All those who seem stupid, they are, and also so are half of those who do not, Quevedo.
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) = x^{2} + y^{2}, e^{x+y}, etc.
Partial derivatives are derivatives of a function of multiple variables, say f(x_{1}, x_{2}, ···, 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 (x_{0}, y_{0}) 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 = y_{0}. 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 = x_{0}
$\frac{dw}{ds}\bigg|_{\vec{u}} = ∇w·\vec{u} = |∇w|·|\vec{u}|cos(θ) = |∇w|·cos(θ)$ where θ is the angle between the gradient and the given unit vector.
The 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.
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). It is denoted or written as $\iint_R f(x, y) dA$
Fubini’s Theorem. 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: $\int \int_R f(x, y)dA = \int_{a}^{b}\int_{c}^{d} f(x, y)dydx = \int_{c}^{d}\int_{b}^{a} f(x, y)dxdy$
Double integrals are crucial in various applications, such as:
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.
When dealing with physical systems or mathematical models, we often encounter situations where we want to find the average value of a function over a certain region. However, if the region has a varying density — meaning that some areas have more “weight” than others, then we calculate a weighted average.
This approach is particularly useful when the density, denoted as δ(x, y), is not uniform across the region R.
The weighted average of a function f with a given density function δ(x, y) over a region R can be computed using a double integral, more specifically, Weighted Average = $\frac{1}{Mass(R)}\iint_R f(x, y)·δ(x, y)·dA$, where:
This formula takes into account how the density function distributes mass across the region, providing a more accurate representation of the average value of f(x, y) than a simple arithmetic mean.
It’s a crucial concept in physics because it simplifies the study of the motion of objects by allowing us to treat them as if all their mass were located at a single point. It is effectively the “balance point” of the object, where if you were to support it at this point, the object would remain perfectly balanced.
For a region R in the XY-plane with a density function δ(x, y), the coordinates of the center of mass (x_{center}, y_{center}) 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$.
Loosely speaking, the moment of inertia about an axis is how hard it is to impact the rotation motion about that axis or a measure of an object’s resistance to changes in its rotational motion about an axis.
When you want to spin something faster or slower, the moment of inertia determines how much effort (torque) is needed to change the object’s rotational speed. The higher the moment of inertia, the harder it is to change the rotational speed of the object.
The moment of inertia is a fundamental property of a rotating object. It quantifies the object’s resistance to changes in its rotational motion. It depends on both the mass distribution of the object and the distance of each mass element from the axis of rotation.
For a region R with density function δ(x, y), the moment of inertia around the origin is calculated by taking the double integral of the object’s mass distribution with respect to the distance from its axis of rotation $I = \iint r^2·dm$=[For a solid with density δ] $ \iint_{R} r^2·δ(x, y)dA.$
Here, r (In Cartesian coordinates, $r^2 = x^2 + y^2$) represents the perpendicular distance of a mass element from the axis of rotation.
If the rotation is about the x-axis, the moment of inertia is: $I_x = \iint_{R} y^2·δ(x, y)dA.$
For rotations about the y-axis, the moment of inertia is: $I_y = \iint_{R} x^2·δ(x, y)dA.$
For rotation about the origin (the polar moment of inertia): $I_0 = I_x + I_y = \iint_{R} (x^2+y^2)·δ(x, y)dA.$
In polar coordinates, where $r^2 =x^2+y^2$ and the area element dA is given by rdrdθ, this integral can be expressed as: $I = \iint_{R} r^2·δ(r, θ)·r·drdθ.$
Kinetic energy is the energy associated with the motion of an object. For an object in linear motion, it can easily be determined by an equation using the mass and velocity of that object. The kinetic energy is one-half the product of the particle’s mass m and the square of its speed, KE = $\frac{1}{2}mv^2$.
Where m is the mass of the object (Kg) and v is its velocity (n/s) -Figure iv a- This equation tells us how much energy an object possesses due to its motion. When something is moving, it has kinetic energy —whether it’s a speeding car, a flying bird, or even a rolling ball.
For a rotating rigid body, the rotational kinetic energy depends on the moment of inertia I and the angular velocity ω, which is the rate at which the object rotates about a fixed axis.
The angular velocity represents how fast an object rotates about a fixed axis, i.e., the rate of change of the angle with respect to time: w = $\frac{Δθ}{Δt}$ where Δθ is the change in angle (radians), Δt is the time taken for this change.
Linear velocity (v) is the speed of an object moving along a straight path.
