Question

Use a double integral to compute the area of the following regions. Make a sketch of the region. The region bounded by \(y=1+\sin x\) and \(y=1-\sin x\) on the interval \([0, \pi]\)

Short answer

Question: Find the area of the region bounded by the curves \(y=1+\sin x\) and \(y=1-\sin x\) on the interval \([0, \pi]\). Solution: The given integral is evaluated as \(\int\int_R dA = \int_{0}^{\pi}\int_{1-\sin x}^{1+\sin x} dy\,dx\). When solved, this gives an area of 4 units².

Step-by-step solution

Sketch the region

Start by plotting the two functions, \(y=1+\sin x\) and \(y=1-\sin x\), on the interval \([0, \pi]\). Observe that these are symmetric with respect to the horizontal line \(y=1\). The two functions intersect at \(x=0\) and \(x=\pi\). ![ Graph the two functions. ][1] Now, identify the region enclosed by the two curves and the interval \([0, \pi]\). We can see that the region forms a symmetric shape around the line \(y=1\).

Set up the double integral

Our goal is to compute the area of the region. The area can be found by integrating the difference of the two functions with respect to \(x\) and \(y\), over the region. First, note that the difference of the functions corresponds to the height between them at a given \(x\). So, we can set up the double integral as follows: $$ \text{Area} = \int\int_R dA = \int_{0}^{\pi}\int_{1-\sin x}^{1+\sin x} dy\,dx $$ In the integral above, the first integral, with limits from \(0\) to \(\pi\), is with respect to \(x\). The second integral, with limits from \(1-\sin x\) and \(1+\sin x\), is with respect to \(y\).

Evaluate the double integral

Now, we'll evaluate the double integral: $$ \int_{0}^{\pi}\int_{1-\sin x}^{1+\sin x} dy\,dx = \int_{0}^{\pi}\left[\int_{1-\sin x}^{1+\sin x} dy\right] dx $$ First, perform the integral with respect to \(y\). For this, we'll just subtract the lower bound of the integral from the upper bound: $$ \int_{0}^{\pi}\left[(1+\sin x) - (1-\sin x)\right] dx = \int_{0}^{\pi} 2\sin x\,dx $$ Next, evaluate the integral with respect to \(x\): $$ \int_{0}^{\pi} 2\sin x\,dx = \left[-2\cos x\right]_0^{\pi} = -2\cos{\pi} + 2\cos{0} = 4 $$ So, the area of the region bounded by the given curves and interval is 4 units².

Key concepts

Area of Regions

Calculating the area of a region defined by curves is a fundamental aspect of calculus. Imagine a space enclosed by lines, like a fenced field. To find its exact area, calculus offers tools like double integrals.
This exercise requires us to find the region bounded by the curves defined by the functions \(y = 1 + \sin x\) and \(y = 1 - \sin x\) within the interval \([0, \pi]\).
### Key Steps: - **Sketch the Curves**: Begin by drawing both functions within the specified interval. - **Identify the Bounded Region**: The area of interest is between the functions on the \(x\) interval \([0, \pi]\).- **Set up the Double Integral**: We integrate the difference between the y-values of the two functions, which gives the vertical height from one curve to the other.For more complex regions, these steps remain crucial. The integration limits and function differences must always accurately reflect the enclosed area's properties.

Sine Function

The sine function, denoted as \(\sin x\), is one of the primary functions in trigonometry. It describes the vertical position (or height) relative to a circle's center, translating into oscillating behavior on a graph.
In this problem, the sine function appears as part of the bounds of the region.
### Characteristics of Sine Function: - **Periodic Nature**: The sine function repeats its values every \(2\pi\) radians. - **Symmetry**: It is an odd function, meaning \(\sin(-x) = -\sin(x)\). This symmetry simplifies understanding its behavior across intervals like \([0, \pi]\).- **Range and Domain**: For any real number \(x\), the sine value is always between \(-1\) and \(1\). Knowing these properties helps visualize how \(y = 1 + \sin x\) and \(y = 1 - \sin x\) fluctuate, thus defining the area between them smoothly.

Integral Calculus

Integral calculus is the study of integrals and their properties. It allows us to calculate areas, volumes, and other quantities using the accumulation of values. When dealing with regions bounded by curves, we often use **double integrals** to systematically determine their areas.
### How Double Integrals Work: - **Multiple Integration**: The double integral spans two dimensions, effectively capturing changes in both \(x\) and \(y\) directions. - **Nested Integrals**: As seen in the solution, it involves integrating with respect to \(y\) first and then \(x\). This allows the description of a 2D area beneath the curves. - **Application**: In our exercise, it helps calculate the area between \(y = 1 + \sin x\) and \(y = 1 - \sin x\). The bounds in these integrals determine the region between these oscillating curves.The integral effectively sums up small rectangular strips under the curve, which stretch across the interval \([0, \pi]\). That summed area gives the total area of interest, illustrating one of calculus' powers in geometric computations.