I am trying to understand the graph of $\int_0^1\int_{8y}^1dxdy$. Upon solving the double integral, I get $\int_0^1{1-8y}dy\Rightarrow-3.$ Which I believe is correct, but I'm mainly putting this work in case it matters for the graph or I did something wrong. But upon graphing the boundary equations, I get the following graph https://www.desmos.com/calculator/xofiu5ggqs.
This is what I don't understand. I have not seen this particular case. I'm not sure what happens when a horizontal bound "cuts off" (for lack of a better phrase), a vertical bound like this. So does this double integral solve for the red region or black region or something entirely different? Also, to my understanding, a double integral will only be negative if the region is below the x-axis as well so perhaps my work is incorrect as well. So there's a lot I don't understand about this graph.
Edit: I saw some posts that were deleted by their poster that stated that the result of $-3$ comes from the difference of the triangle formed by the intersection of $x=8y, y=1$ and the red region. Is it proper for this to happen? In other words, is it not acceptable for a double integral to represent the area of two regions or is it simply a unique case?
Best Answer
As you can see from Desmos the boundary of your integral does not describe any valid region since it is doing the black triangle plus the negative of a triangle to the right of the red region. You'll have to ask your professor to correct the typo.