For one rotation Δθ = 2π radians, Δt = T (the period of the time for one complete rotation), w = $\frac{Δθ}{Δt} = \frac{2π}{T}$. We all know that v = d/t, hence the lineal velocity is v = $\frac{2\pi r}{T} = r \omega$ ⇒ The formula for kinetic energy of an object in rotational motion is: $\frac{1}{2}mv^2 = \frac{1}{2}mr^2w^2$ where w is the angular velocity (in radians per second) and r represents the distance from the axis -Figure iv b-. The moment of inertia I is mr^{2}. The kinetic energy of a rotating rigid body is directly proportional to its moment of inertia and the square of its angular velocity, KE = $\frac{1}{2}Iw^2$
For simplicity, we’ll place the square on the coordinate plane such that its center is at the origin (0,0), and its sides are aligned with the coordinate axes. The square has a side length of 2a, so its vertices are at (−a, −a), (a, −a), (−a, a), and (a, a), and the area of the square is 4a^{2}.
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) represents the function whose average value we want to compute, and R represents the region over which we are averaging the function. In our problem, the function of interest is the distance of a point (x, y) from the center, given by f(x, y) = $\sqrt{x^2+y^2}$.
Set Up the Integral for the Average Distance
The average_distance d_{average} is then $d_{average} = \frac{1}{4a^2} \iint_R f(x,y)dA = \frac{1}{4a^2} \iint_R \sqrt{x^2+y^2}dxdy$. Here, 4a^{2} is the area of the square, and the integral computes the sum of distances of all points within the square from the center.
Because the square is symmetric about the x-axis, y-axis, and the origin, we can simplify our calculations by considering only the integral of the triangular region in the first quadrant where x ≥ 0, y ≥ 0, 0 ≤ y ≤ x ≤ a and then multiply the result by 8 to account for the entire square:
Average_distance = $\frac{8}{4a^2} \iint_R f(x,y)dA = \frac{2}{a^2} \iint_R \sqrt{x^2+y^2}dxdy = \frac{1}{2a^2} \int_{0}^{a} \int_{0}^{a} \sqrt{x^2+y^2}dxdy.$
To evaluate the integral, it’s convenient to switch to polar coordinates. In polar coordinates: x = rcos(θ), y = rsin(θ), dA = rdrdθ, so the integral becomes:
Average_distance = $\frac{2}{a^2}\iint_R r·rdrdθ = \frac{2}{a^2}\int_{0}^{\frac{π}{4}} \int_{0}^{\frac{a}{cos(θ)}}r^2drdθ$.
Evaluate the Inner integral with respect to r: $\int_{0}^{\frac{a}{cos(θ)}}r^2dr = \frac{r^3}{3}\bigg|_{0}^{\frac{a}{cos(θ)}} = \frac{1}{3}\frac{a^3}{cos^3(θ)} = \frac{a^3}{3cos^3(θ)}$.
Next, substitute the result into the outer integral and evaluate with respect to θ: Average_distance = $\frac{2}{a^2}\iint_R r·rdrdθ = \frac{2}{a^2}·\frac{a^3}{3} \int_{0}^{\frac{π}{4}} sec^3(θ)dθ = \frac{2a^3}{3a^2} \int_{0}^{\frac{π}{4}} sec^3(θ)dθ = \frac{2a}{3}\int_{0}^{\frac{π}{4}} sec^3(θ)dθ$.
Let’s use integration by parts: ∫udv = uv -∫vdu where $u = sec(θ), dv = sec^2(θ)dθ, du = sec(θ)tan(θ)dθ, v = ∫sec^2(θ)dθ = tan(θ)$
$\int sec^3(θ)dθ = sec(θ)tan(θ) -\int tan^2(θ)sec(θ)dθ$
Since tan^{2}(θ) = sec^{2}(θ)-1, we can rewrite the integral as: $\int sec^3(θ)dθ = sec(θ)tan(θ) -\int (sec^2(θ)-1)sec(θ)dθ = sec(θ)tan(θ) -\int sec^3(θ) + \int sec(θ)dθ$.
Therefore, $2\int sec^3(θ)dθ = sec(θ)tan(θ) + \int sec(θ)dθ$.
We know the integral of sec(θ), $\int sec(θ)dθ = ln|sec(θ)+tan(θ)| + C.$
$\int sec^3(θ)dθ = \frac{1}{2}sec(θ)tan(θ) + \frac{1}{2}ln|sec(θ)+tan(θ)| + C$.
