It ain’t about how hard you can hit. Its about how hard you can get hit, and how much you can take, and keep moving forward, Rocky Balboa
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.
Double integrals are a powerful tool in calculus for finding areas, volumes, and other quantities over a two-dimensional region. Sometimes, evaluating these integrals in Cartesian coordinates (using x and y) can be complex or cumbersome. In such cases, converting to polar coordinates can simplify the problem.
Polar coordinates are particularly useful when dealing with regions that have circular symmetry, such as disks, annuli, or sectors Instead of describing a point in the plane using Cartesian coordinates (x,y), polar coordinates describe a point using:
Basically, in polar coordinates, we replace x and y with $r\cos \theta$ and $r \sin \theta$, respectively, and adjust the limits of integration accordingly.
Consider the integral $\int \int_{x^2+y^2≤1, x, y≥0} (1-x^2-y^2)dx$ over the region, x^{2}+y^{2} ≤ 1, x, y ≥ 0. Here, R is the region in the first quadrant bounded by the unit circle.
To convert from Cartesian coordinates (x, y) to polar coordinates (r, θ), use the following relationships: x = rcosθ, y = rsinθ.
We fix θ, r varies from 0 to 1. Then, θ ranges from 0 to ^{π}⁄_{2}. Recall that the sine function (sin(Δθ)) relates the angle (Δθ) to the ratio of the length of the side opposite the angle to the length of the hypotenuse in a right triangle, and when $Δ\theta$ is very close to 0, we can make an approximation: $\sin(Δ\theta) \approx Δ\theta$, hence $ΔA \approx Δr·r\sin(Δ\theta) \approx Δr·rΔ\theta$ ⇒[In the infinitesimal scale] dA = dr·rdθ, $\int \int fdA = \int (\int f·rdr)dθ$.
In our example, the integral becomes, $\iint_R (1-x^2-y^2) dxdy = \iint_R (1-r^2)rdrdθ$
For the given region R (first quadrant of the unit disk. It is defined by the inequalities x^{2} +y^{2} ≤ 1, x ≥0, and y ≥ 0), the limits for r range from 0 to 1, and for θ range from 0 to ^{π}⁄_{2}.
$\int \int_{x^2+y^2≤1, x, y≥0} (1-x^2-y^2)dx$ =[x = rcosθ, y = rsin(θ), f = 1 -x^{2}-y^{2} = 1 -r^{2}] $\int_{0}^{\frac{π}{2}} (\int_{0}^{1} (1-r^2)rdr)dθ =$ [🚀]
We first evaluate the integral with respect to r, keeping θ fixed:
$\int_{0}^{1} (1-r^2)rdr = \frac{r^2}{2}-\frac{r^4}{4}\bigg|_{0}^{1} = \frac{1}{2}-\frac{1}{4} = \frac{1}{4}$
Next, we integrate with respect to θ:
= [🚀] $\int_{0}^{\frac{π}{2}} \frac{1}{4}dθ = \frac{1}{4}θ\bigg|_{0}^{\frac{π}{2}} = \frac{1}{4}·\frac{π}{2} = \frac{π}{8}$.
The volume under the surface z = 1 −x^{2} -y^{2} over the region in the first quadrant bounded by the unit circle is: ^{π}⁄_{8}.
Here’s a step-by-step approach for converting and evaluating integrals in polar coordinates:
The surface -x^{2} -y^{2} + z^{2} = 1 can be rewritten as z^{2} = 1 + x^{2} +y^{2} ⇒ $z = \sqrt{1+x^2+y^2}$. We want to find the volume of the solid enclosed between the surface defined by the equation $z = \sqrt{1+x^2+y^2}$ (it represents a paraboloid that opens upwards, centered on the z-axis) and the plane z = 2.
To find the region where the two surfaces intersect, set z = 2 in the first equation: 3 = x^{2} + y^{2}. So, the region of integration in the xy-plane is the disk with a radius of $\sqrt{3}$ centered at the origin.
Using polar coordinates where x = rcos(θ) and y = rsin(θ), the equation x^{2}+y^{2} = 3, becomes 3 = r^{2} ↭ r = $±\sqrt{3}$ ↭ so r ranges from 0 to $\sqrt{3}$.
In polar coordinates, the equation $z = sqrt{1+x^2+y^2}$ becomes $z = \sqrt{1+r^2}$. The volume we seek is the region between $z = \sqrt{1+r^2}$ amd z = 2. Therefore, the volume V is given by the double integral: $V = \int_{0}^{2π} \int_{0}^{\sqrt{3}} (2-\sqrt{1+r^2})rdrdθ$ where R is the disk, θ ranges from 0 to 2π and 0 ≤ r ≤ $\sqrt{3}$.
