$\iint_{D} 4 x y-y^{3} \; dA$, $ D$ is the region bounded by
$y=\sqrt{x}$ and $y=x^{3} .$
How do we find the bounds for the outer integral? In the solution they sketch a graph and the outer bounds are where $y=\sqrt{x}$ and $y=x^{3}$ intersect. I have two questions:
- Algebraically this can be solved by solving $\sqrt{x} =x^{3}$ which is essentially the problem of finding the roots of $x^3-\sqrt{x}$, right? How can this be done?
- Why is it obvious that the bound of the outer integral is the intersection of those two terms?
Best Answer
Initially, keep in mind that the solution you saw is not the only one possible because you can first integrate in relation to $x$ and then in $y$ or the opposite.
As you can see for a sketch of the graphics of the functions involved, a graphic is above another. However, these graphics meet in two distinct points. Such points are essential to know the limits of the region in question and, consequently, of the integral. To find out these points, just solve the equation $\sqrt{x}=x^{3}$. You can raise both sides of this equation to the square and get the values 0 and 1 (for we are disregarding the possible complex roots).
To solve a double integral in a given region, you must know the limits of this region. The limits of the region are determined by the two curves (the graphs of the two functions). In the internal integral you consider the two functions involved, taking into account which curve is above and which is below (to determine which will be the lower limit and what will be the upper limit of the integral). The external integral will only be the numbers that correspond to the (numerical) limits of the region. If you do the opposite, leaving the expressions of the curves (as functions) in the external integral, the final result of your double integral will depend on the $x$ and $y$ variables and will not be a number, which would be an incongruity.