$\int_{0}^{\frac{π}{4}} sec^3(θ)dθ = \frac{1}{2}sec(θ)tan(θ) + \frac{1}{2}ln|sec(θ)+tan(θ)|\bigg|_{0}^{\frac{π}{4}}$
At $θ = \frac{π}{4}, sec(θ) = \sqrt{2}, tan(θ) = 1.$ At $θ = 0, sec(θ) = 1, tan(θ) = 0.$
$\frac{1}{2}\sqrt{2}·1+\frac{1}{2}ln(1+\sqrt{2}) = \frac{\sqrt{2}}{2}+\frac{1}{2}ln(1+\sqrt{2}).$
d_{average} = $\frac{2a}{3}(\frac{\sqrt{2}}{2}+\frac{1}{2}ln(1+\sqrt{2})) = \frac{a}{3}(\sqrt{2}+ln(1+\sqrt{2}))$
Calculate the total mass. The total mass M of the region R is given by the double integral of the density function over the region: Mass(R) = $\iint_R δ(x, y)dA$.
Since the region R is the unit square, where 0 ≤ x ≤ 1 and 0 ≤ y ≤1, the mass can be computed as:
$\int_{0}^{1} \int_{0}^{1} xydydx$.
Compute the inner integral: $\int_{0}^{1} xydy = x\frac{y^2}{2}\bigg|_{0}^{1} = \frac{x}{2}$.
Compute the outer integral: Mass(R) = $\iint_R δ(x, y)dA = \int_{0}^{1} \int_{0}^{1} xydydx = \int_{0}^{1} \frac{x}{2}dx = \frac{x^2}{4}\bigg|_{0}^{1} = \frac{1}{4}$. Thus, the total mass M of the unit square with this density function is $\frac{1}{4}$.
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 δ(x, y), the coordinates of the center of mass (x_{center}, y_{center}) 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$.
Calculating the Center of Mass: $x_{\text{centreMass}} = 4\int_{0}^{1} \int_{0}^{1} x(xy)dydx, y_{\text{centreMass}} = 4 \int_{0}^{1} \int_{0}^{1} y(xy)dydx$.
$x_{\text{centreMass}} = 4\int_{0}^{1} \int_{0}^{1} x(xy)dydx = 4\int_{0}^{1} \int_{0}^{1} x^2ydydx$
Inner integral: $\int_{0}^{1} x^2ydy = x^2\frac{y^2}{2}\bigg|_{0}^{1} = \frac{x^2}{2}$
Outer integral: $x_{\text{centreMass}} = 4\int_{0}^{1} \int_{0}^{1} x(xy)dydx = 4\int_{0}^{1} \frac{x^2}{2}dx = 2\int_{0}^{1}x^2 = 2\frac{x^3}{3}\bigg|_{0}^{1} = \frac{2}{3}$.
Since this density function is symmetric in x and y, $x_{\text{centreMass}} = y_{\text{centreMass}} = \frac{2}{3}$. So, the center of mass is $(\frac{2}{3}, \frac{2}{3})$
Calculating the Moment of Inertia
The moment of inertia quantifies how the mass is distributed with respect to an axis, influencing the object’s rotational behavior.
For a region R with density function δ(x, y), the moment of inertia around the origin is calculated by taking the double integral of the object’s mass distribution with respect to the distance from its axis of rotation $I = \iint r^2·dm$=[For a solid with density δ] $ \iint_{R} r^2·δ(x, y)dA = \int_{0}^{1} \int_{0}^{1} r^2xydxdy = \int_{0}^{1} \int_{0}^{1} (x^2+y^2)xydxdy = \int_{0}^{1} \int_{0}^{1} x^3y+xy^3dxdy$
Inner integral: $\int_{0}^{1} x^3y+xy^3dx = y\frac{x^4}{4}+y^3\frac{x^2}{2}\bigg|_{0}^{1} = \frac{1}{4}y+\frac{1}{2}y^3$.
Outer integral: $I = \int_{0}^{1} \int_{0}^{1} x^3y+xy^3dxdy = \int_{0}^{1} \frac{1}{4}y+\frac{1}{2}y^3dy = \frac{y^2}{8}+\frac{y^4}{8}\bigg|_{0}^{1} = \frac{1}{8}+\frac{1}{8} = \frac{1}{4}$. So, the moment of inertia about the origin is $\frac{1}{4}.$
If the rotation is about the x-axis, the moment of inertia is: $I_x = \iint_{R} y^2·δ(x, y)dA = \int_{0}^{1} \int_{0}^{1} y^2xydydx = \int_{0}^{1} x(\int_{0}^{1} y^3dy)dx$
Inner integral: $\int_{0}^{1} y^3dy = \int_{0}^{1} y^3 = \frac{y^4}{4}\bigg|_{0}^{1} = \frac{1}{4}$.
