solve-the-given-problems-by-integration-find-the-moment-of-inertia-with-respect-to-the-y-axis-of-a-4

Question

Solve the given problems by integration.
Find the moment of inertia with respect to the $y$ -axis of a flat plate that covers the first-quadrant region bounded by $y=e^{x^{3}}, x=1$ and the axes.

Ahmed Ibrahim · Accepted Answer

problem. It is required to find the moment of inertia about ex acts off a sim plate haunted boys a problem X equal to Y minus y square and the line X plus y equals zero If the dentist e equal Tow X plus why we restored by sketching the graph So we have two equation. 1st 1 is X plus y equals to zero, which is something like that, and the thick and one ese x equal toe. Why minus why square, which is this curve cutting boy eggs at one. And if we and substitute by or sold the two questions off the car, we can get this point and why equal tow to So now he&#x27;s asking about this region and we restored our solution. Boy. Completing M So am I will be equal toe integration from zeal to two and integration in wider veteran from negative y to y minus y square for our dynasty, which is X plus y. The ex knew I, which will be equal toe integration from zero to toe for Why are four over too minus two Y or three loss two way over, too. Do you want, which will be equal toe eight over 15. Now we can calculate I Iove x nursey about X x, which is digression from 0 to 2 and the fame integration in my direction. From why to why minus y square for y square and or dynasty, which is exit lous y the ex the wine. So it will be equal toe integration from 0 to 2 or wipe or six overto minus to wipe or five. Unless to Roy or four. Do you, Roy, which will be equal toe 64 over one or five. Thank you.

Yiming Zhang · Answer

to find the movement of the miniature with First, find the mess. This object When did the density equals to delta? And, uh, it&#x27;s, uh, shit. Patrick objects. So we have 0 to 3 d woody s and that&#x27;s for the mass. And this gives us nine Delta. So that&#x27;s the massive the sob bob gets. And for the moment and in the XX is the moment of inertia by the former owner. We have the density from 0 to 3 and 0 to 3. So it is in the UAE access. We got a why swear T Woody x And this is 27 intelligent and by the symmetry, because this object is pretty symmetry symmetrical. So we have the movements of being Hirscher. Only y axis equals through that that it&#x27;s on the X axis. So the same number $27. That&#x27;s it.

Yiming Zhang · Answer

So here we&#x27;re gonna find the moment when they&#x27;re sure with respect. Why? Let&#x27;s see. Um So first we find the region. We have access from a pipe to pie and every equation we had White goes toe at the X equals two science away X squared exes from party to pie. Here we have the appointment of ex so wise from zero to there&#x27;s a positive choice. The lower ground zero in the upper about mistresses function of X, which is that&#x27;s sinus over X square, the Y. And this is with respect. Why? So we hear McCloy, we multiply X squared and I&#x27;m in trouble over this equation, we&#x27;re gonna have 1/2 times pie to two pi. And inside is where have one minus kowtowing to eggs. Here we have the X, Then the answer for that is half party

Eric Mockensturm · Answer

who asked to determine the moment of the second area moment of the and the radius of gyration of the shaded area. With respect to the X axis, Mr are shaded area looks something like like what I&#x27;ve sketched here and what we&#x27;re gonna do is break it up into into three parts. So the first part, it will be the center beam here or the center section. So this part here and then to would be this upper flan and three will be this lower flan so we can base, weaken, take, find the area moments for each section and then add them up to find for the total. And we want to do it about, um, this X axis here, which is in the center of this vertical member. So the, um, formula for the area moment of a rectangular shape about its central aid, which this is that, um, is 1/3 be one h one cube. So be one. Is this thickness And H is this distance here, and we&#x27;re giving those in the problem. And those there they are. And now for two, we can take the area moment about its center of mass. All right. It&#x27;s, um, sense. Right? And then we need to shift it using the parallel axis deer. I&#x27;m from here to here. And so that distance ties. So then we need the area of this, which is B to H two and then this distance here. Um, and, um, that&#x27;s deep. Well, call that D to so we know, we know a, um that&#x27;s totally And so here&#x27;s the values for, um, for here. And then we have the third part, which is this part here. And so we have the center. We have the center, the area moment about the central right, and then plus the area times this distance from the central to our X axis. I call that the three. And these were all given in the problem. Calculate the total area of this cross section and that we get 8.16 centimeters squared. Um, and then we can calculate the area moment about X, and we just need to add all these things up. So this is I x one I x two and I x three. And they seem reasonable because, you know, this is gonna pretty high area moment about excuse. It&#x27;s tall this one because it&#x27;s far away from the X axis and this one is bigger than this one and because it&#x27;s much wider. So that looks reasonable. And if we had everything up, we get 39.0 centimeters to the fourth. And then if we want to find the radius of gyration, we just need to take the square root of this divided by this and we get 2.19 centimeters.

Eric Mockensturm · Answer

you can ask to find the area moment and the radius A generation of the shaded region with respect to the X axis. Um, and our our shape looks something like this where X axis goes through the center of this member here and were given all the dimensions that we need in the in the problem. And so I divided it into four regions, 12 in blue and 12 and red. Now we need to find the area moment of each of those regions about this X axis, and we can just start doing that, you know, we can see that the, um in red here, the X axis goes through the central right, So we just get the area. Moment, for that part is simply this expression here that will be used many times. And then for the second Red Member, we have it&#x27;s area moment about its center of mass. Plus, it&#x27;s area times the distance from this, uh, century the center it of this to the central to this and that I called to and again we can get all of these dimensions from the schematic in the in the book and then for um, the radius, Uh, the area moment. About of X one, that Red First Member or the Blue First Member. It turns out that the X axis passes through its central aid and same with this one and that since these two are the same they have they&#x27;re gonna have the same area moment about that excess. And so that&#x27;s just, um 1 12 b one h one where we used blue for a different be one on H one that appear. And we have those values. And we can, um, plug everything in and we get this thing here is to point, um, 67 inches to the fourth. This here is 25 point three inches to the fourth and weaken. This is much bigger than this, because this is farther away from the X axis. Um, and then these two guys both have area moment of nine inches to the four. So we gotta multiplying by two. So we have two of them, and then if we add everything up, we get 46 inches to the fourth, then the radius of gyration is easy, because we can calculate the area of all these rectangles and had them up when we get 18 inches squared, and then we take the square root of this divided by this and we get 1.6 inches.

Problem

Solve the given problems by integration. Using Eq…

Problem 36 Easy Difficulty

Solve the given problems by integration.
Find the moment of inertia with respect to the $y$ -axis of a flat plate that covers the first-quadrant region bounded by $y=e^{x^{3}}, x=1$ and the axes.

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