We can do something similar along the \(y\)-axis to find our \(\bar{y}\) value. Did the Golden Gate Bridge 'flatten' under the weight of 300,000 people in 1987? The centroid of an area can be thought of as the geometric center of that area. Example: We divide $y$-moment by the area to get $x$-coordinate and divide the $x$-moment by the area to get $y$-coordinate. If the shape has a line of symmetry, that means each point on one side of the line must have an equivalent point on the other side of the line. This is exactly what beginners need. Using the first moment integral and the equations shown above, we can theoretically find the centroid of any shape as long as we can write out equations to describe the height and width at any \(x\) or \(y\) value respectively. So all I do is add f(x) with f(y)? ???\overline{x}=\frac{(6)^2}{10}-\frac{(1)^2}{10}??? The area, $A$, of the region can be found by: Here, $a$ and $b$ shows the limits of the region with respect to $x-axis$. Step-by-step math courses covering Pre-Algebra through Calculus 3. math, learn online, online course, online math, calculus iii, calculus 3, calc iii, calc 3, multivariable calculus, multivariable calc, multivariate calculus, multivariate calc, multiple integrals, double integrals, iterated integrals, polar coordinates, converting iterated integrals, converting double integrals, math, learn online, online course, online math, linear algebra, systems of unknowns, simultaneous equations, system of simultaneous equations, solving linear systems, linear systems, system of three equations, three simultaneous equations. The coordinates of the centroid are (\(\bar X\), \(\bar Y\))= (52/45, 20/63). Solve it with our Calculus problem solver and calculator. Well explained. If your isosceles triangle has legs of length l and height h, then the centroid is described as: (if you don't know the leg length l or the height h, you can find them with our isosceles triangle calculator). Taking the constant out from integration, \[ M_x = \dfrac{1}{2} \int_{0}^{1} x^6 x^{2/3} \,dx \], \[ M_x = \dfrac{1}{2} \Big{[} \int_{0}^{1} x^6 \,dx \int_{0}^{1} x^{2/3} \,dx \], \[ M_x = \dfrac{1}{2} \Big{[} \dfrac{x^7}{7} \dfrac{3x^{5/3}}{5} \Big{]}_{0}^{1} \], \[ M_x = \dfrac{1}{2} \bigg{[} \Big{[} \dfrac{1^7}{7} \dfrac{3(1)^{5/3}}{5} \Big{]} \Big{[} \dfrac{0^7}{7} \dfrac{3(0)^{5/3}}{5} \Big{]} \bigg{]} \], \[ M_y = \int_{a}^{b} x \{ f(x) g(x) \} \,dx \], \[ M_y = \int_{0}^{1} x \{ x^3 x^{1/3} \} \,dx \], \[ M_y = \int_{0}^{1} x^4 x^{5/3} \,dx \], \[ M_y = \int_{0}^{1} x^4 \,dx \int_{0}^{1} x^{5/3} \} \,dx \], \[ M_y = \Big{[} \dfrac{x^5}{5} \dfrac{3x^{8/3}}{8} \Big{]}_{0}^{1} \], \[ M_y = \Big{[}\Big{[} \dfrac{1^5}{5} \dfrac{3(1)^{8/3}}{8} \Big{]} \Big{[} \Big{[} \dfrac{0^5}{5} \dfrac{3(0)^{8/3}}{8} \Big{]} \Big{]} \]. Using the area, $A$, the coordinates can be found as follows: \[ \overline{x} = \dfrac{1}{A} \int_{a}^{b} x \{ f(x) -g(x) \} \,dx \]. Please enable JavaScript. Try the given examples, or type in your own Centroid Of A Triangle When we find the centroid of a two-dimensional shape, we will be looking for both an \(x\) and a \(y\) coordinate, represented as \(\bar{x}\) and \(\bar{y}\) respectively. So, lets suppose that the plate is the region bounded by the two curves \(f\left( x \right)\) and \(g\left( x \right)\) on the interval \(\left[ {a,b} \right]\). To do this sum of an infinite number of very small things, we will use integration. Stack Exchange network consists of 181 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Centroids / Centers of Mass - Part 1 of 2 The location of the centroid is often denoted with a C with the coordinates being (x, y), denoting that they are the average x and y coordinate for the area. Accessibility StatementFor more information contact us atinfo@libretexts.org. Where is the greatest integer function f(x)= x not differentiable? I feel like I'm missing something, like I have to account for an offset perhaps. ?