Problem 29 - Calculus and its applications

Question

Find the volume generated by rotating the area bounded by the graphs of each set of equations around the \(y\) -axis. $$ y=\sqrt{x^{2}+1}, x=0, x=3 $$

Step-by-Step Solution

Answer

Volume = \(\pi \left( \frac{10 \sqrt{10}}{3} - \sqrt{10} + \frac{2}{3} \right)\) cubic units.

Step 1: Identify the region to be rotated

We have the curve given by the equation \(y = \sqrt{x^2 + 1}\) from \(x=0\) to \(x=3\). This curve represents a segment of a hyperbola that we want to rotate around the \(y\)-axis.

Step 2: Set up the integral for volume using the disk method

In the disk method, the volume of the solid of revolution about the \(y\)-axis is given by\[V = \pi \int_{y_1}^{y_2} [g(y)]^2 \, dy\]where \([g(y)]^2\) is the radius of each disk perpendicular to the \(y\)-axis. In our case, solving for \(x\) gives \(x = \sqrt{y^2 - 1}\), and our bounds are from \(y = \sqrt{0^2 + 1} = 1\) to \(y = \sqrt{3^2 + 1} = \sqrt{10}\).

Step 3: Transform the integral in terms of y

The integral for the volume becomes\[V = \pi \int_{1}^{\sqrt{10}} (y^2 - 1) \, dy\]This is obtained by computing \((\sqrt{y^2 - 1})^2 = y^2 - 1\) as the function describing the radius of the disks.

Step 4: Compute the integral

Carry out the integral step-by-step:1. Integrate \(y^2\) with respect to \(y\):\(\int y^2 \, dy = \frac{y^3}{3}\)2. Integrate \(-1\) with respect to \(y\):\(\int -1 \, dy = -y\)Combined, we have:\[V = \pi \left[ \frac{y^3}{3} - y \right]_{1}^{\sqrt{10}} \]

Step 5: Evaluate the integral at the bounds

Substitute \(y=\sqrt{10}\) and \(y=1\) into the integrated form:\[V = \pi \left( \frac{(\sqrt{10})^3}{3} - \sqrt{10} - \left( \frac{1^3}{3} - 1 \right) \right)\]Calculate each part:- \((\sqrt{10})^3 = 10^{3/2} = 10 \cdot \sqrt{10}\)- Evaluate: \[V = \pi \left( \frac{10 \sqrt{10}}{3} - \sqrt{10} - \left( \frac{1}{3} - 1 \right)\right)\]This simplifies further:

Step 6: Final simplification

Simplify the expression to find the volume:Subtract and simplify each part:\[V = \pi \left( \frac{10 \sqrt{10}}{3} - \sqrt{10} + 1 - \frac{1}{3}\right) \]Combine like terms:\[V = \pi \left( \frac{10 \sqrt{10}}{3} - \sqrt{10} + \frac{2}{3} \right)\] Which gives us the final volume.

Key Concepts

Disk Method

The disk method is a useful technique for finding the volume of a solid of revolution. When you want to find the volume generated by rotating a region around an axis, you imagine slicing the solid into thin, disk-like slices. Each slice is perpendicular to the axis of rotation.

Here's how it works:

• You calculate the radius of each disk, which is the distance from the axis of rotation to the edge of the region being rotated.
• For our example, when rotating around the \(y\)-axis, the radius of each disk depends on the \(x\)-value, expressed in terms of \(y\), as \(x = \sqrt{y^2 - 1}\).
• The volume of each thin disk is approximately \(\pi \times \text{radius}^2 \times \Delta y\), where \(\Delta y\) is the thickness.
• To find the total volume, you integrate these disks from the start to the end of the region in the \(y\)-direction.

This method helps to visualize and compute the volume of more complex shapes by breaking them down into simpler parts.

Solid of Revolution

A solid of revolution is a three-dimensional shape formed by rotating a two-dimensional region around an axis. These fascinating objects often resemble familiar items like cones, cylinders, or doughnuts, depending on the original region's shape.

To create a solid of revolution:

• Pick a region in the plane that is bounded by one or more curves.
• Decide on an axis of rotation. This could be the \(x\)-axis, \(y\)-axis, or any other line.
• As the region rotates around the chosen axis, it sweeps out a three-dimensional shape.
• For our exercise, the region bounded by \(y=\sqrt{x^2 + 1}\), \(x=0\), and \(x=3\) is rotated around the \(y\)-axis to form this solid.

Understanding the concept of a solid of revolution is crucial in solving problems related to volume, as it lays the foundation for applying techniques like the disk method.

Definite Integral

The definite integral is a fundamental tool in calculus. It allows us to compute the accumulated quantity, like area under a curve or volume of a solid, over an interval.

Here's how a definite integral works in the context of volume:

• The integral sums up an infinite number of infinitesimally small quantities, like the volume of each disk in the disk method.
• The bounds of the integral define the interval over which you conduct this summing process. For our volume calculation, the bounds were from \(y = 1\) to \(y = \sqrt{10}\).
• The function inside the integral represents the quantity we're summing, such as the area of disks. In our solution, \([g(y)]^2 = y^2 - 1\) represents this.
• By evaluating the definite integral with given bounds, you calculate the total volume of the solid.

Understanding definite integrals helps to tackle a wide range of problems in physics, engineering, and beyond, wherever accumulation is involved.