Problem 50 - Calculus: Early Transcendentals

Question

The boundaries of the shaded region are the \( y \)-axis, the line \( y = 1 \), and the curve \( y = \sqrt[4]{x} \). Find the area of this region by writing \( x \) as a function of \( y \) and integrating with respect to \( y \) (as in Exercise 49).

Step-by-Step Solution

Answer

The area is \( \frac{1}{5} \).

Step 1: Identify the Functions and Boundaries

The problem involves finding the area between the curve \( y = \sqrt[4]{x} \) and the line \( y = 1 \). The shaded region is bounded by the y-axis (\( x = 0 \)) and extends to the point where \( y \) equals 1 on the curve.

Step 2: Rewriting x as a Function of y

We start by expressing \( x \) in terms of \( y \) using \( y = \sqrt[4]{x} \). To do this, raise both sides to the power of 4: \( y^4 = x \). This gives us the function \( x = y^4 \).

Step 3: Determine Integration Limits

The integration will be with respect to \( y \), from where the curve starts at the y-axis (\( y = 0 \)) to where it meets the line \( y = 1 \).

Step 4: Set Up the Integral

The area under the curve from \( y = 0 \) to \( y = 1 \) can be found by integrating the function \( x = y^4 \) with respect to \( y \). The integral to solve is \( \int_{0}^{1} y^4 \, dy \).

Step 5: Calculate the Integral

Compute the integral \( \int y^4 \, dy \). Use the power rule for integration: \[ \int y^n \, dy = \frac{y^{n+1}}{n+1} + C \]. Applying this rule gives us: \( \int y^4 \, dy = \frac{y^5}{5} \).

Step 6: Evaluate the Definite Integral

Evaluate \( \frac{y^5}{5} \) from \( y = 0 \) to \( y = 1 \). Substituting the limits, we have \[ \frac{1^5}{5} - \frac{0^5}{5} = \frac{1}{5} - 0 = \frac{1}{5}. \]

Step 7: Conclude the Result

The area of the shaded region, as bounded by the functions and limits described, is \( \frac{1}{5} \).

Key Concepts

area under a curve

The concept of finding the area under a curve is quite foundational in calculus. It essentially involves calculating the space enclosed between a curve and a line or axis. In this exercise, you are asked to find the area between the curve given by the equation \( y = \sqrt[4]{x} \), and the line \( y = 1 \), with respect to the y-axis.

To visualize this, imagine drawing the curve from the y-axis to where it meets the line \( y = 1 \). The shaded area is located beneath this part of the curve and above the line or axis that acts as its boundary.

• The curve is \( y = \sqrt[4]{x} \), which is rewritten to express \( x \) as a function of \( y \).
• The other boundary is the horizontal line \( y = 1 \).
• The area will be calculated from where the curve begins at the y-axis, up to the line \( y = 1 \).

When calculating areas beneath a curve, integrating the function over specified limits helps to obtain this measure. This method transforms a graphical problem into a calculable one, providing an exact numerical answer for the area, which in this case is \( \frac{1}{5} \).

function of y

In many integration problems, when aiming to find an area with respect to the y-axis, it's often beneficial to express the variables in terms of \( y \). For the function originally given as \( y = \sqrt[4]{x} \), it means rewriting \( x \) in terms of \( y \).

By manipulating the original equation, you can derive \( x = y^4 \). This step is crucial for setting up your integral correctly, as it delineates how \( x \) values change between the specified y boundaries.

• Start by isolating \( x \): given \( y = \sqrt[4]{x} \), apply the fourth power to both sides.
• This results in \( y^4 = x \), a basic expression for \( x \) as a function of \( y \).

This function effectively reverses the roles of \( x \) and \( y \) to suit integration with respect to \( y \). In practice, this boils down to accurately characterizing how \( x \), the horizontal component of your graph, behaves when the vertical position \( y \) changes from \( 0 \) to \( 1 \).

definite integral limits

Definite integral limits are the boundaries between which you perform the integration. They are significant as they dictate the start and endpoint of the area you are calculating.

In the given problem, you integrate the function \( x = y^4 \) from \( y = 0 \) to \( y = 1 \). These limits are chosen because:

• \( y = 0 \) is where the curve intersects with the y-axis, marking the start of the area of interest.
• \( y = 1 \) is where the curve reaches the line \( y = 1 \), which effectively caps the top of the shaded area.

Thus, these boundaries encapsulate the area beneath the curve within the specified region in terms of \( y \). Evaluating this integral provides the total area contained between these two limits, offering an exact numerical result. As calculated in the exercise, the integral \( \int_{0}^{1} y^4 \, dy \) yields an area of \( \frac{1}{5} \). This methodical process gives you the definite areal measurement, vital in understanding and applying calculus to similar problems.