\overline{y}=\frac{1}{20}\int^b_a\frac12(4-0)^2\ dx??? Can you still use Commanders Strike if the only attack available to forego is an attack against an ally? Find centroid of region bonded by the two curves, y = x2 and y = 8 - x2. To find $x_c$, we need to evaluate $\int_R x dy dx$. Checking Irreducibility to a Polynomial with Non-constant Degree over Integer. If an area was represented as a thin, uniform plate, then the centroid would be the same as the center of mass for this thin plate. ?? Writing all of this out, we have the equations below. We can find the centroid values by directly substituting the values in following formulae. To find the centroid of a triangle ABC, you need to find the average of vertex coordinates. Try the free Mathway calculator and The region bounded by y = x, x + y = 2, and y = 0 is shown below: To find the area bounded by the region we could integrate w.r.t y as shown below, = \( \left [ 2y - \dfrac{1}{2}y^{2} - \dfrac{3}{4}y^{4/3} \right]_{0}^{1} \), \(\bar Y\)= 1/(3/4) \( \int_{0}^{1}y((2-y)- y^{1/3})dy \), = 4/3\( \int_{0}^{1}(2y - y^{2} - y^{4/3)})dy \), = 4/3\( [y^{2} - \dfrac{1}{3}y^{3}-\dfrac{3}{7}y^{7/3}]_{0}^{1} \), The x coordinate of the centroid is obtained as, \(\bar X\)= (4/3)(1/2)\( \int_{0}^{1}((2-y)^{2} - (y^{1/3})^{2}))dy \), = (2/3)\( [4y - 2y^{2} + \dfrac{1}{3}y^{3} - \dfrac{3}{5}y^{5/3}]_{0}^{1} \), = (2/3)[4 - 2 + 1/3 - 3/5 - (0 - 0 + 0 - 0)], Hence the coordinates of the centroid are (\(\bar X\), \(\bar Y\)) = (52/45, 20/63). \dfrac{(x-2)^3}{6} \right \vert_{1}^{2}\\ Note that the density, \(\rho \), of the plate cancels out and so isnt really needed. This means that the average value (AKA the centroid) must lie along any axis of symmetry. Looking for some Calculus help? Show Video Lesson & = \int_{x=0}^{x=1} \left. The variable \(dA\) is the rate of change in area as we move in a particular direction. We will find the centroid of the region by finding its area and its moments. is ???[1,6]???. First well find the area of the region using, We can use the ???x?? \left( x^2 - \dfrac{x^3}{3}\right) \right \vert_1^2 = \dfrac15 + \left( 2^2 - \dfrac{2^3}3\right) - \left( 1^2 - \dfrac{1^3}3\right) = \dfrac15 + \dfrac43 - \dfrac23 = \dfrac{13}{15} To embed this widget in a post on your WordPress blog, copy and paste the shortcode below into the HTML source: To add a widget to a MediaWiki site, the wiki must have the. The moments measure the tendency of the region to rotate about the \(x\) and \(y\)-axis respectively. Remember the centroid is like the center of gravity for an area. example. \end{align}, To find $y_c$, we need to evaluate $\int_R x dy dx$. & = \left. Find the centroid of the region bounded by the curves ???x=1?? where $R$ is the blue colored region in the figure above. Hence, we get that The two curves intersect at \(x = 0\) and \(x = 1\) and here is a sketch of the region with the center of mass marked with a box. On this page we will only discuss the first method, as the method of composite parts is discussed in a later section. Once you've done that, refresh this page to start using Wolfram|Alpha. point (x,y) is = 2x2, which is twice the square of the distance from The midpoint is a term tied to a line segment. We have a a series of free calculus videos that will explain the Send feedback | Visit Wolfram|Alpha Copyright 2005, 2022 - OnlineMathLearning.com. Assume the density of the plate at the The region we are talking about is the region under the curve $y = 6x^2 + 7x$ between the points $x = 0$ and $x = 7$. If you plot the functions you can get a better feel for what the answer should be. In the following section, we show you the centroid formula. Centroid - y f (x) = g (x) = A = B = Submit Added Feb 28, 2013 by htmlvb in Mathematics Computes the center of mass or the centroid of an area bound by two curves from a to b. As the trapezoid is, of course, the quadrilateral, we type 4 into the N box. \dfrac{y^2}{2} \right \vert_{0}^{2-x} dx\\ Chegg Products & Services. Centroids / Centers of Mass - Part 2 of 2 rev2023.4.21.43403. Now the moments, again without density, are, \[\begin{array}{*{20}{c}}\begin{aligned}{M_x} & = \int_{{\,0}}^{{\,1}}{{\frac{1}{2}\left( {x - {x^6}} \right)\,dx}}\\ & = \left. ?