find-the-areas-bounded-by-the-indicated-curves-using-a-vertical-elements-and-b-horizontal-element-10

Question

Find the areas bounded by the indicated curves, using (a) vertical elements and (b) horizontal elements.
$$y=x^{3}, x=0, y=3$$

Lucas Finney · Accepted Answer

for this problem we are asked to find the area bounded by the curves. Y equals x cubed, X equals zero and y equals three. The first step that I&#x27;m going to take here is to just create a rough little sketch of our curves. So I&#x27;ll have first the Y equals X cubed. We&#x27;ll do in the blue here, you know that that looks something like this, then I&#x27;ll do the X equals zero that corresponds to the y axis. And then also we want along when Y equals three. So it&#x27;ll be something over here. So we want to find the area inside of this. So for the first part of the problem we are asked to do this using uh using vertical elements. So we&#x27;ll need to figure out where the X or what the X value is when y equals three. So we can figure out what to integrate X over. Well if we have that, Y equals X cubed and Y equals three. So we would have that three equals X cubed And so X is going to equal three to the power of 1/3. So To find the area will take the integral from zero 2, 3 to the power of 1/3 of now. Our upper function or upper line is going to be three. So it will be three minus are lower line, it will be three minus X cubed dx. Now integrating three will give us three X and integrating negative X cubed will give us x power of 4/4 Evaluating that from zero up to three speaker, 1/3. We can express this in the form, the most simplified form. Uh Nine times three to the power of 1/3 divided by four. Now for part B. And I&#x27;ll note that the numerical value there is approximately 3.245 now for part B were asked to do the same but using um horizontal elements instead. So what we first need to do is we know why is a function of X means then we have to translate this into X as a function of why? So we&#x27;ll have x equals white power of 1/3 is our function and we are going from y equals zero up to y equals three. So The integral will be from 0 to 3 of why? It&#x27;s the power of 1/3 D. Y. Well I&#x27;ll note that the lower bound on the function would be the line corresponding to X equals zero or y equals zero. So anti derivative of this would be wide to the power of 1/3 plus one. To be white power for over three divided by 4/3. So it&#x27;s the same thing as multiplying by 3/4. We&#x27;re evaluating that from zero up to three. So we can see that we&#x27;ll get three times three to the power of 4/3 over four which is equivalent to nine times three to the power of 1/3 over four. And so we can see that we have the same result either way.

Dillon Huddleston · Answer

s so in this case, we can integrate with the state. Why so X equal to two I squared is a problem with access upper proble with access been paused about X axis eso integrating between ask equal to the square excluded Got equal to zero wishes wth e Why access and why Call the three s o. We can integrate just two I squared zero is too. I swear has entered your motive 2/3. Why cute? This could be interpreted from the intersection of X zero with exes to I square that why zero and why&#x27;s three is a buyer in question? So we&#x27;re really right between zero to me. Of course, this zero value did y zero. So me on you consider the upper limit three So three tube is 27 divided by three is on. I am multiplied by 2 18 and we can see that this value of 18

Bobby Barnes · Answer

we want to find the area enclosed between the breach and X is equal to two by squared excessive zero and why is equal to three? So the area were interested in will be located right here. The part that I&#x27;m shading in lack right now. So there&#x27;s two ways that we could go about this. We could eat there, get everything in terms of why, I mean, in terms of X and then solve. Or we could do this in terms of X goes, we could think about it in the same way in terms of the upper minus the lower bound. But the upper is going to be the first to the right, and the lower is gonna be the first to the left. So if we start on the left, they keep on going. We first enter our region at X equals the rooms. That&#x27;s gonna be our lower function. And we keep on going out until we pass through, and then we pass through this red so we can actually do this room. So since that point, there&#x27;s just gonna be zero, and we know our upper bound in this case is three. For why so it&#x27;s going to be, too. Why squared? Minus Cyril, which we&#x27;re just leave us with our function so we could go ahead and integrate this now to find all the area in that region we have shaded. So that&#x27;s gonna be 2/3. Why cute from power rule 0 to 3. Then we can plug in three. So it&#x27;s gonna be 27 divided by three, which is nine times to 18. Logan zero, we get zero. So that area, the region should be 18.

