Integral Calculus Application - CENTROID OF A PLANE REGION

Hinzugefügt: 2026-05-11

[00:00:03]This topic is about centroidids of plane regions. Now for this topic you have to imagine that let's say for example you have you have a shape let's say a rectangle you know very well that the centrid lies exactly in the middle.

[00:00:34]So let's say this is one the the grid calibration is one unit. Therefore we know that the centrid is here.

[00:00:50]This is what we call as part y and this is what we call as bard x.

[00:01:00]Now if the area is a little bit more complex, let's say you have this area.

[00:01:13]Okay. Where do we exactly find the centrid of this L-shaped region? So one technique would be to separate this in different shapes. So this would be area one, area one, and this would be area two.

[00:01:46]Okay?

[00:01:48]And then we'll combine them together and look for the area. Now how do we find the area one and area two? So the centrid for the total area first is we look for the total area for this region. So we find area 1 equal to 4 by 1. So that is four square units.

[00:02:27]Next is for area two that would be two by one. So that would be two square units.

[00:02:42]Now the total area for both is we add both of them that would be six square units. Now why is this important?

[00:02:57]It's important because when we solve for um so when we solve for barred x and bard y we have to use a total time b x and that would be equal to area 1 x1 plus area 2 x2. And for bar y, it would be area total bar y equal to area 1 y1 + area 2 y 2.

[00:03:44]So this means that with this formula we are able to get bard x equal to area 1 x1 plus area 2 x2 all over area total. So you just divided area total on both sides. Same goes for bar y. So in able to get that that would be area 1 y1 plus area 2 y2 divided by area total. So let's try solving for this one. So let's say for example that your axis is here that is your origin.

[00:04:44]Okay. So if we're going to get the bard x this is the bard x for area one and this is the bard y for area two.

[00:05:02]This is X. And then next for area 2, this would be the bard X. That's the location. And this is the Y for area 2.

[00:05:19]Okay.

[00:05:23]Y 2 X2 X1 Y1.

[00:05:29]Okay.

[00:05:30]Now by just simply um just by simply substituting the values here for to solve for bard x total for the entire thing bard x. So that's the centrid of the entire thing because each area actually has um its own um bar y or centrid. So if you combine them both the combined centrid for the two areas will be away from it will be away from the centrid of each um region. Okay. So by substituting we just use four that's area one. So this one and then x1 would be just 0.5.

[00:06:34]Okay, this is in reference to the origin and then plus area 2 which is just two square units divided by from x2 is this distance from here to here. So that is two. Okay. So you just by looking at the drawing itself you already know that it's two divided by the area total which is six square units.

[00:07:09]Okay. So b x would be equal to b x would be equal to one unit.

[00:07:23]Okay. Next is let's do for the same thing for for bard y. B y is just equal to area one which is just four and then bard y is from here to here. So that is two units going up plus area 2 which is two and then area uh I mean the bar y for area 2 is just 0.5 going up divided by the total area which is six. [snorts] So by equating that or by um simplifying that what we have would be we have 1.5 units.

[00:08:22]Okay. So with this value the centr of the entire thing is one unit going from x and then one unit going 1.5 unit going up. So that is located here.

[00:08:45]So this is the centr of the entire thing. So bard y which is 1.5 and bard x bard x [snorts] which is just one unit.

[00:09:03]Okay.

[00:09:04]So I hope you understand that. Now, this one's a bit easy because um we already know just by knowing actually the centrids of specific shapes, it would be easier for us to immediately find the centrid of such um units. But what if we are presented with a more complicated shape that will um make us uh have difficulty solving for such centrids. So in a centrid of a plane area region. So the general formula is similar to that of a um of the example above. It would be so general let me write that down. That is general formula.

[00:10:14]So a x is equal to integral of x c. So that is your centrid in c for each um um differential area.

[00:10:28]And a bar y is your centrid in y for each differential area. So unlike the previous um examples or previous topics in here you have two rounds of computation because first one is you have to find the area by the use of um integration. Okay. So if you if if it turns out that you were already wrong in computing for the area, you you are not you won't be able to finish the whole problem with a correct um um answer because the area itself already needs integration. So there's like three rounds of um solving for integration here or using integration. Okay. Let's start with problem one. Okay. Locate the centrid of a region bounded by x^2 = 8 y. So since it's x squ, you know very well that the shape is facing up or the the curve is going up.