-values as the boundaries of the interval, so ???[a,b]??? The location of centroids for a variety of common shapes can simply be looked up in tables, such as this table for 2D centroids and this table for 3D centroids. Related Pages Untitled Graph. The centroid of the region is at the point ???\left(\frac{7}{2},2\right)???. Please submit your feedback or enquiries via our Feedback page. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. $( \overline{x} , \overline{y} )$ are the coordinates of the centroid of given region shown in Figure 1. Short story about swapping bodies as a job; the person who hires the main character misuses his body. Centroid of the Region bounded by the functions: $y = x, x = \frac{64}{y^2}$, and $y = 8$. Here, you can find the centroid position by knowing just the vertices. The region you are interested is the blue shaded region shown in the figure below. That's because that formula uses the shape area, and a line segment doesn't have one). Let us compute the denominator in both cases i.e. However, if you're searching for the centroid of a polygon like a rectangle, a trapezoid, a rhombus, a parallelogram, an irregular quadrilateral shape, or another polygon- it is, unfortunately, a bit more complicated. We will integrate this equation from the \(y\) position of the bottommost point on the shape (\(y_{min}\)) to the \(y\) position of the topmost point on the shape (\(y_{max}\)). How to combine independent probability distributions? Sometimes people wonder what the midpoint of a triangle is but hey, there's no such thing! However, we will often need to determine the centroid of other shapes; to do this, we will generally use one of two methods. Find The Centroid Of A Bounded Region Involving Two Quadratic Functions. where (x,y), , (xk,yk) are the vertices of our shape. Well first need the mass of this plate. How to determine the centroid of a triangular region with uniform density? In order to calculate the coordinates of the centroid, we'll need to calculate the area of the region first. y = x 2 1. 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Lists: Plotting a List of Points. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. In our case, we will choose an N-sided polygon. ?? Find the exact coordinates of the centroid for the region bounded by the curves y=x, y=1/x, y=0, and x=2. For \(\bar{x}\) we will be moving along the \(x\)-axis, and for \(\bar{y}\) we will be moving along the \(y\)-axis in these integrals. Embedded content, if any, are copyrights of their respective owners. $$M_y=\int_{a}^b x\left(f(x)-g(x)\right)\, dx$$, And the center of mass, $(\bar{x}, \bar{y})$, is, If the area under a curve is $A = \int f(x) {\rm d}\,x$ over a domain, then the centroid is, $$ x_{cen} = \frac{\int x \cdot f(x) {\rm d}\,x}{A} $$. $a$ is the lower limit and $b$ is the upper limit. . Uh oh! Get more help from Chegg . & = \left. If the shape has more than one axis of symmetry, then the centroid must exist at the intersection of the two axes of symmetry. various concepts of calculus. ?\overline{x}=\frac{1}{5}\int^6_1x\ dx??? As discussed above, the region formed by the two curves is shown in Figure 1. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. That is why most of the time, engineers will instead use the method of composite parts or computer tools. ?, we need to remember that taking the integral of a function is the same thing as finding the area underneath the function. ?