Applications of Double Integrals III

Core Concepts

Double integrals are crucial tools in multivariable calculus, allowing us to calculate fundamental geometric and physical quantities.

• Calculating Areas. The area of a region R in the plane can be found by integrating the function $f(x, y) = 1$ over R, $\iint_R 1 \, dA$.
• Calculating Volumes. The volume of a region under a surface z = f(x, y) over a region R 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 f(x, y) over a region R is found using the double integral, divided by the area of R. $F_{\text{average}} = \frac{1}{\text{Area of } R} \iint_R f(x,y)dA$ where f(x, y) represents the function of interest and R denotes the region in the XY-plane.
• The weighted average of a function f with a given density function $\delta(x, y)$ over a region R can be computed using a double integral, more specifically, $\text{Weighted Average} = \frac{1}{\text{Mass}(R)} \iint_R f(x, y) \cdot \delta(x, y)dA$, where:

1. $\text{Mass}(R) = \iint_R \delta(x, y)dA$ is the total mass of the region R, calculated by integrating the density function over the entire region.
2. $dA$ represents a small differential area within the region R.
3. $f(x, y)$ is the function whose average we are calculating, with the density function $\delta(x, y)$ acting as the weight.

• Center of Mass. The center of mass of an object is the hypothetical “balance point” where its entire mass can be thought of as being concentrated for the purpose of analyzing its motion.

For a region $R$ in the xy-plane with density function $\delta(x, y)$, the coordinates of the center of mass $(\bar{x}, \bar{y})$ are found by calculating the moments about the y-axis and x-axis, respectively: $x_{\text{centreMass}} = \frac{1}{\text{Mass}(R)} \iint_R x \cdot \delta(x, y) dA$, $y_{\text{centreMass}}= \frac{1}{\text{Mass}(R)} \iint_R y \cdot \delta(x, y) dA$

Solved exercises

• Find the total mass and the center of mass of a triangle figure with vertices $(0, 0)$, $(2, 1)$, and $(0, 3)$ with the density function $\delta(x, y) = x + y$.

1. Understanding the region.
   The line that joins the vertices $(0, 3)$ and $(2, 1)$ is calculated as: $y-1 = \frac{1-3}{2-0}(x-0)↭ y -1 = -x +2, y = 3 -x.$
   Similarly, the line that joins the vertices (0, 0) and (2, 1) is $y-0 = \frac{1-0}{2-0}(x-0) ↭ y = \frac{1}{2}x$.
   Refer to Figure A for a visual representation and aid in understanding it.

2. Calculate the total mass:
   $M(R) = \iint_R \delta(x, y) dA = \int_{0}^{2} \int_{x/2}^{3-x} (x+y) dy dx$
   Calculate the inner integral: $\int_{\frac{x}{2}}^{3-x} (x+y)dy = xy + \frac{y^2}{2}\bigg|_{\frac{x}{2}}^{3-x}$
   $x(3-x) + \frac{(3-x)^2}{2} -(x\frac{x}{2} + \frac{1}{2}(\frac{x}{2})^2) = 3x -x^2 + \frac{1}{2}(9-6x+x^2)-\frac{1}{2}x^2-\frac{1}{8}x^3 = 3x -x^2 + \frac{9}{2} -3x + \frac{1}{2}x^2 - \frac{1}{2}x^2-\frac{1}{8}x^3 = \frac{9}{2}-\frac{9}{8}x^2$.
   Calculate the outer integral: $M(R) = \iint_R \delta(x, y) dA = \int_{0}^{2} \frac{9}{2}-\frac{9}{8}x^2dx = \frac{9}{2}x -\frac{9}{24}x^3\bigg|_{0}^{2}$.
   $\frac{9}{2}2 -\frac{9}{24}8 = 9-3 = 6.$
   So, the total mass M(R) of the triangular region is 6 units.

