Find the total area of the region enclosed by the curve x = y²...

Question

Find the total area of the region enclosed by the curve x = y²/³ and the lines x = y and y = -1.

Accepted Answer

Hello there. Today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Find the total area of the region enclosed by the curves, X equals Y to the power of 3 divided by 2, and X is equal to 2Y. OK, so it appears for this particular prong, we're asked to determine what the total area of the region enclosed by the curves is, provided that we're told one curve is X is equal to Y 3 divided by 2. And X is equal to 2 Y. So, ultimately we're asked to focus our attention on these two different curves and we're asked to use these curves to find the total area of the region that's enclosed within these curves. So now that we know what we're ultimately trying to solve for, our first step that we need to take based on our prior knowledge is we need to find the limits of integration. So we need to find the Y values because once again we are determining the limits of integration. So let us start to solve for our limits of integration by finding the upper limit. And in order to find the upper limit, as we should recall, we need to set the two curves equal to each other in order to find their intersection point. So we need to take y to the power of 3 divided by 2, and we need to set it equal to 2y. Now we need to take our expression and set it equal to 0, so we need to take y to the power of 3 divided by 2 minus 2y is equal to 0. We now need to factor out our variable y, and when we do, we will have y multiplied by the parentheses of y to the power of 1 divided by 2 minus 2. Is equal to 0. Fantastic. And the reason why we needed to factor out our y is because it will help us determine our two intersection points. So we will determine that y will be equal to 0 and y to the power of 1/2 is equal to 2 will mean that y is equal to 4. So that means without a doubt that the region is bounded between Y is equal to 0 and Y is equal to 4. So once again, this is the area, or rather the region that is bounded between Y equals 0 and Y equals 4. So, our second step now is we need to determine the right and left boundaries, and based on our prior knowledge, we need to recall that for any y that is in the interval 0.4, which is an open interval because there's parentheses. So any y variable in the interval parentheses 0.4, we will determine which function is on the right, which will mean that it will have the larger x value, and in order to do this, we need to test. So let's note that we will test y is equal to 1. So you can go ahead and write that x is equal to 2y. And when we rearrange our expression, or rather plug in our value of 1 in for Y, rather, we will find that x is equal to 2 multiplied by 1, which is equal to 2, and similarly, when X is equal to y to the power of 3 divided by 2 is equal. So once again, we'll note that X is equal to y to the power of 3 divided by 2, and when we plug in a value of 1 in for Y, we'll find that X is equal to 1 of 3 divided by 2, which is simply equal to 1. And we need to note that since 2 is greater than 1, that will mean that the right boundary, which I'll call RB for short, that will mean that the right boundary is X is equal to 2Y, and the left boundary, which I'll call LB, will be X is equal to Y 3 divided by 2. Fantastic. So, our third step is we need to set up our integral, and we will determine that A, the area is equal to the integral evaluated from A to B of parentheses of FY minus G Y. DX and we need to note that FY is the right function and G Y is the left function. So using this formula, we can state that the area can be written as. For this particular problem, A, the area is equal to the integral evaluated from 0 to 4 of 2Y minus y to the power of 3 halves DY. And our fourth step now is we need to evaluate our integral. So we need to take the integral evaluated from 0 to 4 2Y minus Y 3 divided by 2. Dy, which will be equal to parentheses of 2 y to the power of 2 divided by 2 minus y to the power of 3 divided by 2. Plus one. Divided by. 3 divided by 2. Plus 1 and we need to note once again that this expression will be evaluated from 0 to 4. OK, we are on a roll here. So when we begin our evaluation, we will find that it's equal to y to the power of 2. Minus Y to the power of 5 divided by 2 divided by 5 divided by 2, and we need to evaluate our expression from 0 to 4. Which will be equal to parentheses of y2 minus 2 multiplied by Y 5 divided by 2, divided by 5, evaluated from 0 to 4. And this will be equal to parentheses of 4 to 2 minus 2 multiplied by parentheses of 4, and then it's going to be 2 multiplied by 4, and 4 will be to the power of 5 divided by 2. All divided by 5 minus parentheses of parentheses of 0 to the power of 2 minus 2 multiplied by parentheses of 0 to the power of 5 divided by 2 divided by 5, and when we simplify this expression and plug it into our calculator and we write it as a fraction, we will find that our final answer is 16. Divided by 5 and our units will be in square units. And that's it. We've officially solved for this problem, and it's in square units because we're dealing with the area, and that's it. We've solved for this problem. Hooray, we did it. So going back to look at the problem itself to recap what we've learned, we can safely say without a doubt that our final answer once again. Without a doubt will be 16 divided by 5 square units, and that's it. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Find the total area of the region enclosed by the curve x = y²/³ and the lines x = y and y = -1.

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