Problem 50 - Calculus Single Variable

Question

In each of Exercises 49-52, calculate the volume \(V\) of the solid that is obtained by rotating the planar region between the curves \(y=x^{2}\) and \(y=x^{1 / 2}\) about the given line. $$ x=-1 $$

Step-by-Step Solution

Answer

The volume is \(\frac{29\pi}{30}\).

Step 1: Identify the Bounded Region

First, sketch the graphs of the curves \(y=x^2\) and \(y=\sqrt{x}\). The region is bounded above by \(y=\sqrt{x}\) and below by \(y=x^2\) from \(x=0\) to \(x=1\), where the curves intersect.

Step 2: Set Up the Disk Method Formula

We revolve around the line \(x=-1\), so the distance to the axis of rotation changes the integral setup. The volume \(V\) of the solid is given by the formula: \[ V = \, \pi \int_{a}^{b} [(outer \, radius)^2 - (inner \, radius)^2] \, dx \] Here, \( a = 0 \) and \( b = 1 \), with the outer radius being \(1 + \sqrt{x}\) and the inner radius being \(1 + x^2\).

Step 3: Write the Integral Expression

Substitute the expressions for the outer and inner radii into the formula: \[ V = \, \pi \int_{0}^{1} [(1 + \sqrt{x})^2 - (1 + x^2)^2] \, dx \] Simplify: \[ V = \, \pi \int_{0}^{1} [1 + 2\sqrt{x} + x - (1 + 2x^2 + x^4)] \, dx \] Further simplify to: \[ V = \, \pi \int_{0}^{1} [2\sqrt{x} + x - 2x^2 - x^4] \, dx \]

Step 4: Integrate the Expression

Calculate the definite integral:\[ V = \, \pi \left[ \frac{4}{3}x^{3/2} + \frac{1}{2}x^2 - \frac{2}{3}x^3 - \frac{1}{5}x^5 \right]_{0}^{1} \] Evaluate from 0 to 1: \[ V = \, \pi \left( \frac{4}{3} + \frac{1}{2} - \frac{2}{3} - \frac{1}{5} \right) \]

Step 5: Simplify the Result

Combine like terms inside the parentheses: \[ \frac{4}{3} + \frac{1}{2} = \frac{8}{6} + \frac{3}{6} = \frac{11}{6} \] \[ \frac{11}{6} - \frac{2}{3} = \frac{11}{6} - \frac{4}{6} = \frac{7}{6} \] \[ \frac{7}{6} - \frac{1}{5} = \frac{35}{30} - \frac{6}{30} = \frac{29}{30} \] Thus, the volume \( V \) is: \[ V = \, \pi \times \frac{29}{30} = \frac{29\pi}{30} \]

Step 6: Conclusion: Determine the Volume

The volume of the solid obtained by rotating the region between the curves \(y=x^2\) and \(y=\sqrt{x}\) around the line \(x=-1\) is: \[ V = \frac{29\pi}{30} \]

Key Concepts

Disk Method

When calculating the volume of a solid of revolution, one effective technique is the disk method. This approach is handy when rotating a region around an axis, such that thin slices or disks of the solid are perpendicular to the axis of rotation. To visualize this, imagine pancakes stacked across the rotation axis. The volume of each disk is determined by its radius and thickness. The disk method uses the formula for the volume of a cylinder, adjusted to accommodate the changing radius in calculus. The volume for a small slice is calculated by:

• determining the area of a disk face (\( \pi (radius)^2 \)
)
• multiplying it by a small thickness (\( dx \) or \( dy \)
)

Summing these volumes leads to an integral setup that spans the entire region being rotated. This integral gives us the total volume of the solid.

Definite Integral

A definite integral is a crucial concept in calculus, especially for calculating areas and volumes. It sums up areas beneath a curve over a given interval.To compute a definite integral, follow these steps:

• Identify the function \( f(x) \).
• Determine the limits of integration: the starting point (\( a \)) and ending point (\( b \)).
  
• Set up the integral: \( \int_{a}^{b} f(x) \, dx \).
• Evaluate it to find the accumulated area or volume.

During integration, you handle curves that represent a larger radius minus a smaller radius for solids of revolution. Once solved, definite integrals yield precise numerical values, unlike indefinite integrals that give general antiderivative forms.

Curve Intersection

Identifying where curves intersect is often the first step in problems involving bounded regions. It helps establish the limits of integration.For intersection:

• Set the expressions equal: \( y_1 = y_2 \).
• Solve the equation to find relevant \( x \)-values.
  

In our exercise, curves \( y = x^2 \) and \( y = \sqrt{x} \) meet where both equations hold true. Solving \( x^2 = \sqrt{x} \) provides the points of intersection. Typically resulting in a smaller interval, such as \( x = 0 \) and \( x = 1 \) for this example. These points become the limits of integration and help outline the region to be rotated.

Axis of Rotation

An axis of rotation is the line around which a shape is revolved to generate a solid. Its position impacts how we calculate the radii within the disk method.The axis could be a standard axis or a shifted line, such as \( x = -1 \) in our problem. When the axis is not typical (such as the y-axis), it modifies the way we calculate the distances:

• The outer radius shifts to account for the distance from this new line to the outer curve.
• Similarly, adjust the inner radius for the rotation axis.
  

In particular, the radii become \( 1 + \sqrt{x} \) and \( 1 + x^2 \) for both curves, respectively, given \( x = -1 \) as the axis. Adjusting for these factors ensures a correct setup of the integral formula for the volume.