Problem 49 - Calculus: Early Transcendentals

Question

The area of the region that lies to the right of the \( y \)-axis and to the left of the parabola \( x = 2y - y^2 \) (the shaded region in the figure) is given by the integral \( \displaystyle \int^2_0 (2y - y^2) \, dy \). (Turn your head clockwise and think of the region as lying below the curve \( x = 2y - y^2 \) from \( y = 0 \) to \( y = 2 \).) Find the area of the region.

Step-by-Step Solution

Answer

The area of the region is \( \frac{4}{3} \).

Step 1: Understand the Problem

Identify that the integral \( \int_{0}^{2} (2y - y^2) \ dy \) represents the area of the region to the right of the \(y\)-axis and to the left of the curve \(x = 2y - y^2\) from \(y = 0\) to \(y = 2\).

Step 2: Interpret the Integral

The given integral \( \int_{0}^{2} (2y - y^2) \ dy \) implies that the area can be found by evaluating the definite integral of the function \( 2y - y^2 \) with respect to \( y \) between the limits \( y = 0 \) and \( y = 2 \).

Step 3: Integrate the Function

Find the antiderivative of the function. For \( f(y) = 2y - y^2 \), the antiderivative is \( F(y) = y^2 - \frac{y^3}{3} + C \), where \( C \) is a constant of integration.

Step 4: Evaluate the Definite Integral

Use the antiderivative to evaluate the definite integral from 0 to 2: \[\int_{0}^{2} (2y - y^2) \ dy = \left[ y^2 - \frac{y^3}{3} \right]_{0}^{2} = \left[ 2^2 - \frac{2^3}{3} \right] - \left[ 0^2 - \frac{0^3}{3} \right]\]

Step 5: Calculate the Values

Calculate the values: \( [2^2 - \frac{2^3}{3}] - [0 - 0] = 4 - \frac{8}{3} = \frac{12}{3} - \frac{8}{3} = \frac{4}{3} \) Thus, the area is \( \frac{4}{3} \).

Key Concepts

Definite Integral

A definite integral is a powerful tool in calculus used to compute the accumulation of quantities and the area under a curve between two specific points. In the original exercise, the definite integral \( \int_{0}^{2} (2y - y^2) \, dy \) is used to calculate the area of a shaded region bounded by the curve \( x = 2y - y^2 \).
The process of finding a definite integral involves two key parts:

• Identifying the interval: In this problem, the interval is from \( y = 0 \) to \( y = 2 \). These are the lower and upper limits of the integral.
• Calculating using the antiderivative: Once the antiderivative of the function is determined, it is evaluated at these limits to find the accumulated area between them.

This definite integral provides not just a static value, but also a geometrical insight into the nature of the graphed curve and the space it encompasses within those bounds.

Antiderivative

An antiderivative, often called an indefinite integral, is a function whose derivative is the given function. In simpler terms, it is the reverse process of differentiation. For the function \( f(y) = 2y - y^2 \), the antiderivative is found to be \( F(y) = y^2 - \frac{y^3}{3} + C \).
Finding the antiderivative is a crucial step in evaluating definite integrals:

• The constant \( C \) can be ignored for definite integrals as it cancels out when evaluating the limits.
• Instead of the rate of change (derivative), an antiderivative gives the accumulated value, leading to area calculations.

Once you have the antiderivative, applying the fundamental theorem of calculus becomes easier, allowing us to use it to evaluate the definite integral and thus find the total area as required.

Area under a Curve

Calculating the area under a curve involving a definite integral is one of the most significant applications of integral calculus. For instance, the original exercise focuses on finding the area under the curve of \( x = 2y - y^2 \) from \( y = 0 \) to \( y = 2 \).
This involves turning our normal perspective of the curve into a region between certain limits:

• The curve does not have to be above the axis; it merely acts as a boundary.
• The orientation and limits of integration matter; here, it is along the \( y \)-axis.

By integrating the function across the specified range, the curve's geometric confines are quantified. This approach is crucial not just in mathematics, but also in physics and engineering, where such calculations can relate to the displacement, area, and energy levels.