} = \frac{15}{20} = \frac{3}{4}$
5. Conclusion. The volume under the surface $f(x, y) = 3x + 2y$ over the region R bounded by the curves $y = x^2$ and $x = y^2$ is $\frac{3}{4}$ cubic units.

Method B: Horizontal Slice

1. Finding the Region $R$. The corresponding intersection points are $(0, 0)$ and $(1, 1)$.
2. Method B: Horizontal Slices (Integrate with respect to $x$ first)
   For horizontal slices, $y$ ranges from 0 to 1. For any fixed $y$, $x$ ranges from the left curve $x = y^2$ to the right curve $x = \sqrt{y}$ (since $y = x^2$ implies $x = \sqrt{y}$).
   Using horizontal slices, the volume V is given by $V = \int_{0}^{1}\int_{y^2}^{\sqrt{y}} (3x +2ydx)dy$.
3. Evaluating this order will also yield exactly $\frac{3}{4}$ cubic units, confirming our result.

• Total Mass of a Lamina with Variable Density.

Let’s explore the process of finding the total mass of a flat object (also known as a lamina) when its density varies across its surface. The concept is fundamental in physics and engineering, particularly when dealing with non-uniform materials.

For a flat lamina, the density might not be uniform, meaning that different points on the lamina may have different densities. Suppose the density of the flat object is described by the function $\delta(x, y)$, where $\delta(x, y)$ represents the mass per unit area (mass density) at any point $(x, y)$ within the region $R$. The mass $\Delta m$ of this small piece can be approximated by multiplying the density $\delta(x, y)$ at that point by its area $\Delta m \approx \delta(x, y) \cdot \Delta A$. In the limit as the area $\Delta A$ becomes infinitesimally small, the mass of that tiny piece of the lamina becomes $dm = \delta(x, y) \cdot dA$, where $dA$ represents the infinitesimal area element at (x, y) within the region R.

To find the total mass of the flat object (lamina) over the region (R), we need to sum up the contributions of all these infinitesimal mass elements across the entire region. Mathematically, we integrate the density function over the region: $\boxed{\text{Mass} = \iint_R \delta(x, y) \cdot dA}$ where $dA$ represents an infinitesimal area element.

Example. Find the mass of a lamina with vertices at $(0, 0)$, $(0, 3)$, and $(3, 0)$ if its density function is $\delta(x, y) = xy \text{kg/m}^2$.

1. Setting up the double integral. To find the total mass of the lamina, we need to integrate the density function $\delta(x, y) = xy$ over the triangular region. Mass = $\iint_R \delta(x, y) \cdot dA = \iint_R xydA$
2. Determine the Limits of Integration. The region R is a triangle with vertices at $(0, 0)$, $(0, 3)$, and $(3, 0)$.
   The region is bounded by the x-axis ($y=0$), the y-axis ($x=0$), and the line connecting $(0, 3)$ and $(3, 0)$. The slope of this line is $-1$, and its y-intercept is $3$, giving the equation $y = 3 - x$. Therefore, for any fixed x from 0 to 3, y ranges from 0 (the x-axis) up to 3 -x (the line). Refer to Figure 2 for a visual representation and aid in understanding it.
3. Set up the integral. Given the above limits, the double integral for the mass becomes: Mass = $\int_{0}^{3} (\int_{0}^{3-x} xydy)dx$
4. Compute the inner integral (with respect to y): $\int_{0}^{3-x} xydy = x\frac{y^2}{2}\bigg|_{0}^{3-x} = \frac{x(3-x)^2}{2}$. This represents the mass contribution from a thin vertical strip at a fixed x position (denoted as x₀).
5. Evaluate the outer integral (with respect to x) to find the total mass: $\int_{0}^{3} \frac{x(3-x)^2}{2}dx = \int_{0}^{3} \frac{9x}{2}-3x^2 +\frac{x^3}{2} = \frac{9x^2}{4}-x^3+\frac{x^4}{8}\bigg|_{0}^{3} = \frac{81}{4}-27+\frac{81}{8} = \frac{162}{8}-\frac{216}{8}+\frac{81}{8}=\frac{27}{8}.$

Conclusion. The total mass of the triangular lamina is: $\frac{27}{8}kg$ (or mass units).

