Problem 23 - Calculus: Early Transcendentals

Question

Sketch the region enclosed by the given curves and find its area. \( y = \sqrt[3]{2x} \) , \( y = \frac{1}{8}x^2 \) , \( 0 \le x \le 6 \)

Step-by-Step Solution

Answer

The total area is calculated by evaluating the definite integrals from \(x = 0\) to \(x = \sqrt[5]{16}\) and from \(x = \sqrt[5]{16}\) to \(x = 6\).

Step 1: Identify Intersection Points

To find where the curves intersect, set the equations equal to each other: \[ \sqrt[3]{2x} = \frac{1}{8}x^2 \]Solve for \(x\) by cubing both sides: \[ (2x) = \left(\frac{1}{8}x^2\right)^3 \]Simplify and solve this equation to find the intersection points. This can be simplified to: \[ 8(2x) = x^6 \]Thus, \[ 16x = x^6 \]Factoring gives: \[ x(x^5 - 16) = 0 \]So, \(x = 0\) or \(x^5 = 16\), which gives \(x = \sqrt[5]{16}\).

Step 2: Set Up the Integral for Area

The area between the two curves from \(x = 0\) to \(x = 6\) is found by integrating the difference of the functions. The general formula for the area is:\[ A = \int_{a}^{b} (f(x) - g(x)) \, dx \]where \( f(x) \) is the upper function and \( g(x) \) is the lower function.

Step 3: Determine the Upper and Lower Functions

In the region from \(x = 0\) to \(x = \sqrt[5]{16}\), \(y = \sqrt[3]{2x} \) is the upper curve, and \(y = \frac{1}{8}x^2\) is the lower curve. From \(x = \sqrt[5]{16}\) to \(x = 6\), \(y = \frac{1}{8}x^2\) becomes the upper curve and \(y = \sqrt[3]{2x}\) becomes the lower curve due to their intersection around \(x = \sqrt[5]{16}\), which can be determined by testing points.

Step 4: Calculate the Area in Two Segments

Calculate the area first from \(x = 0\) to \(x = \sqrt[5]{16}\) where \(y = \sqrt[3]{2x} \) is the upper function:\[ A_1 = \int_{0}^{\sqrt[5]{16}} \left(\sqrt[3]{2x} - \frac{1}{8}x^2 \right) \, dx \]Calculate the area from \(x = \sqrt[5]{16}\) to \(x = 6\) where \(y = \frac{1}{8}x^2 \) is the upper function:\[ A_2 = \int_{\sqrt[5]{16}}^{6} \left(\frac{1}{8}x^2 - \sqrt[3]{2x} \right) \, dx \]Evaluate these integrals to find the total area.

Step 5: Evaluate the Integrals

To find \( A_1 \): You need to find the integral of both functions \[ \int \sqrt[3]{2x} \, dx = \frac{3}{5}(2x)^{5/3} \]\[ \int \frac{1}{8}x^2 \, dx = \frac{1}{24}x^3 \]Calculate \( A_1\) using the boundaries For \( A_2\), similarly \[ A_2 = \int \frac{1}{8}x^2 \, dx - \int \sqrt[3]{2x} \, dx \]And evaluate with limits to find total area:\[ A = A_1 + A_2 \]

Step 6: Calculate Exact Numerical Area

Evaluate the definite integrals found in Step 5 to find the exact numerical solution for the area. Insert the computed bounds \( 0, \sqrt[5]{16}, 6 \) into each integral and simplify to obtain the numerical area enclosed by the curves.

Key Concepts

Area between curves

In calculus, finding the area between two curves is a common problem that involves integration. The idea is to determine which of the two curves lies above the other within a certain interval and then calculate the net area between them. This requires setting up an integral where the upper curve is subtracted from the lower curve over the specified range.

Given functions such as \( y = \sqrt[3]{2x} \) and \( y = \frac{1}{8}x^2 \), we need to first identify which curve is the upper function and which is the lower function within the interval of interest. This often involves finding the intersection points of the curves as these points tend to form the boundaries for such calculations.

Once the upper and lower functions are determined, setting up the integral becomes straightforward. We use the formula for area:

• \( A = \int_{a}^{b} [f(x) - g(x)] \, dx \), where \( f(x) \) is the upper curve and \( g(x) \) is the lower curve.

In our example, the bounds \( a \) and \( b \) are determined by the intersection points and given conditions, such as \( 0 \le x \le 6 \).

Cubing equations

Cubing equations is a useful algebraic technique often employed to simplify or manipulate equations involving cube roots or powers. It involves raising both sides of an equation to the power of three to eliminate the cube root function. This step is crucial when working with equations like \( y = \sqrt[3]{2x} \) and \( y = \frac{1}{8}x^2 \) to find where the curves intersect.

For example, to solve \( \sqrt[3]{2x} = \frac{1}{8}x^2 \), we cube both sides to get rid of the cube root:

• Cube both sides: \( (2x) = \left(\frac{1}{8}x^2\right)^3 \).

This manipulation simplifies the equation to a solvable polynomial equation, in our case forming \( 16x = x^6 \). This is then solved by factoring or other algebraic methods to find the points of intersection. Cubing is a powerful tool for transforming complex problems into simpler form through elegant mathematical operations.

Definite integral evaluation

Evaluating definite integrals involves calculating the exact area under a curve between specified limits. This process is central to finding the exact area enclosed by two curves as it gives a numeric result representing this area.

When working with definite integrals, once the functions and bounds are identified, we replace them into the integral expression. For instance, if we have a definite integral defined as \( \int_{a}^{b} [f(x) - g(x)] \, dx \), we evaluate this by finding the antiderivatives (indefinite integrals) of \( f(x) \) and \( g(x) \), and then applying the fundamental theorem of calculus, which states:

• \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \), where \( F(x) \) is the antiderivative of \( f(x) \).

By plugging the upper and lower bounds into these antiderivatives, we achieve the exact area. This precise computation is crucial for understanding the extent of the space enclosed between the curves.

Intersection points calculation

Calculating the intersection points of curves is a necessary step when determining the area between them. These points indicate where the curves cross and usually form the boundaries of integration. The intersection can be found by setting the equations of the curves equal to each other and solving for \( x \).

For example, with the curves \( y = \sqrt[3]{2x} \) and \( y = \frac{1}{8}x^2 \), we set these equations equal to determine where they intersect:

• \( \sqrt[3]{2x} = \frac{1}{8}x^2 \).
• By cubing both sides, we simplify this to a polynomial: \( 16x = x^6 \).

This then factors to find solutions such as \( x = 0 \) and \( x = \sqrt[5]{16} \), which are points where the curves meet. Identifying these intersections allows us to accurately set up the integral boundaries to compute the area between the curves.