\overline{y}=\frac{1}{A}\int^b_a\frac12\left[f(x)\right]^2\ dx??? To find the average \(x\)-coordinate of a shape (\(\bar{x}\)), we will essentially break the shape into a large number of very small and equally sized areas, and find the average \(x\)-coordinate of these areas. Connect and share knowledge within a single location that is structured and easy to search. ?\overline{x}=\frac{1}{A}\int^b_axf(x)\ dx??? Also, if you're searching for a simple centroid definition, or formulas explaining how to find the centroid, you won't be disappointed we have it all. the point to the y-axis. Specifically, we will take the first rectangular area moment integral along the \(x\)-axis, and then divide that integral by the total area to find the average coordinate. Which one to choose? So, we want to find the center of mass of the region below. Centroids of areas are useful for a number of situations in the mechanics course sequence, including in the analysis of distributed forces, the bending in beams, and torsion in shafts, and as an intermediate step in determining moments of inertia. {\frac{1}{2}\left( {\frac{1}{2}{x^2} - \frac{1}{7}{x^7}} \right)} \right|_0^1\\ & = \frac{5}{{28}} \\ & \end{aligned}& \hspace{0.5in} &\begin{aligned}{M_y} & = \int_{{\,0}}^{{\,1}}{{x\left( {\sqrt x - {x^3}} \right)\,dx}}\\ & = \int_{{\,0}}^{{\,1}}{{{x^{\frac{3}{2}}} - {x^4}\,dx}}\\ & = \left. The centroid of a region bounded by curves, integral formulas for centroids, the center of mass,For more resource, please visit: https://www.blackpenredpen.com/calc2 If you enjoy my videos, then you can click here to subscribe https://www.youtube.com/blackpenredpen?sub_confirmation=1 Shop math t-shirt \u0026 hoodies: https://teespring.com/stores/blackpenredpen (non math) IG: https://www.instagram.com/blackpenredpen Twitter: https://twitter.com/blackpenredpen Equipment: Expo Markers (black, red, blue): https://amzn.to/2T3ijqW The whiteboard: https://amzn.to/2R38KX7 Ultimate Integrals On Your Wall: https://teespring.com/calc-2-integrals-on-wall---------------------------------------------------------------------------------------------------***Thanks to ALL my lovely patrons for supporting my channel and believing in what I do***AP-IP Ben Delo Marcelo Silva Ehud Ezra 3blue1brown Joseph DeStefanoMark Mann Philippe Zivan Sussholz AlkanKondo89 Adam Quentin ColleyGary Tugan Stephen Stofka Alex Dodge Gary Huntress Alison HanselDelton Ding Klemens Christopher Ursich buda Vincent Poirier Toma KolevTibees Bob Maxell A.B.C Cristian Navarro Jan Bormans Galios TheoristRobert Sundling Stuart Wurtman Nick S William O'Corrigan Ron JensenPatapom Daniel Kahn Lea Denise James Steven Ridgway Jason BucataMirko Schultz xeioex Jean-Manuel Izaret Jason Clement robert huffJulian Moik Hiu Fung Lam Ronald Bryant Jan ehk Robert ToltowiczAngel Marchev, Jr. Antonio Luiz Brandao SquadriWilliam Laderer Natasha Caron Yevonnael Andrew Angel Marchev Sam Padilla ScienceBro Ryan BinghamPapa Fassi Hoang Nguyen Arun Iyengar Michael Miller Sandun Panthangi Skorj Olafsen--------------------------------------------------------------------------------------------------- If you would also like to support this channel and have your name in the video description, then you could become my patron here https://www.patreon.com/blackpenredpenThank you, blackpenredpen More Calculus Lessons. There will be two moments for this region, $x$-moment, and $y$-moment. {\left( {x - \frac{1}{4}\sin \left( {4x} \right)} \right)} \right|_0^{\frac{\pi }{2}}\\ & = \frac{\pi }{2}\end{aligned}& \hspace{0.5in} &\begin{aligned}{M_y} & = \int_{{\,0}}^{{\,\frac{\pi }{2}}}{{2x\sin \left( {2x} \right)\,dx}}\hspace{0.25in}{\mbox{integrating by parts}}\\ & = - \left. \dfrac{x^4}{4} \right \vert_{0}^{1} + \left. Center of Mass / Centroid, Example 1, Part 1 Is there a generic term for these trajectories? ?, well use. It's the middle point of a line segment and therefore does not apply to 2D shapes. We will find the centroid of the region by finding its area and its moments. Scroll down The x- and y-coordinate of the centroid read. Counting and finding real solutions of an equation. To calculate the coordinates of the centroid ???