The best place to find a helping hand is at the end of your own arm, Swedish Proverb.
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.
The integral of the constant function f(x,y) = 1 over the region R gives the area of R. Mathematically, this is expressed as: Area of R = $\int \int_{R} 1dA$
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).
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.
The integral of the constant function f(x,y) = 1 over the region R gives the area of R. Mathematically, this is expressed as: Area of R = $\int \int_{R} 1dA$
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.
$=(\frac{-27}{3}+\frac{9}{2}+18)-(\frac{8}{3}+2-12) = \frac{-54+27+108-16+60}{6} = \frac{125}{6}$.
Thus, the area of the region bounded by y = -x^{2} and x + y = -6 is $\frac{125}{6}$.
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: Area of R = $\int \int_{R} 1dA$. Sometimes, it is a little more complicated than that.
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: $A = A_1 + A_2 = \frac{1}{3} + \frac{1}{2} = \frac{2}{6} + \frac{3}{6} = \frac{5}{6}$
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}≤ 1, and x, y ≥ 0, which corresponds to the first quadrant of the disk.
Setting Up the Integral in Cartesian Coordinates. In Cartesian coordinates the double integral would be: $\int \int_{x^2+y^2≤1, x, y≥0} (1-x^2-y^2)dA$
However, due to the circular nature of the region R, converting the integral to polar coordinates simplifies the computation and seems to be a sensitive thing to do.
Convert to polar coordinates. In polar coordinates, the relationship between Cartesian and polar coordinates is: x = rcos(θ), y = rsin(θ), and x^{2} + y^{2} = r^{2.}.
Since rΔθ represents the arclength corresponding to an angle change of Δθ, the area element ΔA in Cartesian coordinates approximates to Δr⋅rΔθ in polar coordinates. Therefore, the area element dA in polar coordinates becomes rdrdθ(Figure ii). f(x, y) = 1-x^{2}-y^{2} = 1 -r^{2}.
The integral now becomes: $\int_{0}^{\frac{π}{2}} \int_{0}^{1} (1-r^2)rdrdθ $
Since we’re dealing with a quarter of the unit disk, the bounds for r range from 0 to 1, and the bounds for θ range from 0 to $\frac{π}{2}$.
Compute the Inner Integral (with respect to r): $\int_{0}^{1} (1-r^2)rdr = \frac{r^2}{2}-\frac{r^4}{4}\bigg|_{0}^{1} =\frac{1}{2}-\frac{1}{4} = \frac{1}{4}$
Compute the Outer Integral (with respect to θ):🚀$\int_{0}^{\frac{π}{2}} \frac{1}{4}dθ = \frac{1}{4}θ\bigg|_{0}^{\frac{π}{2}} = \frac{1}{4}·\frac{π}{2} = \frac{π}{8}$
Basically, to calculate the volume we divide the region into very thin rectangles boxes R_{ij}, each with area ΔA; 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: $V ≈ \sum_{i, j}f(x_i, y_j) \Delta A$. In the limit as the size of the rectangles approaches zero, this sum becomes the double integral: $\iint_R f(x, y) dA$.
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: $V = \iint_R f(x, y) dA$
In this problem, f(x, y) = 8x +6y and the region R is defined by 0 ≤ x ≤ 1, 0 ≤ y ≤ 2x^{2}. Therefore, the volume V can be computed using vertical slices as follows: $V = \int_{0}^{1}\int_{0}^{2x^2} (8x + 6ydy)dx$.
This means x ranges from 0 to 1. For each fixed x (denoted as x_{0}), y ranges from 0 to 2x^{2} (Refer to Figure i for a visual representation).
Compute the Inner Integral (with respect to y): $\int_{0}^{2x^2} 8x + 6ydy = 8xy+3y^2\bigg|_{0}^{2x^2} = 8x(2x^2)+3(2x^2)^2−(8x⋅0+3⋅0^2) = 16x^3+12x^4.$
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 = $\frac{32}{5}$.
Now, let’s verify the result using horizontal slices (integrating first with respect to x):
Double Integral Setup
The volume under the surface z = f(x, y) = 8x +6y over the region R is given by the double integral: $\iint_R f(x, y) dA$. Using horizontal slices, the volume V is given by $V = \int_{0}^{2}\int_{\sqrt{\frac{y}{2}}}^{1} (8x + 6ydx)dy$.
Here, y ranges from 0 to 2. For each fixed y (denoted as y_{0}), x ranges from $\sqrt{\frac{y}{2}}$ to 1 (Refer to Figure ii for a visual representation and aid).
Compute the Inner Integral (with respect to x): $\int_{\sqrt{\frac{y}{2}}}^{1} (8x + 6y)dx = 4x^2+6xy\bigg|_{\sqrt{\frac{y}{2}}}^{1} = 4 + 6y -(2y+\frac{6}{\sqrt{2}}y\sqrt{y}) = 4 + 4y -3\sqrt{2}y^{\frac{3}{2}}$
Compute the Outer Integral (with respect to y): $\int_{0}^{2} (4 + 4y -3\sqrt{2}y^{\frac{3}{2}})dy = 4y +2y^2-\frac{6\sqrt{2}}{5}y^{\frac{5}{2}}\bigg|_{0}^{2} = 8 +8 -\frac{6\sqrt{2}\sqrt{32}}{5} = 16 -\frac{6\sqrt{64}}{5} = 16 -\frac{6·8}{5}= 16 -\frac{48}{5} = \frac{80}{5}-\frac{48}{5} = \frac{32}{5}$. Both methods, using vertical and horizontal slices, yield the same result, $V = \frac{32}{5}$. This consistency confirms the accuracy of our calculation.
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 equation of the plane 3x + 2y −z = 0, we can express z as: z = 3x+2y. 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.
