To solve for the area under the curve $y = 3\sin(x)$ and above the curve $y = 3\cos(x)$ in the interval $[0, 2\pi]$, we first need to determine the points of intersection, then integrate the difference between the two functions over the appropriate intervals.

Step 1: Find points of intersection

The points where the curves intersect are determined by setting $3\sin(x) = 3\cos(x)$, or equivalently: $\sin(x) = \cos(x)$ Dividing both sides by $\cos(x)$ (where $\cos(x) \neq 0$) gives: $\tan(x) = 1$ The solution to $\tan(x) = 1$ in the interval $[0, 2\pi]$ is: $x = \frac{\pi}{4}, \frac{5\pi}{4}$

Step 2: Set up the integrals

Since the curves intersect at $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$, we will integrate the difference between $3\sin(x)$ and $3\cos(x)$ over the intervals:

1. From $x = 0$ to $x = \frac{\pi}{4}$, $3\cos(x)$ is above $3\sin(x)$.
2. From $x = \frac{\pi}{4}$ to $x = \frac{5\pi}{4}$, $3\sin(x)$ is above $3\cos(x)$.
3. From $x = \frac{5\pi}{4}$ to $x = 2\pi$, $3\cos(x)$ is above $3\sin(x)$.

So the total area is: $A = \int_0^{\frac{\pi}{4}} (3\cos(x) - 3\sin(x))\, dx + \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} (3\sin(x) - 3\cos(x))\, dx + \int_{\frac{5\pi}{4}}^{2\pi} (3\cos(x) - 3\sin(x))\, dx$

Step 3: Compute the integrals

Let's calculate each integral separately.

First integral: $\int_0^{\frac{\pi}{4}} (3\cos(x) - 3\sin(x))\, dx$

$\int_0^{\frac{\pi}{4}} 3\cos(x)\, dx = 3\sin(x) \Big|_0^{\frac{\pi}{4}} = 3\left(\frac{\sqrt{2}}{2}\right) - 0 = \frac{3\sqrt{2}}{2}$ $\int_0^{\frac{\pi}{4}} 3\sin(x)\, dx = -3\cos(x) \Big|_0^{\frac{\pi}{4}} = -3\left(\frac{\sqrt{2}}{2}\right) + 3 = 3 - \frac{3\sqrt{2}}{2}$ So the first integral is: $\frac{3\sqrt{2}}{2} - \left(3 - \frac{3\sqrt{2}}{2}\right) = 3\sqrt{2} - 3$

Second integral: $\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} (3\sin(x) - 3\cos(x))\, dx$

This is symmetric to the first one but over the interval where $3\sin(x)$ is above, so it gives the same value: $3\sqrt{2} - 3$

Third integral: $\int_{\frac{5\pi}{4}}^{2\pi} (3\cos(x) - 3\sin(x))\, dx$

This is symmetric to the first integral again: $3\sqrt{2} - 3$

Step 4: Add the results

The total area is the sum of these integrals: $A = (3\sqrt{2} - 3) + (3\sqrt{2} - 3) + (3\sqrt{2} - 3)$ $A = 3(3\sqrt{2} - 3) = 9\sqrt{2} - 9$

Thus, the area under $y = 3\sin(x)$ and above $y = 3\cos(x)$ over the interval $[0, 2\pi]$ is $9\sqrt{2} - 9$.

Would you like further details on any step or have any questions?

Here are 5 follow-up questions to extend the concept:

1. What is the area under $y = \sin(x)$ and above $y = \cos(x)$ in the same interval?
2. How do you calculate the area between two curves when the curves switch dominance more than once?
3. What happens to the area if we change the limits of integration to $[0, \pi]$?
4. How does the method change if we were working with parametric curves instead of functions?
5. How would you set up this problem if the functions involved were not trigonometric?

Tip: Always visualize the functions when solving area problems between curves—it helps identify key points like intersections!