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Area under a region

 

If you want to find the area under a region, you can draw rectangles and add the area of the rectangles together to find an aproximation. If you want to find the exact area, you use an integral bounded by the region. But what about when we introduce a 3rd plane? How do we aproximate the volume under a region when we consider all three planes? The concept is very similar. To aproximate the volume, we use rectangles but add the element of height and sum the rectangles volumes. For the actual volume, we use a double integral (using Fubini's Theorem) bounded by the region.
Consider the region contained in the intersection of \[y=x^\frac{1}{2}\] \(and\) \[y=x^3.\]

We are going to aproximate and find the actual volume of this region under the function \[f(x,y)=xy.\]

We're going to find the aproximated area using squares, so let's zoom in a little bit until we have a 5x5 area for a total of 25 squares to work with.

Do you see the 5x5 section of squares that outline the region contained by the two functions? Each block is 0.2x0.2 wide/long. What we want to do is look at the top right corner of each square in the grid. For any top right corner of a square that is on the line of a function or contained inside the region of the functions, we notate them.

What are the coordinates of these points? We will need them for the function \(f(x,y)=xy.\)

Now, we simply plug these coordinates into the \(f(x,y)=xy\) function and input these numbers into a 5x5 table on google sheets. This table will make it easier to visualize what's going on as we input these numbers as heights on onshape to represent whats happening in 3-d. These are the heights of each square, now turned into rectangular prisms. To find the volume of these prisms, we take the length (0.2) times the width (0.2) of each square and multiply it by the height and take the total sum to get the aproximated voolume of this region, noted below the table.

The representation on Onshape is done in a 1:1 scale using inches as metrics, meaning each square has side length and width 0.2 and the heights are exactly as they are in the table.

To find the actual volume of the region we use a double integral. The outside integral is the integral from x=0 to x=1, or the boundary of the region with respect to the x-axis. The inner integral is the integral from \(y=x^3\) to \(y=x^\frac{1}{2}\) or the area bounded by the bottom equation up to the area bounded by the upper equation. This double integral gives us the volume of the region in all 3 planes by the function \(f(x,y)=xy\) contained or bounded by the functions in the 2-dimensions with respect to the y-axis contained in-between the functions and then the x-axis in-between the inner and outer-most point from the origin along the x-axis. \[\int_{}^{}\int_{D}^{}f(x,y)dA\] \[\int_{}^{}\int_{R}^{}f(x,y)dydx\] \[\int_{a}^{b}\int_{c}^{d}f(x,y)dydx\] \[\int_{a}^{b}\int_{g1(x)}^{g2(x)}f(x,y)dydx\] \[\int_{0}^{1}\int_{x^3}^{x^\frac{1}{2}}xy dydx\]

So, the aproximated volume was 0.1328 cubic units and the actual volume is 0.1041 cubix units. I chose these functions because it would be contained within a 1x1.5 cubic inch volume for easy scaling (in this case no need to scale at all) in Onshape for 3-d printing and representation. I also like that the numbers were small and easy to work with once you wrap your head around how small the region really is. It adds depth and understanding beyond what I've ever known when we look at that region in all 3 planes and incoorperate a function containing x and y terms. We've all seen or could imagine this region on a graph, but the actual scale of it is so much smaller than it seems in a 2-d plane, which is why I like it.

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