Question

Computer plot the graph of \(x^{3}+y^{3}=8 .\) Write the integrals for the following quantities (see Chapter 5 if needed) and evaluate them as \(B\) functions. The first quadrant area bounded by the curve.

Short answer

The area bounded by the curve in the first quadrant is approximately 19.7314.

Step-by-step solution

- Understand the given curve equation

The provided curve is described by the equation \(x^3 + y^3 = 8\). This is an implicit function describing a relationship between \(x\) and \(y\).

- Recognize the area of interest

The task involves finding the area bounded by this curve in the first quadrant, which includes positive values for both \(x\) and \(y\).

- Transform the curve equation for integration

To integrate, we express \(y\) in terms of \(x\). From \(x^3 + y^3 = 8\), solve for \(y^3\): \(y^3 = 8 - x^3\), then \(y = (8 - x^3)^{1/3}\).

- Set up the integral for the area

The area in the first quadrant under the curve can be found by integrating \(y = (8 - x^3)^{1/3}\) with respect to \(x\): \[\text{Area} = \int_{0}^{2} (8 - x^3)^{1/3} \, dx\.\]

- Change variable to standard form

Use the substitution \( u = x^3 / 8 \) to transform the integral into the form involving a Beta function. Hence, \(x = (8u)^{1/3}\) and \(dx = \frac{8}{3}u^{-2/3} du\). Rewriting the boundaries, when \(x = 0\), \(u = 0\), and when \(x = 2\), \(u = 1\).

- Rewrite and simplify the integral

The integral becomes: \[\text{Area} = \int_{0}^{1} 8^{1/3} (1 - u)^{1/3} \cdot \frac{8}{3} u^{-2/3} du\,\]which simplifies to: \[\frac{8 \cdot 8^{1/3}}{3} \int_{0}^{1} (1 - u)^{1/3} u^{-2/3} \, du\]

- Express in terms of Beta function

The integral now reflects the Beta function definition: \[B(x,y) = \int_{0}^{1} t^{x-1} (1 - t)^{y-1} \, dt\]. With \(x = \frac{1}{3}\) and \(y = \frac{2}{3}\), the integral becomes: \[\frac{8 \cdot 8^{1/3}}{3} B \left( \frac{1}{3}, \frac{2}{3}\right).\]

- Compute the Beta function

Using the property of the Beta function \(B(x,y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}\), find: \[B\left( \frac{1}{3}, \frac{2}{3} \right) = \frac{\Gamma(1/3) \Gamma(2/3)}{\Gamma(1)}. \] From tables or advanced calculators, \(\Gamma(1/3) \approx 2.67893\) and \(\Gamma(2/3) \approx 1.35412\), and \(\Gamma(1) = 1\). Thus, \[B\left( \frac{1}{3}, \frac{2/3} \right) \approx 3.67693.\]

- Calculate the area

Substitute back to obtain the area: \[\text{Area} = \frac{8 \cdot 8^{1/3}}{3} \cdot 3.67693 = \frac{8 \cdot 2}{3} \cdot 3.67693 \approx 19.7314.\]

Key concepts

Graph Plotting

Plotting the graph of the equation \(x^3 + y^3 = 8\) gives us valuable insights into the behavior of the curve. This is called an implicit plot because the equation defines a relationship between \(x\) and \(y\) but not explicitly one in terms of the other. For graph plotting, you can use graphing calculators or software tools like Desmos, GeoGebra, or Python’s Matplotlib library.

The curve intersects the axes at points (2, 0) and (0, 2), defining the boundaries on the right and top sides of the first quadrant. The job of finding the area under this curve can be visualized easily once you see the plot. This step helps you understand where the curve lies and how the limits of integration will be set up.

Integration

Integration is an essential concept to understand when working with areas under curves. Here, we'll integrate the function obtained from our transformed curve equation between the appropriate limits to calculate the area.

The integral setup for finding the required area in the first quadrant is: \[ \text{Area} = \int_{0}^{2} (8 - x^3)^{1/3} \ dx \]

This integral sums up infinitesimally small vertical slices of the area under the curve from \(x = 0\) to \(x = 2\).

Gamma Function

The Gamma function \(\Gamma(x)\) is an extension of the factorial function to real and complex numbers. It is defined as: \[ \Gamma(n) = (n-1)! \quad \text{for positive integers} \]

The Beta function \(B(x, y)\) is closely related to the Gamma function and can be computed using the property: \[ B(x, y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)} \]

Using Gamma function values, we calculate: \[ B\left( \frac{1}{3}, \frac{2}{3} \right) = \frac{\Gamma(1/3) \Gamma(2/3)}{\Gamma(1)} \approx 3.67693 \]

This result helps find the final area under the curve.

Variable Substitution

Variable substitution simplifies the integral by converting it into a form that can easily be recognized and evaluated as a Beta function. Here’s how it works step-by-step:

• Use the substitution \( u = x^3 / 8 \) to transform the integral.
• Thus, \( x = (8u)^{1/3} \) and \( dx = \frac{8}{3} u^{-2/3} du \).
• Rewriting the boundaries: when \( x = 0 \), \( u = 0 \), and when \( x = 2 \), \( u = 1 \).

The integral becomes: \[ \text{Area} = \int_{0}^{1} 8^{1/3} (1 - u)^{1/3} \cdot \frac{8}{3} u^{-2/3} \ du \]

This step allows us to put the integral in the standard form for the Beta function, making it easier to solve.