We first integrate with respect to r:
$\int_{0}^{\sqrt{3}} (2-\sqrt{1+r^2})rdr = \int_{0}^{\sqrt{3}} (2r-r\sqrt{1+r^2})dr=$[Substitution u = 1 + r^{2}, du = 2rdr] $\int_{0}^{\sqrt{3}} 2rdr -\int_{0}^{\sqrt{3}}r\sqrt{u}\frac{du}{2r} = \int_{0}^{\sqrt{3}} 2rdr -\int_{0}^{\sqrt{3}}u^{\frac{1}{2}}\frac{du}{2} = r^2-\frac{1}{3}u^{\frac{3}{2}} = r^2-\frac{1}{3}(1+r^2) ^{\frac{3}{2}}\bigg|_{0}^{\sqrt{3}} = 3-\frac{1}{3}4^{\frac{3}{2}}+\frac{1}{3} = 3-\frac{1}{3}\sqrt{4}^{3}+\frac{1}{3} = 3-\frac{8}{3}+\frac{1}{3} = 3-\frac{7}{3} = \frac{9}{3}-\frac{7}{3} = \frac{2}{3}$
V = $\int_{0}^{2π} \frac{2}{3}dθ = \frac{2}{3}θ\bigg|_{0}^{2π} = \frac{4π}{3}.$
The region R is defined by the inequalities 1 ≤ x^{2} + y^{2} ≤ 4 and x ≤ 0.
Using polar coordinates, x = rcos(θ), y=rsin(θ), and dA=rdrdθ, The region in polar coordinates is defined by R = {(r, θ): 1 ≤ r ≤ 2, $\frac{π}{2} ≤ θ ≤ \frac{3π}{2}$} (Refer to Figure A for a visual representation and aid in understanding it).
The bounds for r are from 1 to 2, and the bounds for θ are from $\frac{π}{2}$ to $\frac{3π}{2}$, covering the left half of the plane. Hence, The double integral in polar coordinates becomes:
$\iint_R (x + y) dA = \int_{\frac{3π}{2}}^{\frac{π}{2}} (\int_{1}^{2} (rcos(θ)+rsin(θ))rdr)dθ$
$\int_{1}^{2} (rcos(θ)+rsin(θ))rdr = (cos(θ)+sin(θ))\int_{1}^{2} r^2dr = (cos(θ)+sin(θ))·\frac{r^3}{3}\bigg|_{1}^{2} = (cos(θ)+sin(θ))·(\frac{8}{3}-\frac{1}{3}) = (cos(θ)+sin(θ))·\frac{7}{3}$
$\iint_R (x + y) dA = \frac{7}{3}\int_{\frac{3π}{2}}^{\frac{π}{2}} (cos(θ)+sin(θ))·dθ = \frac{7}{3} (sin(θ)-cos(θ))\bigg|_{\frac{π}{2}}^{\frac{3π}{2}} = \frac{7}{3} (sin(\frac{3π}{2})-cos(\frac{3π}{2})-(sin(\frac{π}{2})-cos(\frac{π}{2}))) = \frac{7}{3}(-1-0-(1-0)) = \frac{-14}{3}$
This negative result simply reflects that the function x+y is predominantly negative over the region R.
Note: The paraboloid is opening downwards because -3x^{2}-3y^{2}, that is, both variables are negative, and then it has shifted up 10 units in the z direction, and it is cut up by the plane z = 4.
First, we find the intersection of the paraboloid z = 10 -3x^{2}-3y^{2} and the plane z = 4: 10 -3x^{2}-3y^{2} = 4 ↭ 3x^{2} + 3y^{2} = 6 ↭ x^{2} + y^{2} = 2, so this describes a circle with radius r = $\sqrt{2}$ centered at the origin in the xy-plane (Figure B).
In polar coordinates, x = rcos(θ) and y = rsin(θ). The region of integration becomes: $0 ≤ r ≤\sqrt{2}, 0 ≤ θ ≤ 2π.$
The height of the solid above the xy-plane (z = 4) at any point (r, θ) is given by the difference between the paraboloid and the plane: 10 -3x^{2}-3y^{2} -4 = 6 -3(x^{2} + y^{2}) = 6 -3r^{2}.