3. Center of Mass. The center of mass of an object is the hypothetical point where its entire mass can be thought of as being concentrated for the purpose of analyzing its motion.
   For a region R in the XY-plane with a density function $\delta(x, y)$, the coordinates of the center of mass $(x_{\text{centreMass}}, y_{\text{centreMass}})$ are given by: $ x_{\text{centreMass}} = \frac{1}{\text{Mass(R)}} \iint_R x \delta(x, y)dA, y_{\text{centreMass}} = \frac{1}{\text{Mass(R)}} \iint_R y \delta(x, y) dA$.

4. Calculating the Center of Mass: $x_{\text{centreMass}} = \frac{1}{6} \int_{0}^{2} \int_{\frac{x}{2}}^{3-x} x(x+y)dydx, y_{\text{centreMass}} = \frac{1}{6} \int_{0}^{2} \int_{\frac{x}{2}}^{3-x} y(x+y)dydx$.
   Finding the Center of Mass (x-coordinate):
   Evaluate the inner integral: $\int_{\frac{x}{2}}^{3-x} x(x+y)dy = x^2y+\frac{xy^2}{2}\bigg|_{\frac{x}{2}}^{3-x} = x^2(3-x)+\frac{x(3-x)^2}{2}-x^2\frac{x}{2}-\frac{x(\frac{x}{2})^2}{2}$
   $3x^2-x^3+\frac{x(9-6x+x^2)}{2}-\frac{x^3}{2}-\frac{x^3}{8} = 3x^2-x^3+\frac{9x}{2}-3x^2+\frac{x^3}{2}-\frac{x^3}{2}-\frac{x^3}{8} = (\frac{-8-1}{8})x^3 +\frac{9x}{2} = \frac{9x}{2}-\frac{9x^3}{8}$.
   Now, evaluate the outer integral: $x_{\text{centreMass}} = \frac{1}{6} \int_{0}^{2} \left(\frac{9x}{2} - \frac{9x^3}{8}\right) dx = \frac{1}{6} \left[\frac{9x^2}{4} - \frac{9x^4}{32}\right]_{0}^{2}$
   $= \frac{1}{6} \left(\frac{9 \cdot 4}{4} - \frac{9 \cdot 16}{32}\right) = \frac{1}{6} \left(9 - \frac{9}{2}\right) = \frac{1}{6} \left(\frac{9}{2}\right) = \frac{3}{4}$.
   Similarly, $y_{\text{centreMass}} = \frac{1}{6} \int_{0}^{2} \int_{x/2}^{3-x} y(x+y) dy dx$.
   Evaluate the inner integral: $\int_{x/2}^{3-x} (xy + y^2) dy = \left[\frac{xy^2}{2} + \frac{y^3}{3}\right]_{x/2}^{3-x} = \left(\frac{x(3-x)^2}{2} + \frac{(3-x)^3}{3}\right) - \left(\frac{x(x/2)^2}{2} + \frac{(x/2)^3}{3}\right)$
   $= \left(\frac{9x - 6x^2 + x^3}{2} + \frac{27 - 27x + 9x^2 - x^3}{3}\right) - \left(\frac{x^3}{8} + \frac{x^3}{24}\right) = \left(\frac{9x}{2} - 3x^2 + \frac{x^3}{2}\right) + \left(9 - 9x + 3x^2 - \frac{x^3}{3}\right) - \frac{4x^3}{24} = 9 - \frac{9x}{2} + x^3\left(\frac{1}{2} - \frac{1}{3} - \frac{1}{6}\right) = 9 - \frac{9x}{2} + 0 = 9 - \frac{9x}{2}$
   Evaluate the outer integral: $y_{\text{center}} = \frac{1}{6} \int_{0}^{2} \left(9 - \frac{9x}{2}\right) dx = \frac{1}{6} \left[9x - \frac{9x^2}{4}\right]_0^2 = \frac{1}{6} (18 - 9) = \frac{9}{6} = \frac{3}{2}$.
   The center of mass is $\left(\frac{3}{4}, \frac{3}{2}\right)$.

