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Integration for over regions in the plane

With the continuous learning of the calculus course, it is no longer limited to a single plane. By performing quadratic integration, we have obtained a calculation method that can calculate the volume. From plane to three-dimensional, this is a new breakthrough. Simply put, it is to perform two integral operations, but it brings us new knowledge and challenges.

I really like the kind of "things" that burst out from an angle in front of you (similar to some skills in some games). So I chose two very simple equations to make them have a certain angle: y=2x and y=0.5x. Since the two equations will extend indefinitely, we also give him an interval. x=2 and y=2. Then the bottom of the entire graph has been determined, as shown in the figure:


Then we will divide this area a little bit more carefully, because it is a 2x2 square, then we will divide it into 100 small squares, so that the length of each side is 0.2. We use the point in the upper right corner as the benchmark. This point is within the angle between the two equations, and it belongs to a part of the volume (the point where the blue line passes is also counted, but the point where the green line passes first does not add to the volume. within.)

Then proceed to the next step. Now that we have determined the bottom, we need to find the height of each small square. I chose the equation:

Because I want to try what the root sign will be in this kind of graph in addition to various squares. And it will not cause the point (2, 2) to be too high and difficult to destroy the beauty. So let's first calculate the points obtained in this way, the height of each square and their volume.

Then the overall picture looks like this
Then after calculation, add all the rectangles together (a total of 55 rectangles, but fortunately the table can be directly calculated: D). So in the end, the volume I got was 2.659021.

So let's calculate its real volume, the way should be:
Then since the entire graph is composed of four sides, I divide the equation into two parts to calculate: x from 0 to 1 and x from 1 to 2. The equation expression is：


It can be seen that the number of 2.298 is much less than 2.659, which does not seem to be very close. I have also performed many calculations and checked whether the size and equation of each volume of the model are wrong, but it seems that this is the case. The main reason should be that the height of the entire volume increases as x and y increase. Think of it as a slope. This is different from the example we did before. The previous curve does not increase all the time, but has an ups and downs function. Then if I define it according to the upper right corner of the small square, every cylinder will It is a little bigger than it should be, instead of being big and small, a very similar value can be obtained by complementing each other. So the volume obtained by this method will be larger than its real volume.

This figure should be very interesting, because its center should be a straight line (x=y) but it is not a plane. Unfortunately, due to the limitation of volume, it cannot be displayed. Although it looks like a plane, it does Not a plane.

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