Find the area of the region described in the following exercis...

Question

Find the area of the region described in the following exercises.

The region in the first quadrant bounded by y=x^2/3 and y=4

Accepted Answer

Hello. In this video, we are going to be determining the area of the region in the first quadrant bounded by Y is equal to X2 divided by 2, and Y is equal to 8. So, in order to approach this problem, let's go ahead and first start by drawing an axis. So, we are trying to find some area in the first quadrant, and with respect to the axis, the first quadrant is going to be the quadrant located to the upper right. Now, the area we are interested in is bounded by two equations. The first equation is Y equal to X2 divided by 2, or it can be rewritten as 1/2 X2, and the second equation is given to us as Y is equal to 8. Now, Y is equal to 8 is quite simple to graph here. It is a horizontal line that passes through the value of 8 on the y axis. But 1/2 X squared is a positive parabola that is located at the origin and opens positively upward. So, since we are only interested in the area that is located in the first quadrant and the area that is bounded by these two curves, the area we're interested in is going to be this bounded area between the three regions. Now, we need to come up with an integral that is going to help us solve for this area, but in order to find this integral, the first thing we need to do is determine the bounds of the integral. So, in order to determine the area, we are going to integrate with respect to X. Now, in this case here we know that our area starts at the origin where X is equal to 0, but where is this area going to end? Well, in order to figure that out, we can find the point of intersection with respect to the X axis. So, in order to solve for this value of X, we can take both of the given equations, Y is equal to 1/2 X2 and Y is equal to 8, set them equal to each other in terms of X and solve. So by doing this, we will have 1/2 X 2 is equal to 8. Now, let's go ahead and solve for X. We will multiply both sides of this equation by 2, leaving us with X2 equal to 16. And by taking the square root of both sides of this equation, we will be left with X's equal to positive and negative 4. Now, in this case here, because 4 is the only value that's located in quadrant 1, we know that this intersection point between the two equations is located where X is equal to 4. And so what this means is that when we set up the integral, we are going to be integrating from 0 to 4. Now, what about finding the actual integrand? Well, in this case here, we always want to take the larger area and subtract the smaller area from it. So in order to find this area within the region, we are going to take the bigger radius, which is the distance from the X-axis to the higher graph, and subtract the area from the X-axis to the lower equation. So in this case here, the bigger equation is Y is equal to 8. So we will have 8 minus the smaller radius, which is going to be the distance from the X axis to this first equation, and again, this is the parabola, so this is going to be 1/2 X squared. So we will subtract 1/2 X squad. From this bigger radius, and this is going to be the integral that allows us to find the area between the three regions. And so now what we need to do is we just need to solve this integral. By taking the antiderivative of each term within the in end, we will be left with 8 X minus 16 X cubed, and this is going to be evaluated from 0 to 4. By using the fundamental theorem part 2, we will plug in our bigger boundary, giving us 8 multiplied by 4, minus 1/6 multiplied by 4 cubed. Then we will subtract the value of the lower boundary, but notice that our lower boundary is 0. If we plug in 0 to every term within our antiderivative, we will just be left with 0. So, all we need to do now is algebraically simplify. 8 multiplied by 4 is going to give us 32. And 4 cubed divided by 6 is going to simplify to be 64 divided by 6, and combining these two values together is going to leave us with 64 divided by 3, which is going to be the area that is bounded by the three curves. And with that being said, the overall solution to this problem is going to be D. So I hope this video helps you in understanding how to approach this type of problem, and we will go ahead and see you all in the next video.

Find the area of the region described in the following exercises.The region in the first quadrant bounded by y=x^2/3 and y=4

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Find the area of the region described in the following exercises.

The region in the first quadrant bounded by y=x^2/3 and y=4