Problem 31 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the area of the surface generated by revolving the given curve about the \(x\) -axis \(y=\sqrt{r^{2}-x^{2}},-r \leq x \leq r\)

Step-by-Step Solution

Answer

The surface area is \(4\pi r^2\).

Step 1: Identify the Surface of Revolution

The given curve is the upper semicircle defined by the equation \( y = \sqrt{r^2 - x^2} \) over the interval \(-r \leq x \leq r\). This semicircle will be rotated about the x-axis to generate a full sphere.

Step 2: Set Up the Formula for Surface Area

The formula for the area \( A \) of a surface of revolution generated by revolving a curve \( y = f(x) \) around the x-axis is given by:\[A = 2\pi \int_{a}^{b} y \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx\]where \(a = -r\) and \(b = r\) for our curve.

Step 3: Differentiate the Function

Compute \( \frac{dy}{dx} \) for \( y = \sqrt{r^2 - x^2} \). Using the chain rule,\[ \frac{dy}{dx} = \frac{-x}{\sqrt{r^2 - x^2}}\]

Step 4: Simplify the Expression Inside the Integral

Substitute \( y \) and \( \frac{dy}{dx} \) back into the formula for surface area:\[A = 2\pi \int_{-r}^{r} \sqrt{r^2 - x^2} \sqrt{1 + \left( \frac{-x}{\sqrt{r^2 - x^2}} \right)^2} \, dx\]Simplifying the expression inside the square root gives:\[1 + \frac{x^2}{r^2 - x^2} = \frac{r^2}{r^2 - x^2}\]Thus, \( \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \) simplifies to \( \frac{r}{\sqrt{r^2 - x^2}} \).

Step 5: Evaluate the Integral

The integral now becomes:\[A = 2\pi \int_{-r}^{r} r \, dx\]This simplifies to:\[A = 2\pi r \left[ x \right]_{-r}^{r} = 2\pi r (r - (-r)) = 4\pi r^2\]

Step 6: Conclusion

The surface area generated by revolving the semicircle \( y = \sqrt{r^2 - x^2} \) about the x-axis is \( 4\pi r^2 \). This is the surface area of a sphere with radius \( r \).

Key Concepts

Integral Calculus

Integral calculus is a fundamental area of mathematics focused on accumulation and areas. When we talk about surfaces of revolution, integral calculus provides a way to calculate areas by slicing a three-dimensional object into infinitesimally small parts and summing them up.

For the given exercise, we used the concept of integration to calculate the surface area of a sphere. The specific integral formula used is: \[ A = 2\pi \int_{a}^{b} y \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \]Key Points of this formula:

• The function \( y = f(x) \) is the curve we revolve around the x-axis.
• The limits \( a \) and \( b \) are the interval endpoints.
• The expression \( \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \) comes from the Pythagorean theorem, considering the slope of the surface formed.

With these elements, integration allows us to sum up these infinite small segments to make up the entire surface area. It is a powerful concept that lets us handle complex curves and shapes.

Differentiation

Differentiation is another essential aspect of calculus that is about rates of change. In this problem, differentiation was crucial for finding the slope of the curve, which affects the surface area calculation when revolutionized around an axis.

To find the surface area of the sphere, we first needed the derivative of the semicircle function given by: \[ y = \sqrt{r^2 - x^2} \]Using the chain rule in differentiation, the derivative, \( \frac{dy}{dx} \), was calculated as: \[ \frac{-x}{\sqrt{r^2 - x^2}} \]Why This Is Important:

• The derivative describes how steep the curve is at any point \( x \).
• This slope affects how accurately our integral formula sums up the small pieces of the entire surface.

Understanding differentiation helps make sense of how changing one quantity affects another, which, in this situation, allows us to measure the curved surface precisely as it turns around the x-axis.

Sphere

A sphere is a perfectly symmetrical three-dimensional object where every point on its surface is equidistant from its center. In this exercise, we revolved a semicircle about the x-axis to form a sphere.

Here are a few important characteristics of spheres related to calculus and geometry:

• A sphere is generated by revolving a semicircle about its diameter.
• The surface area of a sphere is given by the formula \( 4\pi r^2 \), derived from integral calculus.
• The volume of a sphere can also be calculated using calculus (though not covered here), and is \( \frac{4}{3}\pi r^3 \).

In physical and theoretical applications, spheres often represent phenomena like celestial objects or bubbles because they embody minimal surface area for a given volume, which is aesthetically pleasing and often functionally efficient. By using calculus, we can understand and replicate these forms mathematically.