91Ó°ÊÓ

The region of integration is a crucial concept in double integration. It helps us identify where we need to integrate a function in the plane. For the given exercise, the region is bounded by three curves: \( y = \sqrt{x} \), \( y = 2 \), and \( x = 0 \). These curves form a specific area where our function will be integrated.
When graphing these inequalities, \( y = \sqrt{x} \) represents a curve that starts from the origin and increases as \( x \) increases. The line \( y = 2 \) is a horizontal line that cuts across at \( y=2 \). Lastly, \( x = 0 \) is the y-axis.
As you analyze this region, you'll notice it's a triangular area with vertices at points (0,0), (4,2), and (0,2). The upper boundary is the line \( y = 2 \), the right boundary comes from the curve \( y = \sqrt{x} \), and the left boundary is the y-axis. It’s important to recognize this physical region because it dictates the limits of our integration when setting up the double integral.

Changing variables is a technique often used in calculus, especially in integration, to simplify the integrals involving complex functions. In our exercise, we use substitution as a change of variables to make the integration simpler.
In step four of the solution, we substitute \( u = y^3 \). This substitution comes in handy to simplify the power within the sine function and transform the integral into an easier form.
When changing the variable, do not forget to modify the limits of the integration. Originally the limits for \( y \) were from 0 to 2. Once you substitute \( u = y^3 \), these limits become 0 to 8, because when \( y = 0 \), \( u = 0^3 = 0 \) and when \( y = 2 \), \( u = 2^3 = 8 \).
Alongside substituting \( u = y^3 \), we also find \( du \). By deriving \( u = y^3 \), we find \( du = 3y^2 \, dy \), which means \( y^2 \, dy = \frac{1}{3}du \). This relation transforms the integral into a new variable that’s easier to compute.

Integration by substitution is a useful process to adapt complex integrals into simpler ones. It helps especially when direct integration is tough. In double integrals involving functions like \( \sin(y^3) \), substitution can make our life easier.
In the provided solution, once we reach the integral \( \int y^2 \sin(y^3) \, dy \), we use substitution to handle \( y^3 \) in the sine function. Substituting \( u = y^3 \) gives us a direct path to transform the integral. We already found out that \( y^2 \, dy = \frac{1}{3} \, du \), letting us simplify to \( \int \frac{1}{3} \sin(u) \, du \).
The integral is much more straightforward to solve now. The integration of \( \sin(u) \) is a standard one resulting in \( -\cos(u) \). This shows how substitution merely changes one challenging integration into a familiar, solvable equation. Completing this step finalizes our integration process smoothly, delivering us the solution efficiently.