Tyler Gaona · Answer

So it&#x27;s probably easiest to write thes to curves as functions of why, in terms of X, so I&#x27;ll just write thes as Why equals X cubed and here wise equal to three X squared minus four. So to find the area of the region between them, the first thing we should do is figure out where these curves intersect. So for that I have to set them equal, which, if, after moving everything to one side, is equivalent to writing, writing this. Okay, so now let&#x27;s see, how can we solve this equation for X? It&#x27;s not just a quadratic, so we can&#x27;t just plug into the quadratic formula, but let&#x27;s see what we can do. So maybe one thing we can try here if we don&#x27;t have, say, access to a computer, which Khun just solve this for us has made. Let&#x27;s see if we can maybe just find a solution just by looking at it, we might try some easy numbers to plug in. So if you plug in, say, just one, for instance, you get one minus three plus four. Okay, so that&#x27;s one minus three is negative two plus fours, too, so it&#x27;s not zero so that&#x27;s not a solution. But say, if we had a minus one minus three plus for that would be a solution. And you can check if you plug in X is negative one. That&#x27;s what we get. So few plug in negative one. Here you get minus one minus three plus 40 So tell this negative one is a solution. Meaning we can factor like this. So x minus minus one. All right, Hoss. Explosive one. And then we&#x27;re gonna have some quadratic term. Okay, so I know that I have to have an x cubed. So that&#x27;s why no, they&#x27;re coefficient. Here is one. Because when I multiply these, the only way I&#x27;m going to get next cubed is from thes two terms. And then I have some linear term. And then I know also that to get a four here, so I want thes to multiply thes to to multiply out and give me execute minus three X plus four. To get this for is the constant. While I have one times something so that something has to be four. Okay. And so let&#x27;s figure out what this is actually supposed to be, so we can just multiply these out and then compare the coefficients of there What we get when we multiply it with the polynomial we started with. So if I multiply these out, I get next Cube. And now let&#x27;s see what we have on an X squared I have on a X squared plus a one x squared. So write that like this. And what about the linear terms? Well, I have a plus four x one person X and then I just have my plus four for Constant. Okay, so this is what we get if we multiply these two out, and now let&#x27;s compare the coefficients well, appear on the linear term. The A is there is no linear term. So this four plus a should be equal to zero, which just tells you that a should be minus for okay and we can check out, You know, if we look at the quadratic term, a minus four works there also. Okay, So that means then we factor this properly as explosive one x squared minus for X plus for equal zero. OK, And so now we have this quadratic and it looks like we should be on a factor this is exposed. Swon. This looks to me like it&#x27;s going to be X minus two squared. Okay, so we have the intersection. Points are when X is equal to minus one. And then we have a multiple route when X is equal to two. But still it&#x27;s intersecting at minus one, and X equals two. So now that we know where the curves intersect, we can if we can set up an integral to evaluate the area of the region. So it&#x27;s C a first. Let&#x27;s just try to sketch a picture. What&#x27;s going on? So here is negative one. Here&#x27;s to Okay, so then I have X cubed. No, which looks something like that. And then I have three x squared, minus four sets of parabolas, some sort. Okay, so this is what we have. So this proble So it&#x27;s a problem of this three is stretching it a little bit and minus four is taking it down. And since we know execute, I mean, you can just plug in X equals zero here and see that zero is greater than minus four. That tells you the green function is on the bottom. The red function is on the top. And so then the area is going to be the integral from minus 1 to 2 of the red function that I&#x27;ve drawn here X cubed, minus a green function. Okay. And I&#x27;ll leave the evaluation of the center grow up to you. Since we&#x27;ve there are plenty of examples in the previous videos evaluating in a girls like this, the important point is just being able to get here and setting it up. So we found the points of intersection figured out which graph which function is above the other. And then we use that to set up our integral that gives us the area of the region between the curves.

