Problem 51 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The volume is \( \frac{3\pi}{10} \) cubic units.

Step 1: Sketch the Region

First, graph the curves \( y = x^2 \) and \( y = \sqrt{x} \) on the given interval \( 0 \leq x \leq 1 \). The curve \( y = x^2 \) is a parabola opening upwards, while \( y = \sqrt{x} \) is a square root function, starting from point (0,0) and increasing to (1,1). Identify the region bounded between these two curves.

Step 2: Identify Typical Element

Since we are rotating around the \(y\)-axis, we'll use the method of cylindrical shells. A typical element at position \(x\) in the region can be imagined as a vertical slice between the curves \( y = x^2 \) and \( y = \sqrt{x} \).

Step 3: Apply the Disk Element Formula

The volume of a typical disk element when rotating around the \(y\)-axis can be expressed using the disk method formula for integration: \[ V = \pi \int_{0}^{1} (R(x)^2 - r(x)^2) \, dx \] where \( R(x) = \sqrt{x} \) (outer radius) and \( r(x) = x^2 \) (inner radius).

Step 4: Integrate to Find Volume

Substitute \( R(x) \) and \( r(x) \) into the formula to find the volume:\[ V = \pi \int_{0}^{1} ((\sqrt{x})^2 - (x^2)^2) \, dx \]This simplifies to: \[ V = \pi \int_{0}^{1} (x - x^4) \, dx \]

Step 5: Evaluate the Integral

Calculate the integral:\[ V = \pi \left[ \frac{x^2}{2} - \frac{x^5}{5} \right]_{0}^{1} \]Evaluating this from 0 to 1 gives:\[ V = \pi \left(\frac{1}{2} - \frac{1}{5}\right) = \pi \left(\frac{5 - 2}{10}\right) = \pi \left(\frac{3}{10}\right) \]

Step 6: Final Volume Calculation

The calculated volume of the solid formed by rotating the region around the \(y\)-axis is:\[ V = \frac{3\pi}{10} \] cubic units.

Key Concepts

Cylindrical Shells

Understanding the concept of cylindrical shells is essential when dealing with rotational volumes, especially around the y-axis. Picture a vertical slice of your bounded region; when this slice is rotated around the y-axis, it forms a cylindrical shell.
Here’s why cylindrical shells are useful:

• They simplify finding the volume by focusing on these vertical slices.
• Each slice can be measured in terms of its height and circumference, making calculations more straightforward.

To apply the method of cylindrical shells, remember that every shell has a height, a radius (distance to y-axis), and a thickness (an infinitesimal change in x). Integrating these shells gives us the total volume of the solid.

Disk Method

The disk method is another way to determine volumes of solids of revolution. This technique is particularly useful when rotating a region around the x-axis but can also be adapted for the y-axis. Here’s how it works:

• Visualize cutting the solid into many thin, coin-shaped disks stacked along the axis of rotation.
• The volume of each disk is fairly easy to calculate: it revolves around one radius representing the edge of the region.

In this exercise, you have to compare the radii from two functions, which gives the outer and inner disks. By integrating the difference of the areas of these disks, you can find the volume of the solid between these curved surfaces. This method is closely tied to the formula \[ V = \pi \int (R(x)^2 - r(x)^2) \, dx \] where you subtract the inner solid from the outer solid to find the remainder.

Integral Calculus

Integral calculus is the mathematical study that allows us to calculate volumes, like the one in this exercise, using integration. It's deeply connected to the disk and shell methods. Integral calculus enables us to find areas under curves, calculate rates of change, and, importantly, determine volumes of solids of revolution.
When integrating, you're essentially summing infinitely small slices to find a whole:

• Definite integrals compute the volume between specific bounds.
• You can visualize this integration process by seeing how each infinitesimal part contributes to the entire volume.

For the problem at hand, we used \[ \int_{0}^{1} (x - x^4) \, dx \] which efficiently sums up the differences in curved areas across bounded points.

Parabola and Square Root Function

Both the parabola and the square root function play crucial roles in defining the boundaries of the region you'll be rotating. The parabola, expressed as \(y = x^2\), is a classic U-shaped curve. For this exercise, it's plotted from \((0,0)\) to \((1,1)\), opening up and defining one boundary of the area. Meanwhile, the square root function, represented by \(y = \sqrt{x}\), acts as the other bounding curve. It starts steeply increasing from \((0,0)\) and tapers off as \(x\) increases.

• These functions cross each other at the points of intersection along the interval.
• Their difference in shape creates the crescent-like region that we revolve.

Understanding the characteristics of these functions helps in correctly applying integration techniques of the solid’s volume calculation.