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The centroid of a planar region.
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Centroids of simple shapes.
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The centroid of a planar region is a point that plays the role of its center.
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One way to think of it is that if the region were a thin plate,
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one could balance the plate horizontally by supporting it at its centroid.
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If a region has a line of symmetry, then its centroid is on that line.
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If a region has two or more lines of symmetry, then the centroid is the point where they intersect.
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A simple disk has an obvious center that coincides with its centroid.
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Here's one line of symmetry, and here's another.
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Those lines intersect at the centroid.
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In fact, for a disk, every line through the centroid is a line of symmetry.
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Here's an equilateral triangle.
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A line through any vertex in the midpoint of its opposite side is a line of symmetry.
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So here there are three of them.
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The centroid of the triangle is the point where they intersect.
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For a rectangle, lines through opposite corners are lines of symmetry.
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As are, lines through midpoints of opposite sides.
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For a regular pentagon, lines of symmetry are just as an equilateral triangle.
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Each passes through a vertex in the midpoint of the opposite side.
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Other regular polygons have lines of symmetry similar to either the rectangle or the equilateral triangle,
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depending on whether there are an even or odd number of sides.
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We are more concerned, however, with less regular regions.
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A quarter-disc, for instance, has only one line of symmetry.
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The centroid is somewhere on that line, but it's not obvious exactly where.
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Likewise, for a half-disc.
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A region between two curves, such as this, likely has no line of symmetry.
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Yet it will have a centroid, and we'd like to learn how to compute it.
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Moments. Let our region in the plane with area A.
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Think of R as a lamina, that is a thin plate with uniform thickness.
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Assume that throughout this lamina the mass density is one unit of mass per unit area,
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then the mass of the lamina is numerically equal to the area A,
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that is, they're the same, but with different units.
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Let X bar Y bar denote the centroid of R.
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Then R's moment about the Y axis is defined as,
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n sub Y equals X bar, or the distance from the Y axis, times the area A.
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And this moment about the X axis is, n sub X, equals the distance from the X axis,
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or Y bar, times the area A.
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Additivity of moments.
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Suppose that the plane of region R consists of non-overlapping sub regions,
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R1 through Rn, with areas A1 through An,
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and centroid X bar 1 Y bar 1 through X bar n Y bar n.
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Of course, the area of R is the sum of the areas of the individual sub regions.
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Suppose also that X bar Y bar is the centroid of the full region R, and that n sub Y and n sub X
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are its moments about the coordinate axes, so that X bar times A equals n sub Y and Y bar times A equals n sub X.
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Now the moments n sub Y and n sub X are the sums of the moments of the individual sub regions R1 through Rn.
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That is, we can express our moments as n sub Y equals the sum of X bar sub I times A sub I,
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and n sub X is the sum of Y bar sub I times A sub I.
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So notice now that we have two ways of expressing each of these moments.
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n sub Y is equal to X bar times the total area, and it's also equal to the sum of X bar sub I times A sub I.
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The moment about the X axis is equal to Y bar times A, and it's also equal to the sum of Y bar sub I times A sub I.
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So we can now divide each of these by A to reveal X bar and Y bar.
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X bar is the sum of X bar sub I times A sub I divided by A, and Y bar is the sum of Y bar sub I times A sub I divided by A.
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Now let's look at an example.
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Let's find the centroid of the region shown here.
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Now the irregular polygon here is actually composed of two rectangles.
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Now let's think about area.
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The circle there has radius 1, so the area of that disk is pi.
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The rectangle on the left is 4 by 2, so its area is 8.
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The other rectangle is 4 by 3, so its area is 12.
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So the total area is pi plus 8 plus 12 or pi plus 20.
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Now to calculate our moments we need to determine the centroid of each of the three parts of the region.
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To do this we need to pair coordinate axes.
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Notice we've chosen to put our origin at the bottom left corner of the rectangle with area 8.
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That's just a convenient choice where we place the origin is completely arbitrary here.
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The centroid of the circle now is clearly at the point 0.6.
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The centroid of the left rectangle is at 1,2.
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Notice that for the larger rectangle the Y coordinate of the centroid is halfway between 2 and 5,
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or 3 and a half, or 7 halves, and so the centroid there is 4,7 halves.
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Now that we have the individual areas and centralities we can write down our moments.
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The moment about the Y axis is going to be the moment of the disc about the Y axis,
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which is the X coordinate of the centroid times the area, and that's 0 times pi.
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Now let's go to the rectangle with area 8.
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The X coordinate of the centroid is 1, the area is 8, so we have 1 times 8.
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Now for the other rectangle we have X bar equal to 4, and an area of 12, so we have a 4 times 12 term.
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Now doing the arithmetic here we have 0 plus 8 plus 48 or 56.
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Now let's get the moment about the X axis.