Bobby Barnes · Answer

we want to find the area enclosed by these two curves that they provide. So that area that were interested in would be this area right here. And in this case, it probably be better too integrate with respect to X. Since solving for X is gonna be pretty straightforward as well as if we were to start from the very bottom down here and then passed through the region, the blue line is always gonna be on the bottom, and then the red line is always going to be on top as opposed to If we go from left to right at some point, the right ID is on the bottom, blue on top, and then we just have the blue all together on the bottom, so becomes kind of confusing on how to set the intervals. So we&#x27;re gonna let this be our lower and the red be our upper. So we need to solve for y in each case for the red, it&#x27;s just gonna be why is equal to x cute and then the blue we should get why is equal to three X spared minus four now to figure out what our points of intersection here. So it should be like these points right here. We would set our two functions equal. So we have execute is equal to three X squared, minus four. And so you move everything over to one side factor, do all that. And in the end, though, we should end up getting that X is equal to negative one and two. So we know what our bounds of integration are gonna be. So let&#x27;s go ahead and set it up now. So it should be negative one, too. Of so he said, our upper function is execute and our lower function was that quadratic. So three X squared plus four d X. Now we can use powerful to integrate. So the 1/4 ex to the fourth minus X cubed and then plus four X we&#x27;re gonna evaluate this from negative one toe, too. So plug it in Tuesday to to the fourth of 16. Divide by four gives us four and minus eight plus c. Then we subtract when we plug in negative one, which you give us 1/4 must one minus a war. And so well, first these eights cancel. Oh, and then what do we get with? It looks like it gives us 27 over four or that area that we shaded in

Tyler Gaona · Answer

we want to find the area of the region enclosed by these two curves. So eventually we&#x27;re going to wantto maybe sketch a little picture of the graph of these two guys, and we use that to set up an integral which will tell us the area. Maybe the first thing we should do is figure out where these curves intersect. So if we have a point x comma y that lies on both of these curves well, there I values are equal. So I should be both of these equations to be satisfied. So I want to find the points X that satisfied this equation. So we&#x27;ll get a quadratic equation here. If I move everything onto the right hand side, I get X squared minus two X minus three is equal to zero and I can factor this as X minus three exports. One equals zeroes. That tells me that intersections or when X is equal to three and X is equal to negative one. Of course, the Y value will be negative three. So if we try to sketch the graph, we have, let&#x27;s see, Here&#x27;s negative. Negative one. Here&#x27;s three. I would say this is me trying and other color. Maybe red, the line wise. Negative three. Okay. And so we have a point of intersection here, a point of intersection here and now this curve, we wonder, is it? You know, in between these two points is it above or below the line? Why equals negative three? Well, if you just plug in, say, for instance, the point X equals zero. You see that when X equals zero on this curve wise also equal to zero. So it&#x27;s up here, So it&#x27;s some for Avalon type shape. Maybe it looks like that. So we&#x27;re interested in this area between these two. And so we confined thatjust by integrating from negative one, two, three. So our upper function minus our lower function. So let&#x27;s find the anti derivative. You get a three x from this part anti druid of two exits just x squared. Then we would have minus X cubed over three, and we&#x27;LL evaluate this between three and minus one. So then let&#x27;s see what we get. We&#x27;LL have nine plus nine minus three cubed over three What&#x27;s again? Just nine, Okay. And then we subtract what we get when we plug and negative one that&#x27;s minus three close one minus negative, one third. So at the end of the day, those will cancel. We have a nine plus three minus one that&#x27;s eleven eleven plus a third. So eleven is thirty three over three, so this is thirty for over three.

Problem

Find the areas bounded by the indicated curves, u…

Problem 40 Easy Difficulty

Find the areas bounded by the indicated curves, using (a) vertical elements and (b) horizontal elements.
$$y=x^{3}, x=0, y=3$$

Answer

Related Courses

Basic Technical Mathematics with Calculus

Related Topics

Discussion

Top Calculus 2 / BC Educators

Grace H.

Heather Z.

Kristen K.

Joseph L.

Calculus 2 / BC Courses

Integration Review - Intro

Definite vs Indefinite

Recommended Videos

Watch More Solved Questions in Chapter 26

Video Transcript

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

Grace H.

Heather Z.

Kristen K.

Joseph L.

Integration Review - Intro

Definite vs Indefinite

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

Basic Technical Mathematics with Calculus

Grace H.

Heather Z.

Kristen K.

Joseph L.

Integration Review - Intro

Definite vs Indefinite

Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

Grace H.

Heather Z.

Kristen K.

Joseph L.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus