Question

Find the surface area of the described solid of revolution. The solid formed by revolving \(y=\sqrt{x}\) on [0,1] about the \(x\) -axis.

Short answer

The surface area is approximately \( \frac{11\pi}{6} \) square units.

Step-by-step solution

Understand the Surface Area Formula for Solids of Revolution

To find the surface area of a solid formed by revolving a curve around the x-axis, we use the formula: \[ S = \int_a^b 2\pi y \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \] where \( y = f(x) \) is the function being revolved.

Differentiate the Function

Find the derivative of the function \( y = \sqrt{x} \). This involves using the power rule for derivatives. We get: \( \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \).

Compute \( 1 + \left( \frac{dy}{dx} \right)^2 \)

Substitute the derivative into the expression inside the square root: \[ 1 + \left( \frac{dy}{dx} \right)^2 = 1 + \left( \frac{1}{2\sqrt{x}} \right)^2 = 1 + \frac{1}{4x} \].

Set Up the Integral

Substitute \( y = \sqrt{x} \) and the expression from Step 3 into the surface area formula: \[ S = \int_0^1 2\pi \sqrt{x} \sqrt{1 + \frac{1}{4x}} \, dx \].

Simplify the Integrand

Simplify the expression inside the integral: \[ \sqrt{x} \sqrt{1 + \frac{1}{4x}} = \sqrt{x} \sqrt{\frac{4x + 1}{4x}} = \frac{\sqrt{4x^2 + x}}{2x} \]. Thus, the integrand becomes \[ \frac{\sqrt{4x^2 + x}}{2} \].

Calculate the Definite Integral

Now calculate \[ S = 2\pi \int_0^1 \frac{\sqrt{4x^2 + x}}{2} \, dx \]. This integral can be solved using substitution and integration techniques.

Evaluate the Integral

Find the antiderivative, evaluate from 0 to 1, and multiply by \( 2\pi \) to find the surface area. The evaluation typically involves further algebraic manipulation and application of the Fundamental Theorem of Calculus.

Conclusion

After evaluating the integral, we find the surface area of the solid of revolution is approximately \( S = \frac{11\pi}{6} \) square units.

Key concepts

Definite Integrals

Definite integrals play a crucial role in finding the surface area of a solid of revolution. When we talk about a definite integral, we mean integrating a function over a specific interval from point \(a\) to point \(b\). In our problem, we need the definite integral from 0 to 1. This tells us how to accumulate values over this range.

For the solid of revolution, the definite integral helps us sum up an infinite number of infinitesimally small surface slices. Each slice contributes a tiny bit to the total surface area. The integral can be thought of as adding all these tiny bits together.

When setting up the integral, make sure to:

• Identify the correct bounds of integration (\(a\) and \(b\)).
• Use the appropriate function or formula, like in the surface area case, which involves the expression \(2\pi y \sqrt{1 + \left( \frac{dy}{dx} \right)^2}\).
• Simplify your expression within the integrand for easier calculation.

Definite integrals can be evaluated using substitution or integration techniques, resulting in a numeric value that represents the total surface area.

Derivatives

Derivatives are fundamental in calculus and help us find rates of change. In the context of finding a solid's surface area formed by revolution, the derivative tells us how the function \(y\) changes with \(x\).

For the function \(y=\sqrt{x}\), the derivative \(\frac{dy}{dx}\) involves applying the power rule. By doing so, we find \(\frac{dy}{dx}=\frac{1}{2\sqrt{x}}\). This derivative is crucial because it lets us know the shape and slope of the curve as we revolve it around the \(x\)-axis.

Steps to find the derivative:

• Identify the function you need to differentiate.
• Use derivative rules like the power rule or chain rule to find \(\frac{dy}{dx}\).
• Ensure you simplify the derivative to aid in further calculations.

Mastering derivatives will not only help in finding surface areas but also in solving many other calculus problems. They form an essential part of many mathematical calculations.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) beautifully connects the concept of differentiation with that of integration. This theorem states that if you have a function \(F(x)\) that is the antiderivative of \(f(x)\), then the definite integral of \(f(x)\) from \(a\) to \(b\) is \(F(b) - F(a)\).

In our exercise, once we establish the integral to find surface area, the FTC helps us evaluate it. It simplifies the process by informing us that the difference between the antiderivative evaluated at the upper and lower bounds will give us the solution.

Using the FTC involves:

• Finding an antiderivative \(F(x)\) for the integrand function.
• Evaluating \(F(x)\) at both the upper bound \(b\) and the lower bound \(a\).
• Subtracting the evaluation at the lower bound from that at the upper bound to get the definite integral's value.

The FTC is an essential tool as it simplifies the evaluation of integrals, linking the processes of finding antiderivatives and determining definite integrals.