27–33. Multiple regions The regions R₁,R₂, and R₃ (see figure)...

Question

27–33. Multiple regions The regions R₁,R₂, and R₃ (see figure) are formed by the graphs of y = 2√x,y = 3−x,and x=3.

Find the area of each of the regions R₁,R₂, and R₃.

Graph showing regions R₁, R₂, and R₃ formed by curves and lines, with labels indicating their areas.

Accepted Answer

Hello. In this video, we are giving the regions R1, R2 and R3, which are shown in the image below. We want to determine the area of each region. So, let's go ahead and start this problem by determining the area of region 1. Now, region 1 is this small sliver that is trapped between the blue line and the red curve. Now, in order to find this area, we are going to take the area of a definite integral from A to B, of our upper curve, minus our lower curve, which means that we are going to be integrating with respect to X. Now, the upper curve in this case is defined as 4 minus X, and the lower curve, which is the red curve, is defined as 3 square root of X. Furthermore, the values of X at these boundaries are going to be 0 and at 1. So we can define the area of R1 to be the integral from 0 to 1 of 4 minus X minus 3 square root of XD X. Now, here, what we can go ahead and do is we can go ahead and take the antiderivative each term directly. That is going to leave us with 4 X minus 1/2 X squared minus. -2 X3 have power. And this is going to be evaluated from 0 to 1. Now, all we have to do in this case is we just need to plug in the boundaries. If we plug in one, we get 4 minus 1/2 minus 2, and if we were to subtract the value of the lower boundary, which is 0, we would get the value of 0. And combining the like terms together is going to give us an area of 3 halves. So this is going to be the area of the R1 region. Now, let's go ahead and find the area of the R3 region. Now, R3 is going to be calculated in a similar way to R1. We are going to define our area as a definite integral of an upper minus lower curve, and that is going to be with respect to X. Now, the bounds of integration are going to exist at this intersecting point between the red and blue curve, which is at 1, and the very end of this boundary, which is going to be at X's equal to 4. So, we can go ahead and define the area as an integral from 1 to 4. Now, in this case here, the upper curve is going to be this red curve, which is 3 square root of X. Then we will subtract the lower curve, which is the quantity 4 minus X, and integrate with respect to X. Now, what we're going to do is we're going to take the antiderivative of each of the terms. So by tying in the antiderivative of each of the terms, we will be left with 2 multiplied by X to the 3 halves power, minus 4 X. Plus 1/2. X squared And this will be evaluated from 1 to 4. Now, if we were to plug in the boundaries, we will have 2 multiplied by 4, raise the 3 halves power. -4 multiplied by 4. Plus 1/2 multiplied by 4 squared, then we will subtract the value of the lower boundary, which is just going to leave us with 2 minus 4 plus 1/2. Then we will go ahead and simplify, and this is going to simplify to be 8 minus -3 halves, which is going to then simplify further to be 19 divided by 2. So this is going to be the value of the area of the R3 region. Now, let's go ahead and calculate the area of the archer region. Now, if we look at the R2 region, if we want to integrate with respect to X, we will need to split up this area because the upper curve is defined differently before and after positive one. Now the upper curve is defined from 0 to 1 as 3 square root of X and is defined as 4 minus X from 1 to 4. But instead of having to take the sum of two different integrals, why don't we go ahead and integrate with respect to Y instead? So, if we integrate with respect to Y, then the area is going to be defined as our boundaries from A to B, of our rightmost curve, minus our leftmost curve, D Y. Now, in this case here, the rightmost curve is going to be defined as our blue line, Y equal to 4 minus X. But since we are integrating with respect to Y, we are going to rewrite X into terms of Y. By rewriting this, we can represent this function as X is equal to 4 minus Y. And the left curve in this case is going to be Y is equal to 3 square root of X, but if we were to solve for X in terms of Y, we can rewrite this as X is equal to 19th. Why squared Then our boundaries are going to be in terms of Y. At the intersection point here, Y has a value of 3, and at the lowest point here, Y has a value of 0. So what this means is that the area is going to be defined as the integral from 0 to 3 of 4 minus Y minus 19 Y 2 D Y. And all we have to do in this case is integrate. So, if we take the antiderivative of each of the terms directly, we will be left with 4 Y minus 12 Y squared. Minus 1 divided by 27 Y cubed, and this will be evaluated from 0 to 3. Then by plugging in the boundaries, we will be left with 4 multiplied by 3, minus 12 multiplied by 3 squad, minus 1 divided by 27, multiplied by 3 cubed. Then we will subtract the value of the lower boundary, but if we plug in zero into the antiderivative, we just get 0. So we just need to simplify now. This is going to simplify to be 12 minus 9 halves minus 1, which can combine to be 13 halves, and this is going to be the area of the R2 region. So, just to summarize, the area of the R1 region is 3 halves. The area of the R2 region, when we integrated with respect to Y is 13 halves, and the area of the R3 region is 19 halves. This is going to be the solutions to the given problem. And with that being said, the overall solution is going to be a. So I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

27–33. Multiple regions The regions R₁,R₂, and R₃ (see figure) are formed by the graphs of y = 2√x,y = 3−x,and x=3. Find the area of each of the regions R₁,R₂, and R₃.

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27–33. Multiple regions The regions R₁,R₂, and R₃ (see figure) are formed by the graphs of y = 2√x,y = 3−x,and x=3.
Find the area of each of the regions R₁,R₂, and R₃.