Calculate the outer integral: $I_x = \iint_{R} y^2·δ(x, y)dA = \int_{0}^{1} \int_{0}^{1} y^2xydydx = \int_{0}^{1} x(\int_{0}^{1} y^3dy)dx = \frac{1}{4}\int_{0}^{1} x dx = \frac{1}{4} \frac{x^2}{2}\bigg|_{0}^{1} = \frac{1}{4}·\frac{1}{2} = \frac{1}{8}$. So, the moment of inertia about the x-axis is $\frac{1}{8}$.
Understanding the region. The line that join the vertices (0, 3) and (2, 1) is $y-1 = \frac{1-3}{2-0}(x-0)↭ y -1 = -x +2, y = 3 -x.$ 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).
Calculate the total mass: Mass(R) = $\iint_R δ(x, y)dA = \int_{0}^{2} \int_{\frac{x}{2}}^{3-x} (x+y)dydx.$
Calculate the inner integrate: $\int_{\frac{x}{2}}^{3-x} (x+y)dy = xy + \frac{y^2}{2}\bigg|_{\frac{x}{2}}^{3-x} = x(3-x) + \frac{(3-x)^2}{2} -(x\frac{x}{2} + \frac{1}{2}(\frac{x}{2})^2) = 3x -x^2 + \frac{1}{2}(9-6x+x^2)-\frac{1}{2}x^2-\frac{1}{8}x^3 = 3x -x^2 + \frac{9}{2} -3x + \frac{1}{2}x^2 - \frac{1}{2}x^2-\frac{1}{8}x^3 = \frac{9}{2}-\frac{9}{8}x^2dx$
Calculate the outer integrate: Mass(R) = $\iint_R δ(x, y)dA = \int_{0}^{2} \int_{\frac{x}{2}}^{3-x} (x+y)dydx = \int_{0}^{2} \frac{9}{2}-\frac{9}{8}x^2dx = \frac{9}{2}x -\frac{9}{24}x^3\bigg|_{0}^{2} = \frac{9}{2}2 -\frac{9}{24}8 = 9-3 = 6.$ So, the total mass M(R) of the triangular region is 6 units.
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 δ(x, y), the coordinates of the center of mass (x_{center}, y_{center}) 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$.
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 x_{center}. 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 integrate: x_{center} = $\frac{1}{6} \int_{0}^{2}( \frac{9x}{2}-\frac{9x^3}{8})dx = \frac{1}{6}(\frac{9x^2}{4}-\frac{9x^4}{32})\bigg|_{0}^{2}= \frac{1}{6}(\frac{9·2^2}{4}-\frac{9·2^4}{32}) = \frac{1}{6}(9-\frac{144}{32}) = \frac{1}{6}(\frac{18}{2}-\frac{9}{2}) = \frac{1}{6}·\frac{9}{2} = \frac{9}{12} = \frac{3}{4}$
Similarly, for $y_{\text{centreMass}} = \frac{1}{6} \int_{0}^{2} \int_{\frac{x}{2}}^{3-x} y(x+y)dydx = \frac{3}{2}$
The center of mass is $(\frac{3}{4}, \frac{3}{2})$.
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.
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 plane y = 4. Therefore, R is bounded by:
Double Integral Setup. 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: $V = \int_{-2}^{2}\int_{x^2}^{4} (x^2)dydx$. 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: $V = 2\int_{0}^{2}\int_{x^2}^{4} (x^2)dydx$
This means x ranges from -2 to 2. For each fixed x (denoted as x_{0}), y ranges from x^{2} (lower curve) to 4 (upper plane) (Refer to Figure 4 for a visual representation).
Compute the Inner Integral (with respect to y): $\int_{x^2}^{4} (x^2)dy = x^2y\bigg|_{x^2}^{4} = 4x^2-x^4.$
Compute the Outer Integral (with respect to y): $V = \int_{-2}^{2}\int_{x^2}^{4} (x^2)dydx = \int_{-2}^{2}(4x^2-x^4)dx = \frac{4}{3}x^3-\frac{1}{5}x^5\bigg|_{-2}^{2} = (\frac{32}{3}-\frac{32}{5})-(\frac{-32}{3}+\frac{32}{5}) = \frac{128}{15}$
To calculate the moment of inertia of a uniform disk (a disk with constant density δ = 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.
