Problem 1 - Calculus: Early Transcendentals

Question

Compute the area of the surface formed when \(f(x)=2 \sqrt{1-x}\) between -1 and 0 is rotated around the \(x\) -axis.

Step-by-Step Solution

Answer

The surface area is \( \frac{8\pi}{3} (3\sqrt{3} - 2\sqrt{2}) \).

Step 1: Formula for Surface Area

To find the area of the surface created by rotating a function around the x-axis, we use the formula for surface area: \[ S = \int_a^b 2\pi f(x) \sqrt{1 + \left( f'(x) \right)^2} \,dx \] where \(f(x)\) is our function and \([a, b]\) is the interval of rotation.

Step 2: Define the Function and Interval

Given the function \(f(x) = 2 \sqrt{1-x}\) and the interval \([-1, 0]\), these are the inputs for our surface area calculation. We need to find the derivative \(f'(x)\) for use in the formula.

Step 3: Differentiate the Function

Differentiate the function \(f(x) = 2 \sqrt{1-x}\). Using the chain rule, \(f'(x) = \frac{-2}{2\sqrt{1-x}}\times (-1) = \frac{2}{2\sqrt{1-x}} = \frac{1}{\sqrt{1-x}}\).

Step 4: Set Up the Surface Area Integral

Substitute \(f(x) = 2\sqrt{1-x}\) and \(f'(x) = \frac{1}{\sqrt{1-x}}\) into the surface area formula: \[ S = \int_{-1}^{0} 2\pi (2\sqrt{1-x}) \sqrt{1 + \left(\frac{1}{\sqrt{1-x}}\right)^2} \, dx \].

Step 5: Simplify the Integral

Simplify \(\sqrt{1 + \left(\frac{1}{\sqrt{1-x}}\right)^2}\):\[ 1 + \left(\frac{1}{\sqrt{1-x}}\right)^2 = 1 + \frac{1}{1-x} = \frac{1-x + 1}{1-x} = \frac{2-x}{1-x} \]\[ \sqrt{\frac{2-x}{1-x}} = \frac{\sqrt{2-x}}{\sqrt{1-x}} \].

Step 6: Calculate the Surface Integral

Substitute and simplify the integral. The integral becomes:\[ S = \int_{-1}^{0} 2\pi (2\sqrt{1-x}) \cdot \frac{\sqrt{2-x}}{\sqrt{1-x}} \, dx = 4\pi \int_{-1}^{0} \sqrt{2-x} \, dx \].

Step 7: Solve the Integral

To solve \(\int \sqrt{2-x} \, dx\), substitute \(u=2-x\), which gives \(du = -dx\). The bounds change accordingly:\[ \int_{-1}^{0} \sqrt{2-x} \, dx = -\int_{3}^{2} \sqrt{u} \, du = -\left. \frac{2}{3} u^{3/2} \right|_{3}^{2}\].

Step 8: Evaluate the Expression

Evaluate the integral:\[ -\left( \frac{2}{3} \times 2^{3/2} - \frac{2}{3} \times 3^{3/2} \right) \].This simplifies to:\[ \frac{2}{3} (3\sqrt{3} - 2\sqrt{2}) \].

Step 9: Multiply by 4π and Conclude

Finally, multiply the result by \(4\pi\) to obtain the surface area:\[ S = 4\pi \times \frac{2}{3} (3\sqrt{3} - 2\sqrt{2}) \] = \[ \frac{8\pi}{3} (3\sqrt{3} - 2\sqrt{2}) \].

Key Concepts

Surface Area of Revolution

When a function is rotated around the x-axis, it creates a surface of revolution. To find the surface area of this revolution, we use a specific formula:

• Formally, it is expressed as \( S = \int_a^b 2\pi f(x) \sqrt{1 + \left( f'(x) \right)^2} \,dx \).
• This integral helps us measure the 'wrapping' that occurs as the function sweeps through space.
• The term \(2\pi f(x)\) accounts for the circumference of the circle formed by rotating the function around the x-axis.
• The expression \(\sqrt{1 + \left( f'(x) \right)^2}\) compensates for the steepness or slant of the curve itself.

By substituting the given function and its derivative into the formula, we can solve for the surface area al between -1 and 0.

Differentiation

Differentiation is a core principle in calculus that focuses on changing rates. It allows the calculation of derivatives, which can describe how a function behaves at any given point.

• To differentiate a composite function like \(f(x) = 2 \sqrt{1-x}\), we need to use the chain rule.
• The chain rule states: if you have a function composed of an outer function and an inner function, you differentiate the outer function first, then multiply it by the derivative of the inner function.
• For our function, the outer function is \(2\sqrt{u}\) and the inner function is \(u = 1-x\).
• After applying the chain rule, \(f'(x) = \frac{1}{\sqrt{1-x}}\) emerges as the derivative.

Understanding the differentiation of a function and its components is vital for managing complex calculus problems like surface area calculations.

Surface Integral

Surface integrals extend the concept of integration to more complex surfaces in space and can often seem daunting.

• When evaluating the surface area of revolution, we effectively integrate along a curved space which translates into a surface integral.
• Here, we integrate \(S = \int_{-1}^{0} 2\pi (2\sqrt{1-x}) \cdot \frac{\sqrt{2-x}}{\sqrt{1-x}} \, dx \).
• Initially, it might look complicated, but simplifying it to \(4\pi \int_{-1}^{0} \sqrt{2-x} \, dx \) reveals a clearer picture.
• The essence is to ensure that each piece of the formula is understood because integrating piecewise helps capture the whole area.

Through the surface integral, we can calculate the total surface area formed by a rotating curve.

Chain Rule

The chain rule is an essential technique in calculus, especially when dealing with composite functions.

• To apply the chain rule effectively, identify the layers of the function—typically an "outer" and an "inner" function.
• In our example, the outer function is \(2\sqrt{u}\) and the inner function \(u = 1-x\).
• Differentiate the outer function and multiply by the derivative of the inner function: \(\frac{-2}{2\sqrt{1-x}}\times (-1)\).
• When simplified, it resolves to \(f'(x) = \frac{1}{\sqrt{1-x}}\), indicating how the function behaves and changes.

This rule simplifies the differentiation of more complex expressions, essential for tasks like determining surface area through integration.