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We take the Y coordinate of the centroid of the disc and multiply by pi,
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then the Y coordinate of the centroid of the area 8,
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and then add the Y coordinate of the area 12 rectangle times 12, and then calculate the sum, it ends up being 6 pi plus 58.
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Now we're ready to write down X bar and Y bar.
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X bar is M sub Y divided by A, which is 56 over pi plus 20, which turns out to be about 2.42.
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Y bar is M sub X divided by A, that's 6 pi plus 58 over pi plus 20, which is about 3.32.
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Now let's look at where our centroid lies in our region.
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Notice it's just about where we would expect it to be.
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The moments of a narrow strip.
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Suppose we have a strip that's parallel to the Y axis, like the one shown here, and suppose it is width is delta X and its area is delta A.
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Also assume that its horizontal midpoint is X, and the point at the top of the strip is X comma Y top, and the point at the bottom of the strip is X comma Y bottom.
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Now because this is a simple rectangle, it's easy to determine where the centroid is.
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Of course X bar is simply X, and Y bar is the midpoint between Y bottom and Y top, or the average of those two numbers, 1 half of Y top plus Y bottom.
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Also, the area delta A is equal to length times width, or Y top minus Y bottom, times delta X.
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So, the moment about the Y axis is X bar times area, which is X times Y top minus Y bottom times delta X.
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And the moment about the X axis is Y bar times area, which is 1 half of Y top plus Y bottom times the area, Y top minus Y bottom times delta X.
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Notice that when we multiply those two middle factors out, we end up with 1 half of Y top squared minus Y bottom squared, all times delta X.
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Now let's consider a region bounded by the graphs of two functions of X, between X equals A and X equals B, with Y equals F of X forming the top edge, and Y equals G of X forming the bottom edge.
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Consider a typical vertical strip centered at X with width delta X.
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Noting that the centroid of the strip is located at X comma 1 half of F of X plus G of X. This thin strip has area delta A equals F of X minus G of X times delta X.
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The moment about the Y axis of the thin strip is X bar times the area of the strip, which equals X times F of X minus G of X times delta X.
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The moment of the full region about the Y axis can be approximated by summing the moments of thin strips.
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And then the limit is delta X approaches zero, those sums approach the integral from A to B of X times F of X minus G of X dx.
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The moment about the X axis of the thin strip is Y bar times the area of the strip, which equals 1 half of F of X plus G of X times the area delta A.
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Delta A is F of X minus G of X times delta X. And when we multiply out F of X plus G of X times F of X minus G of X, we write this as 1 half of F of X squared minus G of X squared times delta X.
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Now the moment of the full region about the X axis can be approximated by summing the moments of thin strips.
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And then the limit is delta X approaches zero, those sums approach the integral from A to B of 1 half F of X squared minus G of X squared dx.
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Next we divide each of the moments by the area. X bar equals M sub Y divided by A and Y bar equals M sub X divided by A.
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So to calculate the two coordinates of the centroid, we need to calculate three integrals, the area and the two moments, and then divide each of the moments by the area.
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Now let's look at an example. Find the centroid of the region bounded by the graphs of Y equals X and Y equals X to the fourth.
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Here's the picture of the two curves and the region bounded by them. Notice that it lives between X equals zero and X equals one.
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Now let's think about a vertical strip here located at X. We'll first integrate the length of that strip over the interval to obtain the area.
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The integral from zero to one of X minus X to the fourth, the X, is equal to 1 half X squared minus 1 fifth X to the fifth, evaluated between zero and one, which ends up being 1 half minus 1 fifth, minus zero, and 1 half minus 1 fifth is equal to three tenths.
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Next we'll calculate the moment about the Y axis. That's going to be the integral of the X coordinate of our strip times the area of the strip, which is M sub Y equals the integral from zero to one of X times X minus X to the fourth, the X.
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Notice that's nearly the same as the area integral except there's an extra factor of X. We'll multiply out the integrand to get X squared minus X to the fifth. And so now we integrate and get 1 third X cubed minus 1 sixth X to the sixth, evaluated between zero and one, which equals 1 third minus 1 sixth, minus zero, or 1 sixth.
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Next we calculate the moment about the X axis. That's going to be the integral of the Y coordinate of the centroid of the strip times the area of the strip, which ends up being 1 half the integral from zero to one of the square of X minus the square of X to the fourth.
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So we have 1 half the integral zero to one of X squared minus X to the eighth, DX, which equals 1 half of 1 third X cubed minus 1 ninth X to the ninth, evaluated between zero and one, which is 1 half of 1 third minus 1 ninth, minus zero, which is 1 half of 2 ninths or 1 ninths.
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So now we're ready to calculate X bar and Y bar. X bar is the moment about the Y axis divided by the area. That's 1 sixth divided by 3 tenths, which equals 10 divided by 18 or 5 ninths.
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Y bar is the moment about the X axis divided by the area. That's 1 ninth divided by 3 tenths, which is 10 divided by 27.