• Calculate the volume enclosed by the cylinders $z = x^2$, $y = x^2$, and the planes $z = 0$, $y = 4$.

1. Understanding the problem.
   The surface $z = x^2$ represents a parabolic cylinder extending along the y-axis, while $y=x^2$ defines another parabolic cylinder extending along the z-axis. The plane z = 0 is the xy-plane, and y = 4 is a vertical plane that cuts through the region.
   Note: $z = x^2$ represents a parabolic cylinder. Because the variable $y$ is entirely missing from the equation, $y$ can take on any real value. This means the 2D parabola drawn in the $xz$-plane is projected and extended infinitely along the $y$-axis, creating a trough-like (a long, shallow, open container used to hold food or water for livestock such as pigs, cows, or horses) surface.
   Refer to Figure 4 for a visual representation and aid in understanding it.
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: $V = \iint_R f(x, y) dA$.
   Given the surfaces, the region $R$ in the xy-plane is defined by the intersection of the curves $y = x^2$ and the line $y = 4$. Therefore, $R$ is bounded by:
   $\star$ $x$ ranges from $-2$ to $2$, because $y = 4$ and $x^2 = 4$ give $x = \pm 2$.
   $\star$ For each fixed $x$, $y$ ranges from $x^2$ (the lower curve) to $y = 4$ (the upper boundary).
   The function $f(x, y) = x^2$ gives the height of the surface above each point $(x, y)$ in the region $R$. Therefore, the volume V can thus be expressed as: $V = \int_{-2}^{2}\int_{x^2}^{4} (x^2)dydx$.
3. Due to the symmetry of the problem (the surface is symmetric about the y-axis), we can compute the volume for $x$ from $0$ to $2$ and then double the result: $V = 2\int_{0}^{2}\int_{x^2}^{4} (x^2)dydx$
4. Compute the Inner Integral (with respect to y): $\int_{x^2}^{4} (x^2)dy = x^2y\bigg|_{x^2}^{4} = 4x^2-x^4.$
5. Compute the Outer Integral (with respect to x): $V = \int_{-2}^{2}\int_{x^2}^{4} (x^2)dydx = \int_{-2}^{2}(4x^2-x^4)dx = \frac{4}{3}x^3-\frac{1}{5}x^5\bigg|_{-2}^{2} = (\frac{32}{3}-\frac{32}{5})-(\frac{-32}{3}+\frac{32}{5}) = \frac{128}{15}$

• Find the moment of inertia of a disk of radius $a$, $\delta = 1$ about its center.

1. Recall.
   To calculate the moment of inertia of a uniform disk (a disk with constant density $\delta = 1$) of radius $a$ about its center, we need to consider the distribution of mass around the axis of rotation (the center of the disk).
   The moment of inertia essentially tells us how difficult it is to rotate the disk about its center.

2. Set up the problem. The moment of inertia $I$ about the center of the disk is given by the integral: $I = \iint_{R} r^2 \delta(x, y) dA$
   Since the density $\delta(x, y)$ is uniform and equal to $1$, this simplifies to: $I = \iint_{R} r^2 dA$
   In polar coordinates, the area element $dA$ is expressed as $r dr d\theta$, resulting in: $I = \iint_{R} r^2 (r dr d\theta) = \iint_{R} r^3 dr d\theta$ where $r$ ranges from $0$ to $a$ (the radius of the disk), and $\theta$ ranges from $0$ to $2\pi$ (a full circle).

