Problem 40 - Single Variable Calculus

Question

In Exercises \(9-40\), sketch the region bounded by the graphs of the given equations and find the area of that region. $$ x=\sin y+\cos 2 y, \quad x=0, \quad y=0, \quad y=\frac{\pi}{2} $$

Step-by-Step Solution

Answer

The region is bounded by the points \((0,0)\), \((0, \frac{\pi}{2})\), and \((1,0)\), and is enclosed by the equations \(x = \sin y + \cos 2y\), \(x = 0\), \(y = 0\), and \(y =\frac{\pi}{2}\). By integrating the equation \(x = \sin y + \cos 2y\) with respect to \(y\), from \(0\) to \(\frac{\pi}{2}\), we find that the area of the region is 1 square unit.

Step 1: Identify the points of intersection

In order to find the points of intersection between the given equations, we need to solve the systems of equations formed by taking two equations at a time. Let's consider the following systems: 1. \(x = \sin y + \cos 2y\) and \(x = 0\) 2. \(x = \sin y + \cos 2y\) and \(y = 0\) 3. \(x = \sin y + \cos 2y\) and \(y =\frac{\pi}{2} \) Now let's solve them: 1. Setting \(x = 0\) in the first equation, we get \(0 = \sin y + \cos 2y\). Now we'll find the values of \(y\) at which this equation holds true. Notice that the equation is already true for \(y=0\) and \(y=\frac{\pi}{2}\), so we found two intersection points: \((0,0)\) and \((0,\frac{\pi}{2})\). 2. Setting \(y = 0\) in the first equation, we get \(x = \sin 0 + \cos 2(0) = 1\). Thus, the intersection point between these two equations is \((1, 0)\). 3. Setting \(y = \frac{\pi}{2}\) in the first equation, we get \(x = \sin(\frac{\pi}{2}) + \cos(\pi) = 1 -1 = 0\). The intersection point is \((0, \frac{\pi}{2})\), but we already identified this intersection in the first pair of equations. So, the points of intersections are \((0,0)\), \((0, \frac{\pi}{2})\), and \((1,0)\).

Step 2: Sketch the region

Now we will sketch the region bounded by the given equations. We already found the intersection points, so we will plot these points and draw the lines for the other equations. 1. The \(x\) axis corresponds to \(y = 0\)and the \(y\) axis corresponds to \(x = 0\). The curve \(x = \sin y + \cos 2y\) starts at \((0,0)\) and intersects the \(y\) axis (\(x = 0\)) at \((0, \frac{\pi}{2})\). 2. The curve then reaches the \(x\) axis (\(y = 0\)) again at \((1,0)\). Now, we connect these points with smooth curves, making sure that the curve follows the equation \(x = \sin y + \cos 2y\).

Step 3: Calculate the area of the region

To calculate the area enclosed by these equations, let's integrate the equation \(x = \sin y + \cos 2y\) with respect to \(y\), from \(0\) to \(\frac{\pi}{2}\): Area = \(\int_{0}^{\frac{\pi}{2}} (\sin y + \cos 2y) dy = [\cos y + \frac{1}{2} \sin 2y]_0^\frac{\pi}{2} = (\cos(\frac{\pi}{2}) + \frac{1}{2} \sin (\pi)) - (\cos (0) + \frac{1}{2} \sin (0)) = (0 + 0) - (1 + 0) = -1\). However, we are interested in the absolute value of the area, so the actual area of the region is 1 square unit.

Key Concepts

Points of Intersection

When finding the area between two curves, identifying the points of intersection is the first essential step. These points tell us where the curves intersect each other, forming the boundaries of the enclosed region. In this exercise, we had three equations to consider:

• Equation 1: \(x = \sin y + \cos 2y\)
• Equation 2: \(x = 0\)
• Equation 3: boundary lines given by \(y = 0\) and \(y = \frac{\pi}{2}\).

To find the points of intersection, we substitute one equation into another, and solve for variable \(y\). By doing this, we successfully determine that the intersection points are \((0,0)\), \((0, \frac{\pi}{2})\), and \((1, 0)\). These points will act as the endpoints of our region, making them crucial for the integration step.

Integration

Integration is the key technique used to calculate the area between curves. In calculus, integration allows us to sum up an infinite number of infinitesimally small rectangles under a curve to find the total area. For this problem, the area of the region is found by integrating the function \(x = \sin y + \cos 2y \) with respect to \(y\) over the interval \([0, \frac{\pi}{2}]\). By performing integration, we essentially capture the total vertical distance between the curve and the \(y\)-axis. The definite integral set up is:\[\int_{0}^{\frac{\pi}{2}} (\sin y + \cos 2y) \, dy\]Solving this integral provides the area of the bounded region. After solving, we found out the area was -1, highlighting the importance of interpreting results. Since area cannot be negative, we took the absolute value to correctly interpret it as 1 square unit.

Trigonometric Functions

Trigonometric functions, namely sine and cosine, play a pivotal role in this exercise. The equation \(x = \sin y + \cos 2y\) is a representation involving these trigonometric functions, which frequently appear in calculus due to their periodic properties.

• The sine function \(\sin y\) fluctuates between -1 and 1 as \(y\) varies.
• The cosine function \(\cos 2y\) oscillates faster due to the "2" inside the function, modifying the input frequency.

Together, these functions determine the shape of the curve in question as \(y\) changes from \(0\) to \(\frac{\pi}{2}\). Trigonometric functions can create non-linear and sometimes complex shapes, thus the intersection with simple lines like \(x = 0\), \(y = 0\), and \(y = \frac{\pi}{2}\) helps delineate the area for calculation.

Boundaries in Calculus

In calculus, boundaries are vital when discussing regions and areas. They define the limits within which we perform our integrations. For this problem, several boundaries were used:

• \(x = \sin y + \cos 2y\) represents a curve boundary.
• \(x = 0\) defines a vertical boundary aligning with the \(y\)-axis.
• The horizontal boundaries \(y = 0\) and \(y = \frac{\pi}{2}\) delimit the top and bottom.

These boundaries effectively enclose a finite region on the plane, simplifying the calculation of area into known limits. Without clear boundaries, calculating area would be undefined or more complicated. Thus, clearly identifying these can prevent misinterpretation and ensure calculative accuracy. Recognizing and using these limits accurately in integration helps in solving the problem systematically.