Problem 88 - Thomas Calculus

Question

Area Find the area of the "triangular" region that is bounded from above and below by the curves \(y=\csc x\) and \(y=\sin x,\) \(\pi / 6 \leq x \leq \pi / 2,\) and on the left by the line \(x=\pi / 6\) .

Step-by-Step Solution

Answer

The area is \(\ln(2 + \sqrt{3}) - \frac{\sqrt{3}}{2}\).

Step 1: Graph the Functions

First, graph both functions: - The function \( y = \csc x \) is the reciprocal of \( y = \sin x \), so it is undefined whenever \( \sin x = 0 \), but for \( \pi/6 \leq x \leq \pi/2 \), it is defined and positive. - The function \( y = \sin x \) is a standard sine wave.Sketching these over the interval \( \pi/6 \leq x \leq \pi/2 \), you'll notice \( y = \csc x \) lies above \( y = \sin x \).

Step 2: Set up the Integral

To find the area between two curves, use\[\int_{a}^{b} [f(x) - g(x)] \, dx\]Here \( f(x) = \csc x \) and \( g(x) = \sin x \), over \( \pi/6 \leq x \leq \pi/2 \).

Step 3: Integrate the Function

Compute the integral:\[\int_{\pi/6}^{\pi/2} (\csc x - \sin x) \, dx\]This results in two separate integrals:\[\int_{\pi/6}^{\pi/2} \csc x \, dx - \int_{\pi/6}^{\pi/2} \sin x \, dx\]

Step 4: Evaluate the Integral of \(\csc x \)

The integral of \( \csc x \) is \(-\ln|\csc x + \cot x| \) + C. Evaluate it from \( \pi/6 \) to \( \pi/2 \):\[-\ln|\csc(\pi/2) + \cot(\pi/2)| + \ln|\csc(\pi/6) + \cot(\pi/6)|= -\ln(1) + \ln(2 + \sqrt{3})= \ln(2 + \sqrt{3})\]

Step 5: Evaluate the Integral of \(\sin x \)

The integral of \( \sin x \) is \(-\cos x\). Evaluate it from \( \pi/6 \) to \( \pi/2 \):\[-\cos(\pi/2) + \cos(\pi/6)= 0 + \frac{\sqrt{3}}{2}= \frac{\sqrt{3}}{2}\]

Step 6: Calculate the Area by Subtraction

Subtract the result of Step 5 from Step 4:\[\ln(2 + \sqrt{3}) - \frac{\sqrt{3}}{2}\]

Step 7: Finalize the Answer

The area of the triangular region is given by:\[\ln(2 + \sqrt{3}) - \frac{\sqrt{3}}{2}\]

Key Concepts

Area Between Curves

Finding the area between curves is a common problem in calculus, especially when dealing with functions defined over a specific interval. The goal is to determine the size of the region trapped between these curves. To do this, we typically use definite integrals.

If you have two continuous functions, say \( f(x) \) and \( g(x) \), where \( f(x) \) is above \( g(x) \) within the interval \([a, b]\), the area between the curves is computed using the formula:

• \[ \int_{a}^{b} [f(x) - g(x)] \, dx \]

This formula works by calculating the total area under the top curve \( f(x) \) and subtracting the area under the bottom curve \( g(x) \) over the same interval.

In the exercise, where the curves \( y = \csc x \) and \( y = \sin x \) are given, we apply this principle. The interval of interest is \( [\pi/6, \pi/2] \), and it is important to ensure that \( y = \csc x \) lies above \( y = \sin x \) in this range.

Integral Calculus

Integral calculus is a branch of calculus focused on the accumulation of quantities, such as areas under curves, and the calculation of integrals is at its heart. An integral aggregates parts to find the whole, making it useful for finding areas and volumes.

When dealing with finding the area between curves, we are essentially finding the integral of the difference between two functions over a specified interval. This difference, represented as \( f(x) - g(x) \), is integrated over the interval \([a, b]\).

In the given problem, integral calculus was used twice to find:

• The integral of \( \csc x \), known to be \(-\ln|\csc x + \cot x| + C\).
• The integral of \( \sin x \), which is \(-\cos x + C\).

These integrals need to be evaluated over the interval \( \pi/6 \leq x \leq \pi/2 \), and their respective results are subtracted to find the total area between the curves.

Trigonometric Functions

Trigonometric functions are at the core of many problems in calculus, especially those involving periodic behavior. Here, the functions \( y = \csc x \) and \( y = \sin x \) are a reciprocal pair in trigonometric terms.

The sine function, \( y = \sin x \), is a basic trigonometric function that oscillates between -1 and 1. It forms a smooth curve and is often used to model wave-like phenomena due to its periodic nature.

The cosecant function, \( y = \csc x \), is the reciprocal of the sine function. This means it is undefined whenever \( \sin x = 0 \). Cosecant has a distinct shape, often forming steep peaks and valleys, and lies above the sine function wherever defined within \( [\pi/6, \pi/2] \).

In the problem, understanding the behavior and graphs of these trigonometric functions is crucial for correctly setting up and interpreting the integral. By understanding the interval where these functions are investigated, we ensure the accurate application of calculus to find the area between them.