Question

Use double integrals 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 \([0, \pi]\) using double integrals. Answer: The area of the region bounded by the curves \(y=1+\sin x\) and \(y=1-\sin x\) on the interval \([0, \pi]\) is 4.

Step-by-step solution

Sketch the region

Begin by sketching the region. The curves intersect when \(1+\sin x = 1-\sin x\). Solving for \(x\), we get \(\sin x = 0\) and hence, \(x = 0, \pi\). So, the region we are interested in is as shown below: [asy] size(280); import graph; usepackage("amsmath"); real min = 0; real max = pi; real up(real x) {return 1+sin(x);} real down(real x) {return 1-sin(x);} draw(graph(up,min,max)^^graph(down,min,max), linewidth(1)); draw((min,0)--(max,0),EndArrow); draw((0,-1)--(0,3),EndArrow); label("\(x\)",(max,0),E); label("\(y\)",(0,3),N); fill(graph(up,min,max)--graph(down,max,min)--cycle, rgb(0.9,0.9,0.9)); [/asy] We can split this region into two regions A and B (symmetric about the x-axis) and just find the area of region A and multiply the result by two: [asy] size(280); import graph; usepackage("amsmath"); real min = 0; real max = pi; real up(real x) {return 1+sin(x);} real down(real x) {return 1-sin(x);} draw(graph(up,min,max)^^graph(down,min,max), linewidth(1)); draw((min,0)--(max,0),EndArrow); draw((0,-1)--(0,3),EndArrow); label("\(x\)",(max,0),E); label("\(y\)",(0,3),N); fill(graph(up,min,max)--(pi,1)--cycle, rgb(0.9,0.9,0.9)); label("\(A\)",(pi/2,1),NE); [/asy]

Set up the double integral

To find the area of region A, we set up a double integral as follows: $$ 2\cdot\text{Area}(A) = 2\int\int_{A} dA $$ We can integrate first with respect to \(y\) in the range \([1, 1+\sin x]\) and then with respect to \(x\) in the range \([0, \pi]\). So the double integral becomes: $$ 2\cdot\text{Area}(A) = 2\int_{0}^{\pi}\int_{1}^{1+\sin x} dy dx $$

Evaluate the integral

Now, let's evaluate the double integral: $$ 2\cdot\text{Area}(A) = 2\int_{0}^{\pi}\left[\int_{1}^{1+\sin x} dy\right] dx = 2\int_{0}^{\pi} (\sin x)dx $$ To evaluate the integral with respect to \(x\), we compute: $$ 2\cdot\text{Area}(A) = 2 \left[-\cos x\right]_0^\pi = 2(-\cos \pi+1) = 4 $$ So, the area of the region bounded by \(y=1+\sin x\) and \(y=1-\sin x\) on the interval \([0, \pi]\) is 4.

Key concepts

Integration

Integration is a fundamental concept in calculus which provides a way to calculate things like areas under curves, total accumulated quantities, and other concepts that involve summing up infinitesimal pieces. In the context of finding areas, integration enables us to add up an infinite number of rectangles with infinitesimal width to approximate the area under a curve.

When computing the area between curves, we subtract the integral of the bottom function from the integral of the top function within specific limits. In the exercise provided, the limits are defined by the functions themselves, where integration takes place within a confined space - from one curve to the other and within a set interval on the x-axis.

Calculus

Calculus is a vast field of mathematics that involves studying change and motion. Within calculus, there are two main branches: differential calculus and integral calculus. Differential calculus concerns the rate at which quantities change, while integral calculus concerns the accumulation of quantities, as in the case of determining areas.

The double integrals we talk about in the given problem fall under the umbrella of integral calculus. While a single integral computes the line (1D) or area (2D) under a curve, double integrals extend this idea to calculate volumes under surfaces or area for irregularly shaped plane regions, which is precisely the concept applied to find the area in this exercise.

Symmetric Regions

Symmetric regions can highly simplify the integration process. A region is symmetric with respect to an axis if one half is a mirror image of the other. When dealing with these types of regions, we can often calculate the area or volume of just one side and then double the result to obtain the full measurement.

In our scenario, the region between the curves is symmetrical around the y-axis. This symmetry is recognized by observing that for every point on the upper curve, there is a corresponding point on the lower curve at the same x but with an equal and opposite y value. The use of symmetry not only simplifies the integrals we need to solve but also reduces computation time by taking advantage of the geometric properties of the region.

Area of Bounded Regions

Computing the area of bounded regions is a practical application of double integrals. A bounded region in the plane is one that is completely contained within certain bounds, which in many problems are defined by functions. The double integral then becomes a tool to calculate the size of these regions.

In the given exercise, the area is bounded by the functions \(y=1+\sin x\) and \(y=1-\sin x\) as well as the vertical lines \(x=0\) and \(x=\pi\). By setting up a double integral, we can accurately define this space and evaluate the integral to determine the total area. The process illustrates how calculus is not just about dealing with abstract concepts but also about solving real-world problems that involve space and quantities.