(\overline{x},\overline{y})?? We then take this \(dA\) equation and multiply it by \(y\) to make it a moment integral. Cheap . Find the centroid of the region in the first quadrant bounded by the given curves. y = x, x + y = 2, y = 0 Solution: The region bounded by y = x, x + y = 2, and y = 0 is shown below: Let f (x) = 2 - x or x = 2 - y g (x) = x or x = y/ They intersect at (1,1) To find the area bounded by the region we could integrate w.r.t y as shown below The center of mass or centroid of a region is the point in which the region will be perfectly balanced horizontally if suspended from that point. centroid; Sketch the region bounded by the curves, and visually estimate the location of the centroid. Example: The coordinates of the center of mass are then. Calculus: Secant Line. I've tried this a few times and can't get to the correct answer. ?\overline{x}=\frac{1}{20}\int^b_ax(4-0)\ dx??? ???\overline{x}=\frac15\left(\frac{x^2}{2}\right)\bigg|^6_1??? In this problem, we are given a smaller region from a shape formed by two curves in the first quadrant. Recall the centroid is the point at which the medians intersect. y = 4 - x2 and below by the x-axis. How To Use Integration To Find Moments And Center Of Mass Of A Thin Plate? tutorial.math.lamar.edu/Classes/CalcII/CenterOfMass.aspx, Improving the copy in the close modal and post notices - 2023 edition, New blog post from our CEO Prashanth: Community is the future of AI, Mnemonic for centroid of a bounded region, Centroid of region btw $y=3\sin(x)$ and $y=3\cos(x)$ on $[0,\pi/4]$, How to find centroid of this region bounded by surfaces, Finding a centroid of areas bounded by some curves. However, you can say that the midpoint of a segment is both the centroid of the segment and the centroid of the segment's endpoints. What positional accuracy (ie, arc seconds) is necessary to view Saturn, Uranus, beyond? The fields for inputting coordinates will then appear. How to determine the centroid of a region bounded by two quadratic functions with uniform density? So, we want to find the center of mass of the region below. {\left( {\frac{2}{5}{x^{\frac{5}{2}}} - \frac{1}{5}{x^5}} \right)} \right|_0^1\\ & = \frac{1}{5}\end{aligned}\end{array}\]. \end{align}, Hence, $$x_c = \dfrac{\displaystyle \int_R x dy dx}{\displaystyle \int_R dy dx} = \dfrac{13/15}{3/4} = \dfrac{52}{45}$$ $$y_c = \dfrac{\displaystyle \int_R y dy dx}{\displaystyle \int_R dy dx} = \dfrac{5/21}{3/4} = \dfrac{20}{63}$$, Say $f(x)$ and $g(x)$ are the two bounding functions over $[a, b]$, $$M_x=\frac{1}{2}\int_{a}^b \left(\left[f(x)\right]^2-\left[g(x)\right]^2\right)\, dx$$ For our example, we need to input the number of sides of our polygon. In order to calculate the coordinates of the centroid, well need to calculate the area of the region first. Log InorSign Up. If the area under a curve is A = f ( x) d x over a domain, then the centroid is x c e n = x f ( x) d x A over the same domain. Because the height of the shape will change with position, we do not use any one value, but instead must come up with an equation that describes the height at any given value of x. Formulas To Find The Moments And Center Of Mass Of A Region. We will then multiply this \(dA\) equation by the variable \(x\) (to make it a moment integral), and integrate that equation from the leftmost \(x\) position of the shape (\(x_{min}\)) to the rightmost \(x\) position of the shape (\(x_{max}\)). The coordinates of the center of mass is then. The centroid of a plane region is the center point of the region over the interval ???