MA 391 Composition and Communication

Consider the function: \[ f(x) = x^2 \] The process of integration for this function starts in the form of adding the areas of many rectangles via a Riemann sum. The heights of these rectangles are determined by the value of the function at the relevant x value, and the width is decided by the amount of partitions over the bounds of integration. The process of integration sums up the area of infinitely thin rectangles over the relevant bounds. In this case there is a function of one variable over a domain in \( \mathbb{R} \). Now consider the function: \[ f(x,y) = x^2 + y^2 \] Now we have a function of 2 variables. An integral for this function can now be integrated over \(\mathbb{R}^2\), which can be viewed as an area rather than a line in \( \mathbb{R}\). The other major difference for integration over a two dimensional domain is having two variables to integrate. This means that the integral is taking with respect to an area or \( dA \). For this integral, the Riema...

With the continuous learning of the calculus course, it is no longer limited to a single plane. By performing quadratic integration, we have obtained a calculation method that can calculate the volume. From plane to three-dimensional, this is a new breakthrough. Simply put, it is to perform two integral operations, but it brings us new knowledge and challenges. I really like the kind of "things" that burst out from an angle in front of you (similar to some skills in some games). So I chose two very simple equations to make them have a certain angle: y=2x and y=0.5x. Since the two equations will extend indefinitely, we also give him an interval. x=2 and y=2. Then the bottom of the entire graph has been determined, as shown in the figure: Then we will divide this area a little bit more carefully, because it is a 2x2 square, then we will divide it into 100 small squares, so that the length of each side is 0.2. We use the point in the upper rig...

A function with two variables can be pictured by making the value of the function a third axis. If the function consists of x and y, the value is z. This turns an area over a plane into a 3 dimensional object. Two integrals are needed to integrate two variable functions defined over a region. The first integral is with respect to one of the variables and the second is with respect to the other. The bounds are the area that is being integrated. Sometimes, that requires the bounds of the inner integral to be a function of the outer variable. Finding the antiderivitive of the function is the same as any other function, only you treat the variable not being integrated as a constant. Once you find the antiderivitive, replace the variable that was integrated with the bounds like you normally would and integrate that new function with respect to the other variable. What you are left with is the value of the function integrated over the area.   The function I chose to integrat...

Introduction You may now be familiar with known cross-section which takes the volume between curves. With this method we choose a certain shape to make the curve 3-dimensional to find the volume. But now let's say we want to find the volume of a function that is already 3-dimensional. Imagine we have a candle, and we want to find the volume of it. We would first need the function that gives us the "height" of the candle and we know this would be in terms of both x and y since the candle is 3-dimensional. We would then need the length and width of the base of the candle. Putting this together the height would be the function we integrate, and the width and length would be the bounds of each integral. We know the basics so let's look at an example. We custom make candles and had someone come in saying they wanted a slanted candle with the overall shape of a petal and we came up the function \[f(x,y)=xy+x\] and for the petal we used the functions: ...

We are back again, once again, showing off a method of Integration from Calculus courses. Stepping beyond Calculus two, we will be going into one of the main ideas from a multiple variable calculus course, double integration. When doing a double integral, with two variables, we integrate over a region of $R^2$. The approximation is similar to what we would do for one variable, taking a double sum of heights times the area. In general, where we have a constant base area of $\Delta A$, we have the formula for approximation: $V \approx \sum_{i=1}^{n} \sum_{j=1}^{m} \ f(x_i^*, y_j^*) \ \Delta A $ Following from the process from calculus one, by taking the limit of the double sum as i and j approach infinity, we get the exact volume. The domain I decided to use for this is the curve sketched out by: $f(x,y) = x^2 + y^2$. This curve is a paraboloid, but that by itself is not very interesting. Instead of dealing with just this curve, we will be cutting out a region of a triangle,...

Introduction It’s irrefutable that calculus is an incredible tool-especially when it comes to calculating the area underneath different functions. At this point you may have mastered integration in two-dimensions and be looking for a new challenge. If so, look no further! Integration can also be done in 3-dimensions, adding another layer of useful applications to calculus’ repertoire. The Problem Imagine you are a marine biologist who works for an aquarium. The aquarium has created a design for its shark habitat and brings the plans to you for approval. You want to make sure the sharks won’t be overcrowded in the tank, and want to calculate how many cubic feet of water there will be for each shark. To do that, you need to know the volume of the tank. The depth of the tank at any given \( x \) and \( y \) coordinate is by the following equation: \[ f(x,y)= \frac{1}{4}\ \cos(6y) + \frac{1}{4}\ \sin(2x^2) - \frac{1}{2}\ x+ \frac{1}{2}\ y +2 \] Th...

Many of us have seen performers who spin plates on the ends of sticks. Like the graph showing below. The performers try to keep several of them spinning without allowing any of them to drop. If we look at a single plate (without spinning it), there is a sweet spot on the plate where it balances perfectly on the stick. If we put the stick anywhere other than that sweet spot, the plate does not balance and it falls to the ground. (That is why performers spin the plates; the spin helps keep the plates from falling even if the stick is not exactly in the right place.) Mathmatically, the sweet spot is called center of mass of the plate . Lucky, we could use the calculus knowlege to find it. Ler's satrt with finding center of mass in one-dementional case. Thinking about that there are two kids seat on the two side of a seesaw in the playground. If their weight is different, and they all sit at the end of each side of seesaw, the...