3. Evaluate the inner integral (with respect to the radial coordinate $r$ while holding $\theta$ constant): $\int_{0}^{a} r^3 dr = \left[\frac{r^4}{4}\right]_{0}^{a} = \frac{a^4}{4}$

4. Evaluate the Outer Integral (with respect to the angular coordinate $\theta$):$I = \int_{0}^{2\pi} \frac{a^4}{4} d\theta = \left[\frac{a^4}{4}\theta\right]_{0}^{2\pi} = \frac{a^4}{4} \cdot 2\pi = \frac{\pi a^4}{2}$

5. Conclusion: The moment of inertia $I$ of a uniform disk with radius $a$ about its center is $\frac{\pi a^4}{2}$.

• Find the moment of inertia of a disk of radius $a$, $\delta = 1$ about its left extreme.

1. The moment of inertia of a disk about an axis through its left extreme (rather than through its center) provides a different perspective on how the mass of the disk is distributed relative to this new axis. The disk still has a uniform density $\delta = 1$, but since the axis of rotation has shifted, the calculation changes accordingly.
2. Set up the problem. The moment of inertia $I$ about the left extreme of the disk is given by the integral: $I = \iint_{R} r^2 \cdot dA$ where $R$ is the region representing the disk, $r$ is the distance from any point $(x, y)$ on the disk to the axis of rotation (the left extreme of the disk), and $dA$ is the differential area element within the disk.
   If the leftmost point of a disk with radius $a$ is at the origin $(0, 0)$, its center must sit on the x-axis at $(a, 0)$. The equation for this circle is: $(x -a)^2 + y^2 = a^2$.
   Expand the equation: $x^2 -2ax + a^2 + y^2 = a^2 \implies x^2 + y^2 = 2ax$.
   Convert to Polar Coordinates: Using the standard substitutions $x^2 + y^2 = r^2$ and $x = r\cos(\theta)$: $r^2 = 2a(r\cos(\theta))$.
   Solve for r. Dividing both sides by $r$ (where $r \ne 0$) yields the boundary equation: $r = 2a(\cos(\theta))$. Because the entire disk sits to the right of the y-axis, a ray sweeping through the disk goes from the bottom edge to the top edge, which corresponds to the angular limits $\frac{-\pi}{2}$ to $\frac{\pi}{2}$ The differential area element in polar coordinates is still $dA = rdrd\theta$. Thus, the moment of inertia integral becomes: $\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}(\int_{0}^{2a\cos(\theta)} r^2rdr)d\theta = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}(\int_{0}^{2a\cos(\theta)} r^3dr)d\theta$
3. Evaluate the inner integral with respect to $r$, the radial coordinate, while holding $\theta$ constant: $\int_{0}^{2a\cos(\theta)} r^3 dr = \left[\frac{r^4}{4}\right]_{0}^{2a\cos(\theta)} = \frac{16a^4\cos^4(\theta)}{4} = 4a^4\cos^4(\theta)$
4. Evaluate the outer integral (with respect to $\theta$): $I = \int_{-\pi/2}^{\pi/2} 4a^4\cos^4(\theta) d\theta$
   Breaking down the advanced trigonometric evaluation: To integrate $\cos^4(\theta)$, we apply the power-reduction formula $\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$ twice:
   $\cos^4(\theta) = \left(\frac{1 + \cos(2\theta)}{2}\right)^2 = \frac{1 + 2\cos(2\theta) + \cos^2(2\theta)}{4}$
   Apply the formula again to $\cos^2(2\theta) = \frac{1 + \cos(4\theta)}{2}$: $\cos^4(\theta) = \frac{1}{4} \left(1 + 2\cos(2\theta) + \frac{1 + \cos(4\theta)}{2}\right) = \frac{3}{8} + \frac{1}{2}\cos(2\theta) + \frac{1}{8}\cos(4\theta)$
   Now integrate term by term: $\int \left(\frac{3}{8} + \frac{1}{2}\cos(2\theta) + \frac{1}{8}\cos(4\theta)\right) d\theta = \frac{3}{8}\theta + \frac{1}{4}\sin(2\theta) + \frac{1}{32}\sin(4\theta)$
   Evaluating from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$ causes the sine terms to become zero, leaving only the $\theta$ term: $I = 4a^4 \left[ \frac{3}{8}\left(\frac{\pi}{2}\right) - \frac{3}{8}\left(-\frac{\pi}{2}\right) \right] = 4a^4 \left( \frac{3\pi}{16} + \frac{3\pi}{16} \right) = 4a^4 \left( \frac{6\pi}{16} \right) = \frac{3\pi a^4}{2}$
5. Result Insight. This result ($\frac{3\pi a^4}{2}$) is exactly 3 times larger than the moment of inertia about the disk’s center ($\frac{\pi a^4}{2}$), reflecting the fact that the mass is now farther from the axis of rotation, requiring substantially more effort to rotate.