[a,b]???. Parabolic, suborbital and ballistic trajectories all follow elliptic paths. Calculating the centroid of a set of points is used in many different real-life applications, e.g., in data analysis. Find the centroid of the region in the first quadrant bounded by the given curves y=x^3 and x=y^3 Contents [ show] Expert Answer: As discussed above, the region formed by the two curves is shown in Figure 1. Example: Enter the parameter for N (if required). example. Compute the area between curves or the area of an enclosed shape. \int_R y dy dx & = \int_{x=0}^{x=1} \int_{y=0}^{y=x^3} y dy dx + \int_{x=1}^{x=2} \int_{y=0}^{y=2-x} y dy dx\\ example. It can also be solved by the method discussed above. To use this centroid calculator, simply input the vertices of your shape as Cartesian coordinates. Finding the centroid of a triangle or a set of points is an easy task the formula is really intuitive. To embed this widget in a post, install the Wolfram|Alpha Widget Shortcode Plugin and copy and paste the shortcode above into the HTML source. To find the \(y\) coordinate of the of the centroid, we have a similar process, but because we are moving along the \(y\)-axis, the value \(dA\) is the equation describing the width of the shape times the rate at which we are moving along the \(y\) axis (\(dy\)). Write down the coordinates of each polygon vertex. Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. Parametric Equations and Polar Coordinates, 9.5 Surface Area with Parametric Equations, 9.11 Arc Length and Surface Area Revisited, 10.7 Comparison Test/Limit Comparison Test, 12.8 Tangent, Normal and Binormal Vectors, 13.3 Interpretations of Partial Derivatives, 14.1 Tangent Planes and Linear Approximations, 14.2 Gradient Vector, Tangent Planes and Normal Lines, 15.3 Double Integrals over General Regions, 15.4 Double Integrals in Polar Coordinates, 15.6 Triple Integrals in Cylindrical Coordinates, 15.7 Triple Integrals in Spherical Coordinates, 16.5 Fundamental Theorem for Line Integrals, 3.8 Nonhomogeneous Differential Equations, 4.5 Solving IVP's with Laplace Transforms, 7.2 Linear Homogeneous Differential Equations, 8. Skip to main content. Here is a sketch of the region with the center of mass denoted with a dot. We continue with part 2 of finding the center of mass of a thin plate using calculus. Find the centroid of the triangle with vertices (0,0), (3,0), (0,5). Did you notice that it's the general formula we presented before? The coordinates of the centroid denoted as $(x_c,y_c)$ is given as $$x_c = \dfrac{\displaystyle \int_R x dy dx}{\displaystyle \int_R dy dx}$$ $$y_c = \dfrac{\displaystyle \int_R y dy dx}{\displaystyle \int_R dy dx}$$ We get that Now we can calculate the coordinates of the centroid $ ( \overline{x} , \overline{y} )$ using the above calculated values of Area and Moments of the region. Area of the region in Figure 2 is given by, \[ A = \int_{0}^{1} x^4 x^{1/4} \,dx \], \[ A = \Big{[} \dfrac{x^5}{5} \dfrac{4x^{5/4}}{5} \Big{]}_{0}^{1} \], \[ A = \Big{[} \dfrac{1^5}{5} \dfrac{4(1)^{5/4}}{5} \Big{]} \Big{[} \dfrac{0^5}{5} \dfrac{4(0)^{5/4}}{5} \Big{]} \], \[ M_x = \int_{0}^{1} \dfrac{1}{2} \{ x^4 x^{1/4} \} \,dx \], \[ M_x = \dfrac{1}{2} \Big{[} \dfrac{x^5}{5} \dfrac{4x^{5/4}}{5} \Big{]}_{0}^{1} \], \[ M_x = \dfrac{1}{2} \bigg{[} \Big{[} \dfrac{1^5}{5} \dfrac{4(1)^{5/4}}{5} \Big{]} \Big{[} \dfrac{0^5}{5} \dfrac{4(0)^{5/4}}{5} \Big{]} \bigg{]} \], \[ M_y = \int_{0}^{1} x (x^4 x^{1/4}) \,dx \], \[ M_y = \int_{0}^{1} x^5 x^{5/4} \,dx \], \[ M_y = \Big{[} \dfrac{x^6}{6} \dfrac{4x^{9/4}}{9} \Big{]}_{0}^{1} \], \[ M_y = \Big{[} \dfrac{1^6}{6} \dfrac{4(1)^{9/4}}{9} \Big{]} \Big{[} \dfrac{0^6}{6} \dfrac{4(0)^{9/4}}{9} \Big{]} \]. The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. Next let's discuss what the variable \(dA\) represents and how we integrate it over the area.

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