Question

Consider the region \(R\) bounded by the graph of \(f(x)=\sqrt{x^{2}+1}\) on the interval [0,2] a. Find the volume of the solid formed when \(R\) is revolved about the \(x\) -axis. b. Find the volume of the solid formed when \(R\) is revolved about the \(y\) -axis.

Short answer

To find the volume of the solid formed when the region bounded by the function \(f(x) = \sqrt{x^2 + 1}\) on the interval \([0,2]\) is revolved around the x-axis, we used the disk method to find the volume as \(\frac{14}{3}\pi\) cubic units. When the same region is revolved around the y-axis, the volume of the solid is \(\left(\frac{5\sqrt{5}}{3} - \sqrt{5} + \frac{2}{3}\right)\pi\) cubic units.

Step-by-step solution

Revolve around the x-axis (part a)

Step 1: Set up the integral using the disk method formula The disk method formula when revolving around the x-axis is: $$V = \pi \int_a^b [f(x)]^2 dx$$ Here, \(a=0\), \(b=2\), and \(f(x) = \sqrt{x^2 + 1}\). So, we have: $$V = \pi \int_0^2 [\sqrt{x^2 + 1}]^2 dx$$ Step 2: Integrate the function over the given interval Now integrate the function: $$V = \pi \int_0^2 (x^2 + 1) dx$$ Step 3: Evaluate the integral to find the volume Evaluate the integral: $$V = \pi [ \frac{x^3}{3} + x] \Big|_0^2$$ $$V = \pi [ (\frac{8}{3} + 2) - (\frac{0}{3} + 0)]$$ $$V = \frac{14}{3}\pi$$ So, the volume of the solid formed when the region is revolved around the x-axis is \(\frac{14}{3}\pi\) cubic units.

Revolve around the y-axis (part b)

Step 1: Rearrange the function to isolate x Rearrange the function to isolate x: $$f(x) = \sqrt{x^2 + 1} \Rightarrow f^2(x) = x^2 + 1 \Rightarrow x = \sqrt{f^2(x) - 1}$$ Step 2: Set up the integral using the disk method formula The disk method formula when revolving around the y-axis is: $$V = \pi \int_c^d [g(y)]^2 dy$$ In our case, \(g(y) = \sqrt{y^2 - 1}\) and we need to compute the new integration limits \(c\) and \(d\). Since \(x\) ranges from \(0\) to \(2\), we then have \(y=\sqrt{x^2 + 1}\) ranging \(y\) from \(\sqrt{0^2+1}=1\) to \(\sqrt{2^2+1}=\sqrt{5}\). Thus, our new integration limits are \(c=1\) and \(d=\sqrt{5}\). Step 3: Integrate the function over the new interval Now integrate the function: $$V = \pi \int_1^{\sqrt{5}} (y^2 - 1) dy$$ Step 4: Evaluate the integral to find the volume Evaluate the integral: $$V = \pi [ \frac{y^3}{3} - y] \Big|_1^{\sqrt{5}}$$ $$V = \pi [ (\frac{(\sqrt{5})^3}{3} - \sqrt{5}) - (\frac{1^3}{3} - 1)]$$ $$V = \pi [ \frac{5\sqrt{5}}{3} - \sqrt{5} + \frac{2}{3}]$$ So, the volume of the solid formed when the region is revolved around the y-axis is \(\left(\frac{5\sqrt{5}}{3} - \sqrt{5} + \frac{2}{3}\right)\pi\) cubic units.

Key concepts

Disk Method

The Disk Method is a handy technique in integral calculus used to find the volume of solids of revolution. Imagine you have a region in the plane, and you want to rotate this region around an axis to create a 3D solid. The disk method helps us calculate the volume of such solids by slicing them into thin, disk-like pieces.

• For rotation around the x-axis, these disks are horizontal slices where the radius of each disk is the distance from the x-axis to the function value, that is, the y-value of the function.
• The volume of each thin disk with thickness \( \, dx \) is given by \( \pi [f(x)]^2 \, dx \).
• To find the total volume, you sum up these disk volumes from the starting point to the ending point of the region.

This involves setting up an integral along the axis of rotation. For example, the volume \( V \) of a solid obtained by rotating a region around the x-axis between two points \( a \) and \( b \) is \( \pi \int_a^b [f(x)]^2 \, dx \).
This technique is quite powerful, especially when the curve is a simple function of x.

Integral Calculus

Integral calculus is a major branch of calculus that focuses on the concept of integration. Integration is essentially about accumulation - adding up infinitely many infinitesimally small quantities. In the context of solids of revolution, integration helps us to accumulate the volume of countless tiny disks or washers that make up the solid.

• Integral calculus allows you to compute the area under a curve, which extends to finding the volume under surfaces when considered in three dimensions.
• The integral symbol \( \int \) represents the accumulation process.
• The limits of integration, such as \( a \) and \( b \) in the integral \( \int_a^b \), define the interval over which we accumulate the quantity being integrated.

In our example, the process of integrating gives us the volume of the solid by summing the volumes of the individual disks across the interval between \( 0 \) and \( 2 \) when revolving around the x-axis.

Revolving a Region

Revolving a Region refers to the process of spinning a 2D area, such as the one bounded by a function graph, around a line (usually an axis) to form a 3D object. This transformation generates a new geometric shape whose volume can be calculated using methods like the disk or washer method.

• When a region is revolved around the x-axis, the resulting solid has circular cross-sections perpendicular to the x-axis.
• Similarly, revolving around the y-axis yields circular cross-sections perpendicular to the y-axis, requiring adjustments to the integration process such as using functions of y instead.

The key here is visualizing and correctly setting up the equation to represent the solid's cross-sections.
In exercises, different rotation axes can alter the limits and integrands of the integral used to find the volume.

Definite Integral

A Definite Integral computes the integral value over a specified interval and gives a numerical result. It's different from an indefinite integral that represents a family of functions. In the volume of solids problem, the definite integral holds key importance.

• The definite integral is written as \( \int_a^b f(x) \, dx \), where \( a \) and \( b \) are the endpoints of the interval.
• This integral gives the exact accumulation of the quantity represented by the function \( f(x) \) from \( a \) to \( b \).
• The value of a definite integral corresponds to the total volume if the integrand represents a volume element.

By evaluating the definite integral for the region spun around an axis, as we did using limits \( 0 \) and \( 2 \), we were able to determine the exact volume of the 3D solid.