We need to find the intersections of the two parabolas y = x^{2} (i) and x = y^{2} (ii). Substituting y = x^{2} (ii) into x = y^{2}, we get x = (x^{2})^{2} = x^{4}. Solving x = x^{4}, x(x^{3} -1) = 0, so x = 0 or x = 1. The points of intersection are (0, 0) and (1, 1).
The region R is defined by 0 ≤ x ≤ 1, $x^2 ≤ y ≤ \sqrt{x}$. Therefore, the volume V can be computed using vertical slices as follows: $V = \int_{0}^{1}\int_{x^2}^{\sqrt{x}} (3x + 2y)dydx$.
This means x ranges from 0 to 1. For each fixed x (denoted as x_{0}), y ranges from x^{2} (lower curve) to $\sqrt{x}$ (upper curve) (Refer to Figure 3 for a visual representation).
Compute the Inner Integral (with respect to y): $\int_{x^2}^{\sqrt{x}} (3x + 2y)dy = 3xy + y^2\bigg|_{x^2}^{\sqrt{x}} = 3x\sqrt{x}+x-(3x^3+x^4)=3x^{\frac{3}{2}}+x-3x^3-x^4$
Compute the Outer Integral (with respect to y): V = $\int_{0}^{1}\int_{x^2}^{\sqrt{x}} (3x + 2y)dydx = \int_{0}^{1} (3x^{\frac{3}{2}}+x-3x^3-x^4)dx = \frac{6}{5}x^{\frac{5}{2}}+\frac{1}{2}x^2-\frac{3}{4}x^4-\frac{1}{5}x^5\bigg|_{0}^{1} = \frac{6}{5}+\frac{1}{2}-\frac{3}{4}-\frac{1}{5} = \frac{24}{20} + \frac{10}{20}-\frac{15}{20}-\frac{4}{20} = \frac{15}{20} = \frac{3}{4}$
The volume under the surface z = 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.
The volume under the surface z = f(x, y) = 3x +2y over the region R is given by the double integral: $\iint_R f(x, y) dA$. Using horizontal slices, the volume V is given by $V = \int_{0}^{1}\int_{y^2}^{\sqrt{y}} (3x +2ydx)dy = \frac{3}{4}$.
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 δ(x, y), where δ(x, y) represents the mass per unit area (mass density) at any point (x, y) within the region R. The mass Δm of this small piece can be approximated by multiplying the density δ(x,y) at that point by the area ΔA: Δm ≈ δ(x, y)·ΔA. In the limit as the area ΔA becomes infinitesimally small, the mass of that tiny piece of the lamina becomes dm = δ(x, y)·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: Mass = $\iint_R δ(x, y)·dA$where dA represents an infinitesimal area element.
Example. Finding the Mass of a triangular lamina with vertices (0, 0), (0, 3), and (3,0). The density function is given by ρ(x,y) = xykg/m^{2.}. We need to find the total mass of this lamina.
Conclusion. The total mass of the triangular lamina is: $\frac{27}{8}kg.$
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) 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 to find the average value.
Problem. Finding the average value of the function f(x,y) = 7xy^{2} over the region bounded by the line x = y and the curve x = √y
Define the region R. First, we need to understand the region R where the integration will take place. The line x = y intersects the curve x = $\sqrt{y}$ at points where $y = \sqrt{y} ↭ y^2 = y ↭ y^2-y=0 ⇒ y = 0,\text{or} y-1=0 ⇒ y = 1$. Therefore, the intersection points are (0, 0) and (1, 1) (Refer to Figure 1 for a visual representation and aid in understanding it).
Find the Area of the region R. To calculate the average value, we first need to find the area of the region R. The area is given by integrating the constant function 1 over the region: Area of R = $\int \int_{R} 1dA = \int_{0}^{1} (\int_{y}^{\sqrt{y}} 1dx)dy$
Evaluate the Inner integral = $\int_{y}^{\sqrt{y}} 1dx = \int_{y}^{\sqrt{y}} x \bigg|_{y}^{\sqrt{y}}dx $
= $\sqrt{y}-y$
Substitute into the outer integral to find the area: Area of R = $\int_{0}^{1} (\sqrt{y}-y)dy = \frac{2}{3}y^{\frac{3}{2}}-\frac{y^2}{2}\bigg|_{0}^{1} = \frac{2}{3}-\frac{1}{2} = \frac{4}{6}-\frac{3}{6} = \frac{1}{6}$. So, the area of the region is $\frac{1}{6}$.
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 1/6, f(x, y) = 7xy^{2}, the average value becomes: $F_{\text{average}} = 6·\int_{0}^{1} (\int_{y}^{\sqrt{y}} 7xy^2dx)dy$.
Calculate the inner integral (with respect to x): $\int_{y}^{\sqrt{y}} 7xy^2dx = \frac{7}{2}x^2y^2\bigg|_{y}^{\sqrt{y}} = \frac{7}{2}y^2(y-y^2) = \frac{7}{2}(y^3-y^4).$
Evaluate the outer integral (with respect to y): $F_{\text{average}} = \frac{1}{\text{Area of } R} \iint_R f(x,y)dA = 6·\int_{0}^{1} (\int_{y}^{\sqrt{y}} 7xy^2dx)dy = 6·\int_{0}^{1} \frac{7}{2}(y^3-y^4)dy = 21(\frac{y^4}{4}-\frac{y^5}{5}\bigg|_{0}^{1}) = 21(\frac{1}{4}-\frac{1}{5}) = 21(\frac{5}{20}-\frac{4}{20}) = \frac{21}{20}$
Thus, the average value of the function f(x,y)= $7xy^2$ over the region R is $\frac{21}{20}$.