Question

Find the volume of the solid generated when the region in the first quadrant bounded above by \(y=2\) and on the right by \(y=-\ln x\) is revolved about the \(y\) -axis.

Short answer

The volume is \( 2\pi (1 - 0.375e^{-4}) \) cubic units.

Step-by-step solution

Set Up the Integral

To find the volume of a solid of revolution generated by rotating a region about the y-axis, we use the method of cylindrical shells. The formula is \[ V = 2\pi \int_{a}^{b} x \cdot h(x) \, dx \]where \( h(x) \) is the height of the shell and \( a \) and \( b \) are the bounds. In this case, the height \( h(x) \) is from \( y=2 \) to \( y=-\ln x \). Thus, \( h(x) = 2 + \ln x \).

Determine the Bounds

The bounds are determined by the intersection points of the curves \( y=2 \) and \( y=-\ln x \) in the positive quadrant. Setting the equations equal gives \( 2 = -\ln x \). Solving for \( x \), we get \( \ln x = -2 \) which means \( x = e^{-2} \). The lower bound is \( x = e^{-2} \) and the upper bound is \( x = 1 \) because the region is in the first quadrant.

Compute the Integral

Substitute the bounds and height function into the volume formula:\[ V = 2\pi \int_{e^{-2}}^{1} x (2 + \ln x) \, dx \]The integral can be split into two parts:\[ V = 2\pi \left( \int_{e^{-2}}^{1} 2x \, dx + \int_{e^{-2}}^{1} x \ln x \, dx \right) \]

Integrate the Functions

First, integrate the simpler function:\[ \int 2x \, dx = x^2 \]Thus,\[ \int_{e^{-2}}^1 2x \, dx = [x^2]_{e^{-2}}^1 = 1 - e^{-4} \]For the \( x \ln x \) integral, use integration by parts with \( u = \ln x \) and \( dv = x \, dx \). This yields:\[ \int x \ln x \, dx = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C \]Plugging in the bounds\[ \left[ \frac{x^2}{2} \ln x - \frac{x^2}{4} \right]_{e^{-2}}^1 = 0 - (-0.625e^{-4}) \].

Calculate the Final Volume

Substituting back into the volume formula:\[ V = 2\pi \left(1 - e^{-4} - (-0.625e^{-4}) \right) = 2\pi \left(1 - 0.375e^{-4}\right) \].

Conclusion

Thus, the volume of the solid generated is \[ 2\pi \left(1 - 0.375e^{-4}\right) \] cubic units.

Key concepts

Cylindrical Shells Method

The cylindrical shells method is a powerful technique in calculus for finding the volume of a solid of revolution. When a region in the plane is revolved around an axis, it generates a 3D shape. The cylindrical shells method is particularly useful when this revolution uses the y-axis.

To use this method, we imagine the solid as composed of several thin, hollow cylinders. The formula for the volume is \[ V = 2\pi \int_{a}^{b} x \cdot h(x) \, dx \]where:

• \( x \) is the average radius of the shell, which is the distance from the axis.
• \( h(x) \) is the height of the shell, typically the difference between the curves defining the top and bottom of the region.
• \( a \) and \( b \) are the bounds of the region along the axis of revolution, determined by finding where the curves intersect.

In the exercise, the height \( h(x) \) is the difference between \( y=2 \) and \( y=-\ln x \), leading to \( h(x) = 2 + \ln x \). The solution neatly substitutes these into the formula to solve for volume.

Integration by Parts

Integration by parts is a technique derived from the product rule for differentiation. It's used to solve integrals where standard methods don't apply easily. The formula for integration by parts is:\[\int u \, dv = uv - \int v \, du\]where:

• \( u \) and \( dv \) are chosen from the integrand \( f(x) \).
• \( du \) and \( v \) are derived through differentiation and integration respectively.

In our exercise, integration by parts helps solve the more complicated integral \( \int x \ln x \, dx \). By choosing \( u = \ln x \), hence \( du = \frac{1}{x} \, dx \), and \( dv = x \, dx \) leading to \( v = \frac{x^2}{2} \), we transform the integral into manageable terms. These choices simplify the integration process significantly, as demonstrated in the step-by-step solution.

Definite Integrals

Definite integrals are a fundamental concept in calculus used to calculate the net area under a curve from one point to another. This computation provides the total accumulation of quantities, such as area, displacement, or in our case, volume.

The notation for a definite integral from \( a \) to \( b \) for a function \( f(x) \) is:\[\int_{a}^{b} f(x) \, dx\]After finding the indefinite integral, the process involves evaluating the resulting expression at the upper bound and subtracting the value evaluated at the lower bound: \[ F(b) - F(a) \] This calculation is vital for stepping from integration results to accurate, applicable quantities. In finding the volume of the solid in our problem, the definite bounds \( e^{-2} \) to \( 1 \) were used to calculate the precise volume of the solid after applying integration techniques.

Calculus

Calculus is the branch of mathematics that deals with continuous change. It's fundamental in understanding and calculating aspects of motion, structure, and change in natural and designed systems. Basic concepts include derivatives, which give us the rate of change, and integrals, which are used to compute accumulations like area and volume.

In the context of volumes of solids of revolution, calculus offers tools like integration and the cylindrical shells method to convert geometric and algebraic insights into precise numerical answers. Calculus allows us to navigate the abstract into the concrete, solve problems involving continuous data, and understand the world in a more detailed way. Through its methods, we refine shapes in revolving motions into quantifiable volumes, just as was demonstrated in the exercise's solution for the given region revolved around the y-axis.