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So the centroid of our region is the 0.59ths, 1027ths.
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Now let's look at another example. By the centroid of the quarter disk, in which X squared plus Y squared is less than or equal to R squared, and X and Y are both greater than or equal to zero.
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So our region here is the first quadrant portion of the disk with radius R. So let's notice that that quarter circle is the graph of Y equals the square root of R squared minus X squared.
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Now there's a line of symmetry here, the line Y equals X, and we know that the centroid must live on that line. So Y bar will equal X bar. Also, we don't need to integrate you at the area of the region. That will be 1 fourth of pi R squared.
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Now let's calculate the moment about the X axis. That's going to be 1 half of the integral from 0 to R of the square root of R squared minus X squared minus 0 squared, since the lower edge of the region is the line Y equals 0.
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This simplifies to 1 half the integral from 0 to R of R squared minus X squared DX. So we have 1 half of R squared times X minus 1 third of X cubed evaluated from 0 to R.
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The value at R is 1 half of R cubed minus 1 third R cubed, the value at 0 is 0. So we have 1 half of 2 thirds R cubed, which is 1 third R cubed.
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So X bar and Y bar are both equal to the moment about the Y axis divided by the area. That's going to be 1 third of R cubed divided by 1 fourth of pi R squared, so we end up with 4 thirds of R divided by pi.
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Now that's about 0.424 times R, which tells us roughly where the centroid is along our line of symmetry.
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Now, although it's not necessary, let's calculate the moment about the Y axis to make sure that we get the same answer.
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The moment about the Y axis is the integral from 0 to R of X times the square root of R squared minus X squared DX.
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To do the anti-differentiation here, let's do a substitution. We'll let U equal R squared minus X squared, the quantity under the radical, then DU is minus 2X DX.
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So X DX is equal to minus 1 half of DU. Also, we'll transform the limits of integration.
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When X is equal to 0, U is equal to R squared, and when X is equal to R, U is equal to 0.
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So we have the integral from R squared to 0, but minus 1 half the square root of U to U.
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Now, since we're integrating from right to left, and we have a negative sign, let's switch the limits of integration to get rid of that minus sign.
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Now, an anti-derivative of 1 half U to the 1 half power is going to be 1 third times U to the 3 halves power, and we evaluate from 0 to R squared, obtaining 1 third R cubed.
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Notice that's the same as the moment about the X axis, as it should be by symmetry.
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One more example. Find the centroid of the region, in which Y is greater than or equal to X squared, and less than or equal to 1 eighth, times 7X squared plus 4.
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This inequality describes the region between two parabolas, as shown here.
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Turns out the parabola is intersect at X equals plus and minus 2.
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Here, the Y axis is a line of symmetry, and so by symmetry, X bar will be 0.
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So we only need to calculate Y bar. Now, let's calculate the area. We'll take advantage of symmetry here, and write the area as 2 times the integral from 0 to 2 of 1 eighth, 7X squared plus 4, minus X squared, DX.
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If we factor out 1 eighth, then the quantity in parentheses becomes 7X squared plus 4 minus 8X squared, and so we have 2 times 1 eighth, or 1 fourth, times the integral from 0 to 2 of 4 minus X squared DX.
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This equals 1 fourth of 4X minus 1 third X cubed, evaluated from 0 to 2, which is 1 fourth times 8 minus 8 thirds, which equals 8 over 4 times 2 thirds, or 4 thirds.
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Let's calculate the moment about the X axis. We'll take advantage of symmetry here to again integrate from 0 to 2, so our moment about the X axis will be 2 times 1 half of the integral 0 to 2 of the square of 1 eighth of 7X squared plus 4, minus the square of X squared.
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Now, we'll factor out a 1 eighth squared here, which places a 64 in front of the second term, and then we'll multiply out 7X squared plus 4, quantity squared, and simplify, and the result ends up being 164 times 49 minus 64X to the 4th plus 56X squared plus 4 squared to 16.
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We'll pull the 1 over 64 outside the integral and simplify, and so now we need to integrate 16 plus 56X squared minus 15X of the 4th from 0 to 2.
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Now, we anti-differentiate. We have 1 over 64 times 16X plus 56 times 1 third X cubed minus 15 times 1 fifth, or 3, times X of the 5th evaluated between 0 and 2.
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Now substituting 2 into that anti-derivative gives us 164 times 32 plus 56 over 3 times 8, or 448 over 3, minus 3 times 32 or 96.
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So, we get 2 minus 96 is negative 64, which is negative 192 divided by 3, and so we get 1 over 64 times 448 minus 192 over 3, which is 256 over 192.
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The numerator there is 4 times 64, but the denominator is 3 times 64, so we end up with 4 thirds for the moment about the X axis.
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So, Y bar is 4 thirds divided by 4 thirds or 1, so the centroid of our region is the 0.0 comma 1.
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Notice here that the centroid is not in the region.