Centroid:
Find the centroid of the region cut from the f...

Question

Centroid:

Find the centroid of the region cut from the first quadrant by the curve

y = 1/√(x + 1) and the line x = 3.

Accepted Answer

Welcome back everyone. In this problem we want to find the y coordinate of the centroid of the region r located in the first quadrant that is bounded above by the curve y equals 1 divided by the square root of x + 2 on the right by the vertical line x equals 2 and below by the x axis. A says it's 2 plus root 3 multiplied by LN 3 divided by 6. B says it's 1 plus root 3 multiplied by LN 3 divided by 6. C says it's 3 plus root 2 multiplied by LN 2 divided by 8. And D says it's 2 plus root 2 multiplied by LN2 divided by 8. Now, let's try to do a quick sketch of our region just to get a sense of what we're really looking for, OK. So here, if we do a quick sketch of our X and Y axis. Then we're looking for the region r in the first quadrant that's bounded above by the curve y equals 1 plus the square root of x + 2, which would look maybe something like this. OK, on the right by the vertical line x equals 2, which would be this dotted line here. OK. So that's the line x equals 2 and below by the x axis. So we're looking for a region that looks something like this, OK, maybe looks something like this. This would be the region R, and for this region, we are looking for the Y coordinate of the centroid. In other words, we're looking for some y value Y bar. Now how are we going to figure it out? Well, if we clearly define our region, notice that our region are, OK. Is bounded by 0 and 2 on the X axis, OK. And bounded by 0 and 1 divided by the square root of x + 2 on the y axis, OK. And in this region, if we're going to figure out our Y coordinate Y coordinate of the centroid, recall. Now the y coordinate would be equal to the first moment about the x axis divided by the area of our region. So if we can figure out what both of these are, then we should be able to solve for our y coordinate. Now let's start by calculating our area, OK. Recall that the area of our region would be equal to the definite integral between its limits, its bounds, that's A and B of F of X with respect to x, OK, where FX is our function y. So in this case, the area. OK. Is going to be equal to the integral between 0 and 2 of 1 divided by the square root of x + 2 with respect to x. So let's figure this area out. Let's solve by substitution. So let's let you be equal to x + 2. OK, and that means then the derivative of you with respect to x would be 1, so DU equals DX. So we could rewrite or integral that was in terms of X, OK. As one, the integral of one divided by the square root of you with respect to you. And no that would be the integral of u to the negative a half with respect to u, which would be equal to 2 root u plus c. And since we know what u is, we can substitute it and rewrite it as 2 multiplied by root x plus 2 plus c. Since this is our indefinite integral, that means our definite integral, which is the area. OK is going to be equal to 2 multiplied by root x + 2 evaluated between the bones of 0 and 2. Which is going to be 2 multiplied by root 2 + 2. -2 multiplied by the square root of 0 + 2. And which will simplify it to 2 multiplied by root 4, which would be 4 minus 2 root 2. So this is the area of or region R. Now that we have that. Next, let's try to figure out the first moment about the x-axis. What do we know? Well, recall. OK That our region are, or sorry, or moment, the first moment about the x axis iy. is given by the integral between A and B of a half of fx squared with respect to x, where f x is the height function of the region or the curve, and from A to B is the interval on the x axis over which the region extends. So by this formula then. That means our moment i will be equal to the integral between 0 and 2 of a half of 1 minus the square root of x + 2 or squared with respect to x, OK, which we could rewrite as a half of the integral. OK, of 1 divided by x + 2 with respect to x, because 1 divided by the square root of x + 2 all squared would give us 1 divided by x + 2. OK, and now, if we were to integrate this with respect to x, we should find that it equals the natural log of x + 2. So our integral will be a half of the natural log of x + 2 evaluated between the bones of 0 and 2. So now let's try to simplify. So this is going to be equal to then a half of LN2 + 2 or LN4 minus LN2. Sorry, minus LN0 + 2, which would be LN2, OK, and now if we think about it using what we know about logs, we could rewrite the difference between LN4 and LN2 as LN 4 divided by 2. OK, or. Half of Hellen too. This would be the first moment about the X-axis. So now that we have the first moment about the x axis and we have the area, then we can determine the y coordinate of the centroid because now that means it will be equal to our moment a half of LN2 divided by our area. Which is 4 minus 22. Now if we start to simplify, we could rewrite this as a half, well, we could rewrite this as LN 2. OK, divided by 2, multiplied by 2, multiplied by 2 minus 2. Which would be LN 2 divided by 4 multiplied by 2 minus root 2 and now we can rationalize, so we can multiply this by 2 plus root 2. Divided by 2 plus root 2. Now when we go ahead and multiply here, then we end up with 2 plus root 2 multiplied by ln 2. Divided by 4 multiplied by 4 minus 2, and that simplifies the 2 plus root 2 ln 2. Divided by 8. So this is going to be the Y coordinate of the centroid. If we look back at our answer choices, that means D is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Centroid:Find the centroid of the region cut from the first quadrant by the curvey = 1/√(x + 1) and the line x = 3.

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Centroid:
Find the centroid of the region cut from the first quadrant by the curve
y = 1/√(x + 1) and the line x = 3.