[00:11:54]and x - 2 y + 8 = 0. So there are no exponents on this equation. Therefore, the second equation is just a straight line and it is diagonal. So what we will do first is we should solve for the points of intersection.

[00:12:22]Okay, let me write down first the the given which is x2 = 8 y and x - 2 y + 8 = 0. So these are the two equations given. So the first one is a curve going up and the other one is a straight line diagonal line. So before we can uh find the exponents we must first look for the points of intersection for this two. We can do that by um equating both sides. So I'm going to transform x^2 = 8 y to x^2 over 8 because again when we're equating in terms of y, you would have to simplify the y first.

[00:13:37]Next is this one. transform this that would be x + 8 = y over 2.

[00:13:51]Okay, so these are the equations available. Now we equate them both. That would be x^2 / 8 = x + 8 over 2.

[00:14:13]Next is we transpose the 2 and the 8 at the bottom making it 2 x^2 = 8 x + 64.

[00:14:28]So we just multiplied two on both sides and this one eight on both sides as well. So that's how we're able to get it.

[00:14:40]And then we divide both sides by two to to um let the coefficients um go down a bit. So this will now be x^2 and then the 8 x supposedly will become 4 and then I will transpose it to the left becomes -4x and then the 64 will be negative also and that will be 32 equal to 0.

[00:15:14]Now one way of finding the roots here would be factoring which will give us x - 8 - 8 and x + 4. So again the roots here are you can do trial and error if you want that's up to you but in here we can also find the roots by factoring. So since we already have two values of x, we know very well that already becomes your limits. And then therefore we solve for the ycoordinate since we already have x = 8 as one point of intersection and x = -4 as our other point of intersection.

[00:16:09]Now we substitute x = 8 here to um our first equation. That will give us y = 8. And for x = -4, y will be equal to 2.

[00:16:37]Okay. [snorts] So with that we can already picture out the points of intersection.

[00:16:50]Okay. So let me draw first.

[00:17:01]Let me draw first my there.

[00:17:09]So this will be 2 4 6 8. So this is the first one and then two -2 and -4. So this is my x and this is my y.

[00:17:28]So my y would be I'll make it smaller 2 4 6 8 so that's easier to draw. So then we will have our first point which is x is 8 y is 8. So that would be here.

[00:17:57]And then the next one would be -4 and 2.

[00:18:03]So that would be here. And then since we know that the equation two is just a straight line. So we'll just connect both lines together.

[00:18:18]So we connect both lines together there.

[00:18:22]And then next is since we know that x^2 is = 8 y or the first equation is a curve facing up then it's up to you. You can plot this if you want to be sure of how this will look like. Uh since it's x squar and there's no negative um negative coefficients in the equation, it only means that it is at the point of origin. The vertex is the point of origin. So the shape the the curve will look like this and this. Okay.

[00:19:06]So the question now would be to find the centrid of the entire thing. Now with the centrid you'll have to again similar to what we did you have to make a representative strip that will um that will give you an idea on all the strips. Remember that there are so many strips here.

[00:19:36]Okay.

[00:19:38]Now, by summing up all the strips, you'll be able to get the total area.

[00:19:45]And same goes for the centrid. If we sum up all the strips, you will also get the total centrid of the entire thing. Same as what we did in this example right here. Okay. So now let's start with before we can solve all that is that we have to get the area. Okay. So this is my y top which is x + 8 / 2 while this is y bottom which is x2 over 8 only. Okay. Now if you still remember our previous topic area is equal to integral of um y top minus y bottom times the thickness which is in dx here.

[00:20:56]dx that's dx and then the limits would be from rightmost value which is 8 and the leftmost value which is -4.

[00:21:13]Okay. So the area would be area would be what is our y top that is x + 8 over 2 minus y bottom which is x2 over 8.

[00:21:39]And then don't forget your width which is dx.

[00:21:45]So as you can see that's very easy. But then again take note that before um we are able to solve for the the centrid we always have to start with the area first. So please remember that if you're already wrong in solving or doing the integration on the area part uh do not expect that you can still achieve the correct answer for the centrids. Okay.

[00:22:19]So this is very important and also you will be able to use this as your um for your limits.

