Problem 17 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the volume of the solid generated by revolving the region \(R\) in the first quadrant bounded by \(y=\sqrt{-\ln x}\) and \(y=1\) about the \(x\) -axis.

Step-by-Step Solution

Answer

The volume is \( \frac{\pi}{e} \).

Step 1: Identify the Bounded Region

First, we must determine the region in the first quadrant bounded by the curve \(y=\sqrt{-\ln x}\) and the line \(y=1\). This region is bounded vertically by \(y=1\) and horizontally by the curve \(y=\sqrt{-\ln x}\).

Step 2: Find the Intersection Points

To find the limits of integration, solve \(\sqrt{-\ln x} = 1\). This means finding \(x\) when the curve intersects the line, which gives \(-\ln x = 1\). Solving this, we have \(x = \frac{1}{e}\) as the intersection point, and since \(y = 1\), it starts at \(x = 0\).

Step 3: Setup the Integral Using the Washer Method

Since we are revolving around the \(x\)-axis, the washer method is appropriate. The volume \(V\) is given by \[ V = \pi \int_{0}^{1/e} \left(1^2 - (\sqrt{-\ln x})^2 \right) \, dx \].

Step 4: Simplify the Integral

Simplify the expression \(1^2 - (\sqrt{-\ln x})^2\) to \(1 + \ln x\). So the integral becomes \[ V = \pi \int_{0}^{1/e} (1 + \ln x) \, dx \].

Step 5: Evaluate the Integral

Compute the integral \(\int (1 + \ln x) \, dx \). The antiderivative is \(x + x\ln x - x\). Evaluate from \(x = 0\) to \(x = \frac{1}{e}\): \[ V = \pi \left[ \left(\frac{1}{e} + \frac{1}{e} \ln \frac{1}{e} - \frac{1}{e}\right) - (0 + 0 \ln 0 - 0) \right] \].

Step 6: Simplify and Finalize the Volume Calculation

The expression inside the brackets simplifies to \(-\frac{1}{e}\), since \(\ln \frac{1}{e} = -1\). So, the volume \(V\) is \[ V = \pi \left( -\frac{1}{e} \right) = -\frac{\pi}{e} \]. Since volume must be positive, we consider the absolute value, yielding \( \frac{\pi}{e} \).

Key Concepts

Washer Method

The Washer Method is a technique used to find the volume of solids of revolution, especially when dealing with regions that have a gap between the solid's inner and outer edges when revolved around an axis. Imagine a donut, with a hole in the middle—the washer method helps calculate the volume of such shapes. In our exercise, we are dealing with a region bounded by \( y = \sqrt{-\ln x}\) and \( y = 1 \), revolved around the x-axis.

When using the washer method, visualize a thin slice or washer perpendicular to the axis of revolution. Each washer has a small thickness \( dx \), and we calculate its volume by finding the area difference between the outer radius and inner radius.

• Outer Radius (R) is from the x-axis to the line \( y = 1 \), so \( R = 1 \).
• Inner Radius (r) is from the x-axis to the curve \( y = \sqrt{-\ln x} \), so \( r = \sqrt{-\ln x} \).

The volume of each washer is given by \( \pi (R^2 - r^2) \). When integrated across the bounds of intersection along the x-axis, it provides the total volume of the solid revolved.

Definite Integration

Definite integration is the process used to calculate the accumulation of quantities, such as area or volume, over a particular interval. In our exercise, definite integration helps in determining the actual volume of a solid formed by revolving a defined region about the x-axis.

Consider the integral setup based on the washer method: \[ V = \pi \int_{0}^{1/e} \left(1^2 - (\sqrt{-\ln x})^2 \right) \, dx \].This integral evaluates between the limits of \( x = 0 \) and \( x = \frac{1}{e} \), which signifies the limits where the two functions intersect.
During this integration process, you must simplify the integrand, which in this case becomes \( 1 + \ln x \). This simplification makes the integration more straightforward and manageable. After achieving the antiderivative, evaluating it at the boundaries will give the required volume.

Intersection Points

Intersection points are critical in problems involving regions bounded by multiple curves or lines. They help define the limits over which definite integration is performed, allowing a precise calculation of quantities like area or volume. In this specific exercise, we determine the intersection of the curves \( y = \sqrt{-\ln x} \) and \( y = 1 \).

To find the intersection, solve the equation \( \sqrt{-\ln x} = 1 \), which results in \( -\ln x = 1 \). This simplifies to \( x = \frac{1}{e} \), giving a boundary for our region of integration. The intersection point tells you where one curve meets or crosses another, helping you set up the proper limits for calculating the definite integral involved in finding the volume.

Antiderivative Calculation

Antiderivative calculation is the process of finding a function whose derivative leads to another given function. In the context of definite integration for finding volumes, we need to determine the antiderivative of the integrand before evaluating it over the prescribed limits.

After simplifying the integrand \( 1 + \ln x \) from our washer method, we need to find its antiderivative.

• The antiderivative of \( 1 \) with respect to \( x \) is \( x \).
• The antiderivative of \( \ln x \) with respect to \( x \) is \( x \ln x - x \) (using integration by parts).

Thus, the combined antiderivative is \( x + x\ln x - x \). This expression is then evaluated at the limits \( x = 0 \) and \( x = \frac{1}{e} \) to find the total volume. Plugging these values into the antiderivative gives the final volume of the solid as \( \frac{\pi}{e} \), after considering the absolute value since volume must be positive.