$V = \int_{0}^{2π}(\int_{0}^{\sqrt{2}} (6-3r^2)rdr)dθ$
$\int_{0}^{\sqrt{2}} (6-3r^2)rdr = \int_{0}^{\sqrt{2}} (6r-3r^3)dr = 3r^2-\frac{3}{4}r^4\bigg|_{0}^{\sqrt{2}} = 3·2-\frac{3}{4}·4 = 6 - 3 = 3.$
$V = \int_{0}^{2π}(\int_{0}^{\sqrt{2}} (6-3r^2)rdr)dθ = \int_{0}^{2π} 3dθ = 3θ\bigg|_{0}^{2π} = 6π.$
Conclusion: The volume of the solid bounded by the paraboloid z = 10 -3x^{2}-3y^{2} and the plane z = 4 is 6π.
The curves intersect where: 3x^{2} + 3y^{2} = 4 -x^{2}-y^{2} ↭[Combine like terms] 4x^{2} + 4y^{2} = 4 ↭ x^{2} + y^{2} = 1. This represents a circle of radius r = 1 centered at the origin in the xy-plane. The z-value in this region is z = 3x^{2}+3y^{2} = 3(x^{2} + y^{2}) = 3·1 = 3 (Figure C).
In polar coordinates, x = rcos(θ), y = rsin(θ), and x^{2} + y^{2} = 1. The region of integration in polar coordinates becomes: $0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π.$
The height of the volume at any point (r,θ) is given by the difference between the upper surface z = 4 -x^{2} -y^{2} and the lower surface z = 3x^{2} +3y^{2}.
Height = $4 -x^2-y^2 -(3x^2+3y^2) = 4 -4x^2-4y^2 = 4 -4(x^2+y^2) = 4- 4r^2 = 4(1-r^2)$
The volume V of the solid can be found by integrating the height over the region:
$V = \int_{0}^{2π}(\int_{0}^{1} 4(1-r^2)rdr)dθ$
$\int_{0}^{1} 4(1-r^2)rdr = 4\int_{0}^{1} (r-r^3)rdr = 4(\frac{r^2}{2}-\frac{r^4}{4})\bigg|_{0}^{1} = 4·(\frac{1}{2}-\frac{1}{4}) = 2 - 1 = 1$
$V = \int_{0}^{2π}(\int_{0}^{1} 4(1-r^2)rdr)dθ = \int_{0}^{2π} dθ = 2π.$
The three surfaces intersect as follows:
The region of intersection is where these three shapes overlap. Since we only want the portion of the sphere that actually lies inside the cylinder given by x^{2}+y^{2}=5, this is also the region that we are looking for. The region D is the disk x^{2} + y^{2} ≤ 5 in the xy-plane.
Futhermore, the region that we want the volume for is really a cylinder with a cap that comes from the sphere: x^{2} + y^{2} + z^{2} = 9, x^{2} + y^{2} = 5 ⇒ 5 + z^{2} = 9 ⇒ z^{2} = 4 ⇒ z = 2 (Figure D).
In polar coordinates, x = rcos(θ) and y = rsin(θ). The region of integration becomes: $0 ≤ r ≤ \sqrt{5}$, 0 ≤ θ ≤ 2π.$
The height z is bounded below by the plane z = 0 and above by the sphere. From the sphere’s equation: $z = \sqrt{9-x^2-y^2}= \sqrt{9-r^2}$
The volume V of the solid can be found by integrating the height over the region D: $V =\int \int_{D} \sqrt{9-x^2-y^2}dA = \int_{0}^{2π} (\int_{0}^{\sqrt{5}} \sqrt{9-r^2}rdr)dθ$
$\int_{0}^{\sqrt{5}} \sqrt{9-r^2}rdr = \int_{0}^{\sqrt{5}}r(9-r^2)^{\frac{1}{2}}dr =[u = 9-r^2⇒ du = -2rdr, r =0 ⇒ u = 9, r = \sqrt{5} ⇒ u = 9 -5 = 4] \int_{9}^{4} -\frac{1}{2}\sqrt{u}du = -\int_{9}^{4} \frac{1}{2}\sqrt{u}du = -\frac{1}{2}\frac{2}{3}u^{\frac{3}{2}}\bigg|_{9}^{4}=-\frac{1}{3}(4^{\frac{3}{2}}-9^{\frac{3}{2}}) = -\frac{1}{3}((2^2)^{\frac{3}{2}}-(3^2)^{\frac{3}{2}}) = -\frac{1}{3}(2^3-3^3) = -\frac{1}{3}(8-27)=\frac{19}{3}$
$V =\int \int_{D} \sqrt{9-x^2-y^2}dA= \int_{0}^{2π} (\int_{0}^{\sqrt{5}} \sqrt{9-r^2}rdr)dθ = \int_{0}^{2π} \frac{19}{3}dθ = \frac{19}{3}θ \bigg|_{0}^{2π} = \frac{38π}{3}$
The curve r = cos(4θ) is a polar equation that represents a type of graph known as a polar rose. These are curves that have petal-like structures. The number of petals is determined by the coefficient of θ inside the cosine function. In this case, r = cos(4θ) results in a rose with 2 x 4 = 8 petals because the coefficient 4 is even (Refer to Figure 1 for a visual representation and aid in understanding it).
