MA 391 Composition and Communication

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  If you want to find the area under a region, you can draw rectangles and add the area of the rectangles together to find an aproximation. If you want to find the exact area, you use an integral bounded by the region. But what about when we introduce a 3rd plane? How do we aproximate the volume under a region when we consider all three planes? The concept is very similar. To aproximate the volume, we use rectangles but add the element of height and sum the rectangles volumes. For the actual volume, we use a double integral (using Fubini's Theorem) bounded by the region. Consider the region contained in the intersection of \[y=x^\frac{1}{2}\] \(and\) \[y=x^3.\] We are going to aproximate and find the actual volume of this region under the function \[f(x,y)=xy.\] We're going to find the aproximated area using squares, so let's zoom in a little bit until we have a 5x5 area for a total of 25 squares to work with. Do

Double Integrals over Rectangles over a Non-Rectangular Region Previously we have discussed methods of approximating the volume of a solid with regard to solids of known cross section and solids that are symmetric about an axis, via shells and washers. We are now discussing ways to approximate the volumes of more irregular solids—i.e. those without known cross section and or are not symmetric about an axis. We will first discuss the solid we will be approximating, followed by the method of approximation and calculating an actual volume by double integrals, and conclude with actual calculations of volume for the solid in question. The solid that we will be considering in bounded on the \(xy\)-plane by the function \(y = \left| x \right| \) and \( y = 5 \); which results in the region shown below: Vertically, or with respect to the \(z\)-axis, the solid is bounded by \( f(x,y) = 2y - x^2 \) which is shown in the following 3D image:

  Fubini's Theorem asserts that for a function \(f(x,y)\) the volume under the surface which is bounded by a region \([a,b] \times [c,d]\) can be found either by first integrating with respect to y, or with respect to x. In either case, the other variable is treated like a constant.  This means that if we make the bounds of y between c and d, and the bounds of x between a and b, then the two following integrals are equivalent! \[\int_{c}^{d}\int_{a}^{b}{f(x,y)dxdy}\] \[\int_{a}^{b}\int_{c}^{d}{f(x,y)dydx} \] This really only works easily for continuous functions. The fact that we can integrate x and y (or any collection of independent variables) in any order is surprisingly fundamental to quantum chemistry. For the three-dimensional wavefunction for electronic orbitals for a molecule, to determine the electronic surface, each electron is given a volume integral over the whole space, for each spin. If we couldn't integrate freely in any direction, matter as we know it w

How to think about integrals over regions in the plane When thinking about integrals over regions in the plane, the first thing I think of is a roller coaster. The xy plane is the base of the roller coaster, and as you walk around under it, the roller coaster changes height. This is how functions of the form f (x, y) generally behave; you can pick any "base" or domain for the roller coaster and find the volume of the resulting three-dimensional shape using double integration with respect to x and y. You can also approximate the volume by taking a rectangular region R around your domain and splitting it into a number of boxes, as discussed below. (Image from https://www.google.com/url?sa=i&url=https%3A%2F%2Fwww.architecturaldigest.com%2Fgallery%2Froller-coaster-designs&psig=AOvVaw2XBThbDuZkz6JaLWpHsX9p&ust=1632786586027000&source=images&cd=vfe&ved=0CAsQjRxqFwoTCICUnZ_qnfMCFQAAAAAdAAAAABAD ) How to compute the vol

     Integrating functions of two variables can be difficult to understand at first because we are integrating in the x and y planes. By doing this, it allows for us to calculate the volume under the function in a 3D space, compared to the area of a 2D space, which would be done through a single integral. To solve a double integral, first integrate the function in one region along the bounds (while pretending the other variable is a constant), then integrate on the other region using the bounds for that region. For my example, let’s use the function bounded on    The domain (the bounds) tells us to integrate this function from -3 to 3 in the x direction and -1 to 3 in y direction. This allows us to integrate in the shape of a rectangle focusing on 1x1 blocks for each point. The function F(x,y) will calculate the height for each part of the function. This will form an object that represents the volume of F(x,y). Approximate Volume: To calculate the approximate volume

In the previous blog, we learned how to use single integral to calculate the volume and center of mass. For single integrals, the region over which we integrate is always an interval. ( i.e. a one-dimensional space) In this blog we are intereted in integrating a function of two variables \(f(x,y)\). With functions of two variable we integrated over a region of \(\mathbb{R}^2\) ( i.e. a two-dimensional space). We will start out by assuming that the region in \(\mathbb{R}^2\) is a rectangle which we will denote as \(R=[a,b]\times [c,d]\). This means that the ranges for \(x\) and \(y\) are \(a\leq x\leq b\), and \(c \leq y\leq d\). Also let's assume \(f(x,y) \geq 0\). The graph of \(f\) is a aurface with equation \(z=f(x,y)\) Our goal is to find the volume between surface and \(R\). First, we want to approximate it. We divide up \(R\) into a series of smaller rect

About integration for a function defined on a region If a solid can be modeled by an integrable function of two variables over a general region in the plane, we can use integration to find the volume of the solid. Let’s consider a region \(D\) on the \(x-y\) plane, which is bounded by two vertical lines and two functions of \(x\). The two vertical lines are \(x=a\) and \(x=b\) respectively, and the two functions of \(x\) is \(y=u(x)\) and \(y=v(x)\). The integrable function is \(f(x,y)\). The integration to find the volume of the solid can be calculated as following: \[V=\int^b_a{\int^{v(x)}_{u(x)}{f(x,y)dy}dx}\] The double integral \(\int^{b}_{a}{\int^{v(x)}_{u(x)}{f(x,y)dy}dx}\) of positive \(f(x,y)\) can be regarded as the volume under the surface \(z=f(x,y)\) over the region \(D\). We can use the Riemann sum to approximate the integral: \[\sum_{i,j}{f(x_i,y_j)}\Delta x \Delta y \] Each term in the Riemann sum is the volume of a thin box with base

Introduction From the previous weeks, the topics and concepts that we have learned have built on top of one another. Using the information we have learned thus far, we are now ready to dive deeper into the calculus world. Integration has been straight forward by integrating functions just on the two-dimensional world. What if we added the third-dimension?! Can integration still be done? The answer is yes, and using integration in the third-dimension is slightly more difficult, but can still be understood quite easily. Today, we will be using double integrals to calculate the volume of a region between two surfaces. Double Integrals? As intimidating as it may sound, double integrals are very useful and easy to learn. To compute a double integral, we just have to evaluate the double integral over a region that is either bounded by two horizontal and two functions of x or two vertical lines and two functions of y. Al

When starting to think about 2D integration over a specific region in the plane, I thought about indoor surfing. I knew I wanted to choose a function that increased exponentially because I thought it would give us something better to look at, so I imagined waves in the ocean. Since waves in the ocean are not restricted to a specific region and they continue on forever, I had to come up with a way to modify my example to show the restraints. This was my thought process when thinking of indoor surfing. Although I've never been indoor surfing myself, I like the idea of it. You are surfing in a controlled wave that is in a controlled region or pool area. The photo below shows an example of an indoor surfing arena. From this picture you can see that most indoor surfing places put the wave in more of a square or rectangular region, but I wanted to spruce up my indoor surfing region and make it a quarter circle. When thinking about 2D integration over a regio