Problem 41 - Calculus for Biology and Medicine

Question

In Problems 41-46, find the volumes of the solids obtained by rotating the region bounded by the given curves about the \(x\) -axis. In each case, sketch the region together with a typical disk element. \(y=x^{2}, y=x, 0 \leq x \leq 1\)

Step-by-Step Solution

Answer

The volume is \(\frac{2\pi}{15}\).

Step 1: Sketch the Region and Identify Bounds

First, sketch the region of the area bounded by the curves. You need to graph the equations of the functions: \(y = x^2\) and \(y = x\), and the lines \(x = 0\) and \(x = 1\). The curves will intersect at the points where \(x^2 = x\), which simplifies to \(x(x-1) = 0\), giving \(x = 0\) and \(x = 1\). Thus, plot these functions on a coordinate plane and shade the bounded region between \(x = 0\), \(x = 1\), \(y = x\), and \(y = x^2\). This will visually show the region to rotate.

Step 2: Set up the Disk Method Formula

The volume of the solid can be calculated using the disk method because we are rotating the area about the \(x\)-axis. The formula for the disk method is given by\[ V = _{a}^{b} ig( f(x)^2 - g(x)^2 ig) \, dx \]where \(f(x)\) is the function that represents the outer curve and \(g(x)\) the inner curve. For this problem, the outer radius is \(f(x) = x^{2}\) and the inner radius is \(g(x) = x\), with the interval \([0, 1]\).

Step 3: Integrate to Find the Volume

Substitute the functions into the integral formula:\[ V = _{0}^{1} ig( (x)^2 - (x^2)^2 ig) \, dx = _{0}^{1} (x^2 - x^4) \, dx \]To find the volume, evaluate this integral from 0 to 1.

Step 4: Evaluate the Definite Integral

Calculate the integral:\[ _{0}^{1} (x^2 - x^4) \, dx = _{0}^{1} (x^2) \, dx - _{0}^{1} (x^4) \, dx \]This simplifies to:\[ \left[ \frac{x^3}{3} \right]_{0}^{1} - \left[ \frac{x^5}{5} \right]_{0}^{1}\]Calculate each term:\[ \left( \frac{1^3}{3} - \frac{0^3}{3} \right) - \left( \frac{1^5}{5} - \frac{0^5}{5} \right) = \frac{1}{3} - \frac{1}{5} \]Finally, find a common denominator and evaluate:\[ \frac{5}{15} - \frac{3}{15} = \frac{2}{15} \]

Step 5: Conclude the Volume Solution

The computed integral gives us the volume of the solid formed by rotating the region around the \(x\)-axis. Therefore, the volume of the solid is \( \frac{2\pi}{15} \).

Key Concepts

Disk Method

The disk method is a powerful tool in calculus for finding the volume of a solid of revolution. This technique is used when a region in the coordinate plane is revolved around a line, forming a 3D object. Picture a flat region cut out of cardboard; when it's spun around an axis, it transforms into a shape that has thickness and volume.

Here's what you need to know about the disk method:

• The axis of rotation: In our case, it is the x-axis. The entire 2D region is spun around this line to create a solid.
• A typical disk: Imagine slicing the solid into numerous "disks" perpendicular to the axis of rotation. Each disk resembles a thin, flat cylinder.
• The formula: The volume element of each disk is \( \pi [f(x)]^2 \, dx \), where \( f(x) \) is the distance from the x-axis to the curve (called the radius of the disk). Combine these elements to get the integral formula for the volume \(V = \pi \int_{a}^{b} [f(x)]^2 \, dx \).

In our exercise, we've used this method to calculate the space occupied by the rotating region, bounded by the given curves.

Definite Integral

The concept of a definite integral is central in calculus, particularly when calculating areas and volumes. In this problem, we're finding the volume of a shape formed by rotating curves, and the definite integral plays a critical role.

Important points about definite integrals:

• Integration bounds: These are the points between which we're calculating, in our exercise from \( x = 0 \) to \( x = 1 \).
• The definite integral of a function \( f(x) \) over an interval \( [a, b] \) captures the net "area" under the function across that interval.
• Applications: When you multiply this area by \( \pi \), you ascertain the volume of the solid formed (as done in the disk method).

In the provided solution, we've evaluated the integral \( \pi \int_{0}^{1} (x^2 - x^4) \, dx \) using the fundamental theorem of calculus. This gives the overall volume of the shape rotated around the x-axis.

Rotating Curves

Rotating curves is a fascinating concept where we transform a 2D area into a 3D object by revolving it about an axis. Imagine winding a piece of paper around a pencil. As you rotate, the shadow that follows forms a three-dimensional figure.

In our exercise, we're specifically looking at:

• The functions involved: The curves given by the functions \( y = x^2 \) and \( y = x \) are rotated to form a solid between them.
• Identifying the bounded region: Before rotating, it's crucial to recognize the area enclosed by these curves. This enclosed region outlines the body that will be converted into volume.
• The axis of rotation: Here, we're revolving the region around the x-axis. The curves stretch along the x-axis, and as they rotate, they create a 3D shape "around" this line.

Understanding rotating curves is key to applying methods like the disk method, which in turn helps find the volume measurements of complex shapes that arise from these rotations.