The petals are symmetrically distributed around the origin. Since the cosine function is periodic with period 2π, it repeats its values. For cos(4θ), the period is $\frac{π}{2}$ ($4·\frac{π}{2}=2π.$)
As with rectangular coordinates, we can also use polar coordinates to find areas of certain regions using a double integral. Generally, the area formula will look like $Area = \int_{α}^{β} \int_{h_1(θ)}^{h_2(θ)}1rdrdθ$
To find the area bounded by the entire curve, it’s crucial to understand that the curve is symmetric. This means the area of all petals is the same. Therefore, instead of calculating the area of the entire curve directly, we can calculate the area of one petal and then multiply by the number of petals (which is 8 in this case) to get the total area, and one of the petals is formed when θ ranges from $-\frac{π}{8}$ to $\frac{π}{8}$.
Therefore, the area is: $Area = \int_{α}^{β} \int_{h_1(θ)}^{h_2(θ)}1rdrdθ = 8\int_{-\frac{π}{8}}^{-\frac{π}{8}} \int_{0}^{cos(4θ)}1rdrdθ$
Evaluate the inner interval: $\int_{0}^{cos(4θ)}1rdr = \frac{r^2}{2}\bigg|_{0}^{cos(4θ)} = \frac{cos^2(4θ)}{2}$
Evaluate the outer integral: $Area = \int_{α}^{β} \int_{h_1(θ)}^{h_2(θ)}1rdrdθ = 8\int_{-\frac{π}{8}}^{-\frac{π}{8}} \frac{cos^2(4θ)}{2}dθ$
Use the trigonometric identity: $cos^2(x)=\frac{1+cos(2x)}{2}⇒cos^2(4θ) = \frac{1+cos(8θ)}{4}$
Total Area = $8\int_{-\frac{π}{8}}^{-\frac{π}{8}} \frac{1+cos(8θ)}{4}dθ = 8[\frac{1}{4}θ +\frac{1}{32}sin(8θ)]\bigg|_{-\frac{π}{8}}^{\frac{π}{8}} = 8[\frac{1}{4}\frac{π}{8}+\frac{1}{32}sin(π)-(-\frac{1}{4}\frac{π}{8}+\frac{1}{32}sin(-π))] = 8[\frac{1}{4}\frac{π}{8}+\frac{1}{4}\frac{π}{8}] = 8·\frac{1}{2}·\frac{π}{8} = \frac{π}{2}units^2$.
The equation r = 1−sin(θ) describes a cardioid (Refer to Figure 2 for a visual representation and aid in understanding it), which is a heart-shaped curve. This cardioid is symmetric about the vertical axis (the line θ = π/2).
Since we are working in polar coordinates, the volume element dV at a point (r, θ) in this solid is f(r, θ) × Area_element. In polar coordinates, the area element is rdrdθ. Thus, the volume is given by the double integral:
$V = \int_{R} f(r, θ)·rdrdθ = \int_{0}^{2π} \int_{0}^{1-sin(θ)} r·rdrdθ$
Inner Integral: $\int_{0}^{1-sin(θ)} r·rdr = \frac{r^3}{3}\bigg|_{0}^{1-sin(θ)} = \frac{(1-sin(θ))^3}{3}$
Outer Integral: $V = \int_{0}^{2π} \int_{0}^{1-sin(θ)} r·rdrdθ = \int_{0}^{2π} \frac{(1-sin(θ))^3}{3}dθ = \frac{1}{3} \int_{0}^{2π}(1 -3sin(θ)+3sin^2(θ)-sin^3(θ))dθ$
We can split this into four separate integrals:
The integrals of functions like sine and cosine over a full period, as well as odd powers of sine and cosine, are indeed zero. This happens because these functions are odd functions (they have symmetry about the origin).
Now, substitute these into the expression for V: V = $\frac{1}{3}[2π-0+3π-0] = \frac{1}{3}5π = \frac{5π}{3}$.