[00:22:30]Oh no, I mean for your you will need this for your um centrids later on because you will be using the same uh value. Okay. So using your calculator I will no longer discuss how the integration is done. So area here would be equal to 36 square units. Okay.

[00:22:55]So you already have one step done there.

[00:23:00]Next, let's um let's uh solve for uh the bar y. Okay, let's start with the bar y. Remember that with a bard y it is area total time the bar y equal to [snorts] centr for in terms of y and then the differential area. Okay. Now what is our centrid for yc? That is just this value.

[00:23:42]Okay, that is that value. So how did we get that value?

[00:23:54]So it is just equal to YC is just equal to Y hold on y bottom + y top divided by two.

[00:24:22]Okay. So that is the equation for your centrid in terms or in in the y uh value or for your bar y and then that is just for yc. Now the da would be just copying this. Okay. So that would be y top minus y bottom um times your dx and then your integral which is has having limits of 8 to -4 and then 36 as my total area and Then what's left that it's unknown is your bar y.

[00:25:17]Okay. So next is we we substitute it with the values where in y bottom is x2 over 8 plus y top which is x + 8 / 2 and then both are divided by two. Okay.

[00:25:48]And then multiply by your differential area which is just y hold on y top which is x + 8 over 2 + a minus minus y bottom [snorts] y bottom which is x2 over 8.

[00:26:18]Then the dx with the limits 8 -4 36 bar y.

[00:26:30]So you understand that first before we continue.

[00:26:39]Okay. So after this we can remove the two and move it to the left here. The 1/2 here. Let's move it to the left of the the integral sign making it 36 y = 12 integral of with limits 8 to -4.

[00:27:07]So this will I will simplify this by um simplifying this both sides. So please refer to your um um gem notes on how we multiply binomials.

[00:27:35]Now similar binomials but with different um signs in the middle. One is negative and the other one is positive. So that is now going to be x + 8 over 2^ 2ar + hold on. Yeah. Oh no. Negative.

[00:28:00]x^2 x2 over 8^ 2 dx and then after that we integrate after integrating we divide both sides with 36. So bard y here this will be 1 / 64 this time because we divided to 36 here on the right and then integral of -4. So I will no longer discuss how we're able to simplify this.

[00:28:52]Okay. So I hope you are you still remember some manipulations.

[00:29:03]So your bar y here will be equal to using calculator you will have 16 over 5 or 3.2 units.

[00:29:19]Okay. So please try though to do this manually.

[00:29:23]I will not um for quizzes under my class I would allow the use of calculators but please continue practicing uh doing it manually because again uh we may not necessarily uh we we do not know if it will be allowed during the departmentals. Okay.

[00:29:47]So again please just remember that this is just multiplied distributed on both um both of these and then that's why we're able to have this value. Okay. So just um uh review regarding distribution or multiplication of two binomials. Okay.

[00:30:16]So this is our um final answer.

[00:30:24]Okay. So the last one for this item will now be four uh bard x this time. Now with bard x, we already know that bard y is just um centroid in x for each differential area and then you sum it all up. Now for b x it's just the same centrid in terms of area and then the differential area.

[00:31:05]Okay. Now, what does XC translate into?

[00:31:12]Okay. So, remember that the differential area is just this. This one is the differential area.

[00:31:21]This one, this becomes one differential.

[00:31:26]Oh, hold on. This one be Yeah, that's differential area.

[00:31:32]Okay. So that's the equation for the differential area. Okay. So this will now be 36 x. We already know 36 from our previous computation. Now what does x transl x xc translate to? So we'll just copy da. da is just y top minus y bottom.

[00:32:01]y top - y bottom * dx.

[00:32:08]Then what is xc? xc is just x actually there.

[00:32:17]Okay, this is just x value. This is very similar to the topic about um about the radius for method of shells where in regardless of where that strip would be the value of x also change it will also change. So this will follow. So if if if your x is at -4 that means your centrid will be at -4. So if your x is at -2 the value of your xc here is also -2.

[00:33:02]So therefore your xy here is just x.

[00:33:07]Okay. So it's just x and then of course your limits will be 8 to 4. So for so I'm just going to um write this down here to remind you of for example you might find this confusing.

