Question

Sketch the region and find its area. The region bounded by \(y=2\) and \(y=\frac{1}{\sqrt{1-x^{2}}}\)

Short answer

Answer: The area of the bounded region is \(2\sqrt{3}\) square units.

Step-by-step solution

Finding the Intersecting Points

To find the intersecting points of the two functions, we set them equal to each other and solve for x: \(2 = \frac{1}{\sqrt{1-x^{2}}}\) Now, we solve for x: \(2\sqrt{1-x^{2}} = 1\) \(\sqrt{1-x^{2}} = \frac{1}{2}\) \(1-x^{2} = \frac{1}{4}\) \(x^{2} = \frac{3}{4}\) \(x = \pm\frac{\sqrt{3}}{2}\) The intersecting points are \(\left(\frac{-\sqrt{3}}{2}, 2\right)\) and \(\left(\frac{\sqrt{3}}{2}, 2\right)\).

Sketching the Bounded Region

Sketch the given functions on the coordinate plane: 1. Draw the horizontal line \(y = 2\). 2. Draw the curve \(y = \frac{1}{\sqrt{1-x^{2}}}\), which has a vertical asymptote at \(x = \pm 1\) and approaches the x-axis as \(x\) approaches 0. 3. Mark the intersecting points \(\left(\frac{-\sqrt{3}}{2}, 2\right)\) and \(\left(\frac{\sqrt{3}}{2}, 2\right)\), which define the limits of the bounded region. The bounded region looks like a "cap" shape on the coordinate plane with vertices at the intersecting points.

Setting up the Integral

To find the area of the bounded region, we set up an integral using the difference of the functions: \(Area = \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \left[2 - \frac{1}{\sqrt{1-x^{2}}}\right] dx\)

Evaluating the Integral

Integrate the function within the integral: \(Area = \left[2x - \arcsin{x}\right]_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}\) Plug in the limits of integration: \(Area = \left[2\left(\frac{\sqrt{3}}{2}\right) - \arcsin{\left(\frac{\sqrt{3}}{2}\right)}\right] - \left[2\left(\frac{-\sqrt{3}}{2}\right) - \arcsin{\left(\frac{-\sqrt{3}}{2}\right)}\right]\) Simplify and combine terms: \(Area = \sqrt{3} - \frac{\pi}{3} + \sqrt{3} + \frac{\pi}{3}\) \(Area = 2\sqrt{3}\) Thus, the area of the bounded region is \(2\sqrt{3}\) square units.

Key concepts

Definite Integral

The definite integral is a powerful concept in calculus used to calculate the area under a curve within specific limits. In this exercise, we use it to find the area of a region bounded by two functions. The definite integral provides the accumulation of quantities, such as area, by summing infinitesimal pieces over an interval.

When setting up a definite integral, you take the integral of a function between two points on the x-axis, called limits of integration. In our example, these limits are \(-\frac{\sqrt{3}}{2}\) to \(\frac{\sqrt{3}}{2}\). The integral considers the space between the horizontal line \(y = 2\) and the curve \(y = \frac{1}{\sqrt{1-x^{2}}}\).

Ultimately, by computing the definite integral \(\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \left[2 - \frac{1}{\sqrt{1-x^{2}}}\right] dx\), we obtain the total area of the bounded region.

Area Between Curves

Finding the area between curves involves determining the space enclosed by two or more graphs. This process requires calculating the difference between the top function and the bottom function over a given range.

For this exercise, we are dealing with a bounded region where one function \(y=2\) is constant and the other \(y=\frac{1}{\sqrt{1-x^{2}}}\) forms a curve. To find this area, we subtract the curve from the line over the intersection limits we've already found.

The integral expression \(\int \left[2 - \frac{1}{\sqrt{1-x^{2}}}\right] dx\) captures this subtraction. Here, the top function is \(y=2\) and the curve represents the boundary below it. This method effectively calculates the exact space between curves, producing \(2\sqrt{3}\) as the area.

Intersection of Functions

The intersection of functions refers to points where two graphs meet. These points are crucial when determining the boundaries for area calculations in calculus.

In this exercise, we find the intersecting points by setting the functions equal: \(2 = \frac{1}{\sqrt{1-x^{2}}}\). Solving these equations, we obtain the x-coordinates \(x = \pm\frac{\sqrt{3}}{2}\). Additionally, the y-value remains constant at 2 since the functions meet at this horizontal line.

These intersections, or boundary points, are essential for setting the correct limits on our integral, ensuring that the area calculation is precise and covers only the region of interest.

Integration Techniques

Integration techniques are essential strategies in calculus for solving various types of integrals. Understanding these techniques helps evaluate the area between curves, as illustrated in this exercise.

For this problem, we used basic integration techniques involving the difference of functions. The integral \(\int \left[2 - \frac{1}{\sqrt{1-x^{2}}}\right] dx\) requires knowledge of integrating constants and inverse trigonometric functions.

Specifically, integrating \(\frac{1}{\sqrt{1-x^{2}}}\) results in \(\arcsin(x)\), utilizing a formula from standard integration tables. These strategies allow you to evaluate integrals successfully and accurately find the area within specific bounds. With practice, these techniques become invaluable tools in solving calculus problems efficiently.