• Moment of inertia of a triangular lamina $R$ with vertices $(0, 0)$, $(2, 0)$, and $(2, 2)$, and density $\rho(x,y) = xy \text{ kg}/\text{m}^2$.

1. Recall. The moment of inertia is a measure of how resistant a body is to rotational motion about an axis. In this problem, we are calculating the moment of inertia of a triangular lamina with respect to the x-axis, y-axis, and the origin.

2. Set up the problem. Given the vertices $(0, 0)$, $(2, 0)$, and $(2, 2)$, the triangular lamina is bounded by the line $y = x$ from $x = 0$ to $x = 2$, $0 \le y \le x$. The density function $\rho(x,y) = xy$ means that the density varies across the lamina depending on the position.
   Moment of Inertia about the x-axis ($I_x$): The moment of inertia about the x-axis is calculated by integrating the product of the square of the distance from the x-axis (which is $y^2$) and the density function $\rho(x,y)$ over the entire region $R$ (Figure 3).
   $I_x = \iint_{R} y^2 \rho(x,y) dA = \int_{0}^{2} \int_{0}^{x} y^2 (xy) dy dx = \int_{0}^{2} \int_{0}^{x} xy^3 dy dx$

3. Evaluating the Inner Integral. $\int_{0}^{x} xy^3 dy = x \left[\frac{y^4}{4}\right]_{0}^{x} = x \left(\frac{x^4}{4}\right) = \frac{x^5}{4}$

4. Evaluating the Outer Integral: $I_x = \int_{0}^{2} \frac{x^5}{4} dx = \left[\frac{x^6}{24}\right]_{0}^{2} = \frac{64}{24} = \frac{8}{3}$.

5. Thus, the moment of inertia about the x-axis is $\frac{8}{3} \text{ kg}\cdot\text{m}^2$.

6. Moment of Inertia about the y-axis ($I_y$):
   Similarly, this is calculated by integrating the square of the distance from the y-axis ($x^2$) times the density: $I_y = \iint_{R} x^2 \rho(x,y) dA = \int_{0}^{2} \int_{0}^{x} x^2 (xy) dy dx = \int_{0}^{2} \int_{0}^{x} x^3y dy dx$

7. Evaluate the Inner Integral: $\int_{0}^{x} x^3y dy = x^3 \left[\frac{y^2}{2}\right]_{0}^{x} = \frac{x^5}{2}$

8. Evaluate the Outer Integral: $I_y = \int_{0}^{2} \frac{x^5}{2} dx = \left[\frac{x^6}{12}\right]_{0}^{2} = \frac{64}{12} = \frac{16}{3}$

9. Thus, the moment of inertia about the y-axis is $\frac{16}{3} \text{ kg}\cdot\text{m}^2$.

10. Moment of Inertia about the Origin ($I_o$):
    Also known as the polar moment of inertia, this is the sum of the moments of inertia about the x-axis and the y-axis: $I_o = \iint_{R} (x^2+y^2) \rho(x, y) dA = I_x + I_y = \frac{8}{3} + \frac{16}{3} = \frac{24}{3} = 8 \text{ kg}\cdot\text{m}^2$

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