Conclusion: The moment of inertia I of a uniform disk with radius a about its center is: $\frac{πa^4}{2}$
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 δ = 1, but since the axis of rotation has shifted, the calculation changes accordingly.
Set up the problem (Figure iv e). The moment of inertia I about the left extreme of the disk is given by the integral: I = $ \iint_{R} r^2·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.
Since the axis of rotation is now at the leftmost point of the disk, we need to adjust our coordinates accordingly. In polar coordinates, the disk can be described with the angle θ ranging from $\frac{-π}{2}$ to $\frac{π}{2}$ (representing the entire disk from the bottom to the top along the left side), and the radius r varying from 0 to 2acos(θ), where θ accounts for the distance from the axis to any point on the disk.
The differential area element in polar coordinates is still dA = rdrdθ. Thus, the moment of inertia integral becomes: $\int_{\frac{-π}{2}}^{\frac{π}{2}}(\int_{0}^{2acos(θ)} r^2rdr)dθ = \int_{\frac{-π}{2}}^{\frac{π}{2}}(\int_{0}^{2acos(θ)} r^3dr)dθ$
Evaluate the inner integral with respect to r, the radial coordinate, while holding θ constant: $\int_{0}^{2acos(θ)} r^3dr = \frac{r^4}{4}\bigg|_{0}^{2acos(θ)} = 4a^4·cos^4(θ)$
Evaluate the inner integral with respect to the angular coordinate: I = $ \int_{\frac{-π}{2}}^{\frac{π}{2}}(4a^4·cos^4(θ))dθ$ =[This integral requires a more advanced technique to evaluate, specifically involving trigonometric identities or a reduction formula. It is left as an exercise] $\frac{3}{2}π·a^4$. This result is larger than the moment of inertia about the disk's center, which reflects the fact that the mass is now farther from the axis of rotation, requiring more effort to rotate the disk about this new axis.
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, which is a thin, flat object, with respect to the x-axis, y-axis, and the origin.
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. The density function ρ(x,y) = xy means that the density varies across the lamina depending on the x and y coordinates.
Moment of Inertia about the x-axis 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 ρ(x,y) over the entire region R. $I_x = \iint_{R} y^2·ρ(x,y)dA =\int_{0}^{2}(\int_{0}^{x}y^2·xydy)dx = \int_{0}^{2}(\int_{0}^{x}xy^3dy)dx$ (Figure 3).
Evaluating the Inner Integral. $x\int_{0}^{x}y^3dy = x(\frac{y^4}{4}\bigg|_{0}^{x})$
$= x·\frac{x^4}{4} = \frac{x^5}{4}$
Evaluating the outer integral: $I_x = \int_{0}^{2} \frac{x^5}{4} = \frac{x^6}{24}\bigg|_{0}^{2} = \frac{2^6}{24} = \frac{64}{24} = \frac{8}{3}$. Thus, the moment of inertia about the x-axis is: $\frac{8}{3}kg·pm^2$.
Similarly, the moment of inertia about the y-axis is calculated by integrating the product of the square of the distance from the y-axis (which is x^{2}) and the density function ρ(x, y) over the entire region R.
The formula for I_{y} is: $I_y = \iint_{R} x^2·ρ(x,y)dA$
Set up the double integral as:
$\int_{0}^{2}(\int_{0}^{x}x^2·xydy)dx$
Evaluate the inner integral: $\int_{0}^{x}x^2·xydy = x^3(\frac{y^2}{2}\bigg|_{0}^{x})dx =\frac{x^5}{2}.$
Evaluate the outer integral: $\int_{0}^{2} \frac{x^5}{2} =\frac{x^6}{12}\bigg|_{0}^{2} = \frac{2^6}{12} = \frac{64}{12} = \frac{16}{3}$. Thus, the moment of inertia about the y-axis is: $\frac{16}{3}kg·m^2$
The moment of inertia about the origin (also known as the polar moment of inertia) is the sum of the moments of inertia about the x-axis and the y-axis: $I_o = \iint_{R} (x^2+y^2)·ρ(x, y)dA = \int_{0}^{2}(\int_{0}^{x}(x^2+y^2)·xydy)dx = I_x + I_y = \frac{8}{3} + \frac{16}{3} = \frac{24}{3} = 8kg·m^2$