Problem 42 - Thomas Calculus

Question

Find the volume of the solid generated by revolving each region about the \(y\) -axis. \begin{equation} \begin{array}{l}{\text { The region enclosed by the triangle with vertices }(0,1),(1,0),} \\ {\text { and }(1,1)}\end{array} \end{equation}

Step-by-Step Solution

Answer

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

Step 1: Understand the geometric figure

The region is a right triangle with vertices at \((0, 1)\), \((1, 0)\), and \((1, 1)\). The hypotenuse is the line connecting these points, and this line can be described by the function \(y = -x + 1\).

Step 2: Set up the integral using the disk method

To find the volume of revolution about the \(y\)-axis, we use the formula for the volume of a solid of revolution: \[ V = \pi \int_{a}^{b} [R(y)]^2 \, dy \]First, express \(x\) in terms of \(y\): \(x = 1 - y\) since the triangle is bounded by this line and revolve it around the \(y\)-axis to obtain the radius \(R(y) = 1- y\). The limits of integration are from \(y = 0\) to \(y = 1\).

Step 3: Integrate to find the volume

Calculate the integral:\[V = \pi \int_{0}^{1} (1-y)^2 \, dy\]First, expand \((1-y)^2\) to \(1 - 2y + y^2\). Then integrate term by term:\[V = \pi \left[ y - y^2 + \frac{y^3}{3} \right]_0^1\]Evaluating the integral gives:\[V = \pi \left( (1 - 1 + \frac{1}{3}) - (0 - 0 + 0) \right)\]

Step 4: Calculate the final volume

Simplify the results from the evaluation:\[V = \pi \frac{1}{3}\]So, the volume of the solid generated by revolving the region around the \(y\)-axis is \(\frac{\pi}{3}\).

Key Concepts

Disk Method

When we talk about finding the volume of a solid generated by revolving a region around an axis, the Disk Method is a popular technique. This method involves revolving a shape around an axis to form a three-dimensional solid. Imagine each slice of this solid as a disk or a coin. To find the volume, we add up the volumes of all these infinitesimally thin disks.

The basic formula used is:

• \( V = \pi \int_{a}^{b} [R(y)]^2 \, dy \)

In the original equation, \(R(y)\) represents the radius of the disks formed when the triangle is rotated around the \(y\)-axis. As the triangle spins around the \(y\)-axis, the radius of each disk is essentially the \(x\)-value of the original function, expressed in terms of \(y\). Hence, the radius is \(R(y) = 1 - y\), since \(x = 1 - y\).

Using this method makes it straightforward to transform a two-dimensional region into a solid by revolving it. The integration process calculates how these disks stack up to form a complete solid.

Solid of Revolution

A Solid of Revolution is a three-dimensional object created by revolving a two-dimensional shape around an axis. Think of it like spinning a potter's wheel to shape clay into a vessel. The process turns the flat shape into a full-bodied figure.

In the exercise, we started with a right triangle. By revolving this triangle around the \(y\)-axis, it formed a bell-like solid. To imagine it, picture the triangle spinning on a vertical axis, making a smooth figure that resembles a cone but is truncated at the top. This specific solid is limited on the top because the vertex point at \((0,1)\) and \((1,1)\) doesn't result in a closed form like a complete cone would.

Calculating the volume of this solid requires understanding how each infinitesimal slice (or disk) contributes to the entire volume. The Disk Method, mentioned earlier, helps determine these individual volumes effectively.

Integral Calculus

Integral Calculus plays a vital role in calculating volumes, areas, and other quantities under curves. In our task of finding volumes of solids of revolution, it serves as a crucial tool for accumulation.

When a function is rotated around an axis, we use integral calculus to add up the thin slices that make up the solid's volume. The disk's thickness is denoted by \(dy\), while its volume is determined by its radius squared, multiplied by \(\pi\).

The step-by-step solution used integration to compute:

• First, it expanded \((1-y)^2\) to get a formal expression.
• Then, it integrated \(1 - 2y + y^2\) term by term.
• Finally, it computed definite integrals from \(0\) to \(1\), resulting in the volume \(\frac{\pi}{3}\).

Integral calculus, therefore, transitions the solution from demonstrating theoretical geometry to a concrete numerical result. By executing precise integration here, we achieve an accurate measurement of the solid's volume.