[00:33:33]If it's in dx, your xc will be xc will be just x. While for your y, it will just be y top plus y bottom over two.

[00:33:56]Okay, so that's [snorts] how it's done.

[00:34:01]Next is let's let's rewrite the whole thing. This will now be 8 -4x that what is our y dot? That is 8. Oh, sorry. It's x x + 8 over 2 - x^2 over 8 dx and then 36 b x.

[00:34:40]Okay. Now you're using using your calculator, we divide both sides with 36.

[00:34:47]Okay. So to eliminate 36 here and whatever your value here for the integration you divide it with 36. Your x will be at 2 units.

[00:35:02]Your part x will be at 2 units.

[00:35:05]Therefore your centroid is 2 units from x.

[00:35:13]That means two units from x. So that would be that would be somewhere here 2 units and then your barred y is 3.2.

[00:35:30]So that is somewhere here. Oh perfect.

[00:35:33]It means the sample strip really did have the exact location of the centrid.

[00:35:39]So this is our centrid.

[00:35:46]3 point uh 3.2 units at bar y. This is 3.2 units at at bar y and x 2 units at b x. Okay. So that is the centrid of this entire um shape.

[00:36:09]Okay.

[00:36:11]So again, we did that by just summing up all the supposedly very thin strips of area and then to get the total area and then summing up also um your centrids per strips and giving us that these exact values. Okay.

[00:36:40]Now um what if for example the strips are horizontal? So from example number one the strip was vertical. Okay. So let's try solving for a problems wherein the strips the strips are horizontal.

[00:37:04]Okay. So question here states that find the centr of the area bounded by the curves y^2 = 4x x = 0 and y = 4. So these are our limits uh I mean these are our boundaries. So first that we need to do is we um draw the shape of the curve for example. So the first one there the the first boundary that I have here is for problem number two the boundary that I have here is y^2 is equal to 4x. Okay.

[00:37:56]And another boundary is a horizontal a vertical line which is x equal to zero. So this is just basically saying that the yaxis is its second um boundary. And lastly you have y = 4.

[00:38:18]So this is just a horizontal line which is equal to y = 4.

[00:38:26]So the curve there y^2 is equal to 4x.

[00:38:30]Remember from previous topic I mentioned that if it is y^2 if that's the if the y has the um exponent it means that the equation or the line is facing the positive side or go facing right. The curvature is facing right. Okay. So if it's negative it should be facing the left side. And if if if it's the x that is um has the exponent positive x^2 or x cub then it is facing up.

[00:39:04]I mean if it's x^2 it is facing up positive x^2. If it's x^2 then it is facing down. So [snorts] let me draw first the area that I need to.

[00:39:26]Okay.

[00:39:28]So it says y = y = 4 meaning the the first the second boundary is this.

[00:39:44]It's just a straight line horizontal line y equal to 4. Okay. So again we know that the it's y^2 and there are no other constant. Therefore it falls the curve falls directly on the centrid like this. Okay. So this is their point of intersection. But for for and sorry and another intersection another um boundary is this here which is the y-axis which is indicated here. Now how do we know that it is the point of intersection? We just simplify this making y^2 = 4x. Now if y is equal to 4 therefore your x is just equal to four as well.

[00:40:51]Therefore the point of intersection is 4 and 4. So that is one limit. Another point of intersection here is 04 and another one here which is 0 0.

[00:41:08]Okay. So I'm just um plotting out all the possible values that can be used for our um limits later. Now since again I mentioned that our orientation for the strips this time will be horizontal. Let me draw that representative strip. So this is my strip. Okay.

[00:41:38]Okay. That's my strip. Now if we solve for the area, this will be now your um width or thickness or height and this one will be your width.

[00:41:53]Okay. So differential area or area is just equal to integral of xr - xl * d y because again the strip is already horizontal.

[00:42:14]So that is xr is just your equation this one. And then of course since it says xr your x should be in its simplest form which is y^2 over 4. Okay. So that is y^2 over 4 - 0 xl which is this which is just 0 and then dy.

[00:42:46]Then lastly your integral sign and your limits. Since we are using dy we know that the the um limits will be in terms of y as well which is 4 and 0.

[00:43:08]So once we integrate that this becomes y cub / 12 with limits 4 to zero or your area total is just equal to 5.33 square units.