• Average Value of a function of two variables over a region.

The average value of a function $f(x, y)$ of two variables over a region R in the xy-plane is a fundamental concept in multivariable calculus.

It is given by the formula $F_{\text{average}} = \frac{1}{\text{Area of } R} \iint_R f(x,y)dA$ where $f(x, y)$ represents the function of interest, R denotes the region in the XY-plane over which the average is being computed, and dA denotes an infinitesimal area element within the region R. This formula essentially sums up the values of the function over the entire region and then divides this result by the area of the region (or “footprint”) to find the average value.

Problem. Find the average value of $f(x,y) = 7xy^2$ over the region bounded by $x = y$ and $x = \sqrt{y}$.

1. Define the region R. First, we need to understand the region R where the integration will take place.
   Set $y = \sqrt{y} \implies y^2 = y \implies y(y - 1) = 0$. Therefore, the intersection points are (0, 0) and (1, 1) (Refer to Figure 1 for a visual representation and aid in understanding it).

2. Find the Area of the region R. To calculate the average value, we first need to find the area of the region R.
   For horizontal slices, $y$ ranges from 0 to 1, and $x$ ranges from $y$ (left curve) to $\sqrt{y}$ (right curve).
   The area is given by integrating the constant function 1 over the region: Area of R = $\int \int_{R} 1dA = \int_{0}^{1} (\int_{y}^{\sqrt{y}} 1dx)dy$

3. Evaluate the Inner integral. $\int_{y}^{\sqrt{y}} 1dx = x \bigg|_{y}^{\sqrt{y}}dx = \sqrt{y}-y$

4. Substitute into the outer integral to find the area: $\text{Area} = \int_{0}^{1} (\sqrt{y}-y)dy = \frac{2}{3}y^{\frac{3}{2}}-\frac{y^2}{2}\bigg|_{0}^{1} = \frac{2}{3}-\frac{1}{2} = \frac{4}{6}-\frac{3}{6} = \frac{1}{6}$. So, the area of the region is $\frac{1}{6}$.

5. Calculate the Double Integral of the Function: $\iint_R 7xy^2 \, dA = \int_{0}^{1} \int_{y}^{\sqrt{y}} 7xy^2 dx dy$
   $\text{Inner:} \quad \int_{y}^{\sqrt{y}} 7xy^2 dx = \left[ \frac{7}{2}x^2y^2 \right]_{y}^{\sqrt{y}} = \frac{7}{2}y^2(\sqrt{y})^2 - \frac{7}{2}y^2(y)^2 = \frac{7}{2}(y^3 - y^4)$
   $\text{Outer:} \quad \int_{0}^{1} \frac{7}{2}(y^3 - y^4) dy = \frac{7}{2} \left[ \frac{y^4}{4} - \frac{y^5}{5} \right]_{0}^{1} = \frac{7}{2} \left( \frac{1}{4} - \frac{1}{5} \right) = \frac{7}{2} \left( \frac{1}{20} \right) = \frac{7}{40}$.

6. Find the Average Value of $f(x,y) = 7xy^2$. The average value of a function f(x, y) over a region R is given by $F_{\text{average}} = \frac{1}{\text{Area of } R} \iint_R f(x,y)dA$.
   Given that Area of R is $\frac{1}{6}$, the average value becomes: $F_{\text{average}} = \frac{\frac{7}{40}}{\frac{1}{6}} = \frac{7}{40} \cdot 6 = \frac{42}{40} = \frac{21}{20}$.

7. The average value of $7xy^2$ over this region is $\frac{21}{20}$.

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