[00:43:33]Again computing for the units is just the first part of the um whole process.

[00:43:42]Now if you missed or if you have done this incorrectly the area solving for the area incorrectly then you will mo you will have a wrong answer for the entire problem already or the entire solution will be wrong. Okay, next would be to solve for the bard x.

[00:44:09]So the bard x is just area total * b x equal to integral of x c * differential area.

[00:44:23][snorts] So a is already known. We've already solved that. That's 5.33 part x equal to xc.

[00:44:33]Now your xc is this value that is just xr + x l / 2.

[00:44:46]So xr + xlid 2.

[00:44:54]Okay. But your XL here will just be zero.

[00:44:59]So please uh take note of that. And then your differential area also which is just this we just copy that would be XR - XL times your dy.

[00:45:16]Okay.

[00:45:18]And then of course our limits which is the same four and zero.

[00:45:24]Now this will cancel out because that is just zero.

[00:45:31]This one is zero. This one is also zero which gives us this integral 5.33 x and then integral of 4 xr. Now what's xr there by the way? XR is just y^2 over 4 all over 2. So this is your xc time your xr which is just y^ 2 over 4.

[00:46:12]So this is xr / 2. Okay, XR is just y^2 over 4 and then since you still have divided by two there. So, uh or maybe we can remove this, remove this and transfer it to the outside here and make this 1/2.

[00:46:45]And then of course don't forget your dy.

[00:46:50][snorts] So that is just y^2 over 4 and then square it. So using your calculator I will not simplify that anymore or yeah maybe let's try to simplify that is just 5.33 b x 12 to 0 y raised to the power of 4 over 16 dy. I think that's a lot easier to understand.

[00:47:28]Okay. Now with that you just integrate this using the calculator and divide it.

[00:47:36]We divide the results of the integration by 5.33.

[00:47:40]That will give you a bar x equal to 1.2 units there.

[00:47:57]So again I will no longer show how y square y raised to the^ 4 / 16 uh needs to be um uh I don't need to explain that further because that's very basic integration that would be y raised to the^ of 5 over 16 * 5 and then substitute the y with limits four and zero and then divide it by 5.33 and 1/2 multiplying 1/2 then that's your um answer okay so I hope you do know we'll probably discuss how to input that in the calculator okay so next would be to solve for the ay bar y so Bard Y is equal to uh we multiply it with the area total first and then your centrid YC for each strip and then the differential or area to um multiply it with. So this is the this is an example this is the example of YC.

[00:49:21]Okay, this is example of YC. That's for one YC and this is for area one or differential area if we're talking about the integrals. Okay.

[00:49:37]So that's why I had to discuss that first. [snorts] Okay. Now let's let's write down all the things that we already know. Area is just 5.33 bar y still left unknown and then integral of that we know the limits is already 4 and zero and then your differential area we know this this one I will need to discuss it still so da is just um xr minus xl this is your da or differential area time dy. Now what is xc? Okay, same as what we did here.

[00:50:25]Same as what we did here that the value of your y for horizontal strips is just y. Okay?

[00:50:38]Because where however wherever the location is, let's say for example your strip is here at the top then your y is four. If your strip is here at the bottom then your y is zero. That means it's a variable. It changes together with a location or regard uh depending on what's the value of your y. So with that your y c is just equal to y which is quite easy. Now we change the value of your xr and xl. xl here being zero cuz we know it touches the um axis.

[00:51:23]So this will now be 5.33 y = to integral of 4 and 0 y xr is just y^2 over 4 d y.

[00:51:41]Okay.

[00:51:44]And then let's simplify that. That would be integral of 4 to 0 um y cub over 4 d y and then by using the calculator you do the computation for that and then divide it with 5.33 or the total area you will be able to get a barred y equal to Three units.

[00:52:19]Okay.

[00:52:21]Three units.

[00:52:25]Hold on. Let me there. Three units.

[00:52:37]Okay. So, how does that translate to the actual drawing? It means that oh it is at bar Y is at three units high here.

[00:52:58]So bard Y is here at three and B X is at here.

[00:53:09]B X is at 1.2 which is exactly I actually did the correct um assumption of where the location is.

[00:53:23]Okay.

[00:53:25]So I think that ends the centrid for plain region B X and B Y.

[00:53:38]Thank you.