Problem 34 - Calculus

Question

In \(29-34\) find the volume and draw a typical slice. The base is the triangle with corners \((0,0),(1,0),(0,1) .\) Slices perpendicular to the \(x\) axis are semicircles.

Step-by-Step Solution

Answer

The volume is \(\frac{\pi}{24}\).

Step 1: Understand the Problem

The base of the solid is a right triangle with vertices at \((0,0)\), \((1,0)\), and \((0,1)\). The solid has semicircular slices perpendicular to the \(x\)-axis.

Step 2: Determine Equation of the Line for the Base

The base of the triangle is on the \(x\)-axis from \((0,0)\) to \((1,0)\), and it rises linearly to \((0,1)\). The equation of the line that forms the hypotenuse of the triangle is \(y = 1 - x\).

Step 3: Find the Radius of the Semicircle

Slices are semicircles with diameter along the \(y\)-axis. The diameter of a semicircle at a point \(x\) is the length from the \(x\)-axis to the line \(y = 1 - x\), which is \(1 - x\). So, the radius is \(\frac{1 - x}{2}\).

Step 4: Calculate the Area of a Typical Semicircular Slice

The area \(A(x)\) of a semicircle with radius \(r\) is \(\frac{1}{2}\pi r^2\). Substituting the radius \(\frac{1-x}{2}\), we have:\[ A(x) = \frac{1}{2} \pi \left( \frac{1-x}{2} \right)^2 = \frac{\pi}{8} (1-x)^2. \]

Step 5: Integrate to Find the Volume

Integrate the area function from Step 4 over the interval \(x=0\) to \(x=1\) to find the volume of the solid:\[ V = \int_0^1 \frac{\pi}{8} (1-x)^2 \, dx. \]

Step 6: Solve the Integral

Solve the integral:\[ V = \frac{\pi}{8} \int_0^1 (1-x)^2 \, dx = \frac{\pi}{8} \left[ \frac{1}{3} - \frac{2}{2} \times \frac{1}{2} + \frac{x^3}{3} \right]_0^1 = \frac{\pi}{8} \left[ \frac{1}{3} \right] = \frac{\pi}{24}. \]

Key Concepts

Semicircular slices

Visualizing the solid from this exercise can be a fun challenge! Imagine a triangle lying flat on the plane, with its base running along the x-axis. Perpendicular to this axis, there are semicircular slices that rise and fall as you move from one side of the triangle to the other. Each slice is like a half-circle peeking out from the line, and the size of these semicircles changes based on their position along the x-axis due to the triangular base. Here's how it works:

• The semicircles are perpendicular to the x-axis.
• The diameter of each semicircle is determined by how far you are along the x-axis.
• The height of the triangle at that point determines the size of the semicircle's diameter.

This geometric concept helps us divide the solid into distinct, calculable sections that can lead to finding the overall volume.

Integral calculus

Integrating over a region is the heart of finding volumes in calculus, especially when it comes to understanding solids that aren't basic geometric shapes. To find the volume of the solid in this exercise, we take the area of each slice and sum them up over a certain range using integration. Here's a simple breakdown:

• The area of a typical slice is calculated with the help of the formula for the area of a semicircle.
• This semicircular area is then expressed as a function of x, based on the radius determined from the triangular base.
• The integration of this area function over the range from x=0 to x=1 gives the complete volume of the solid.

By integrating, we add up the infinitesimally small areas to yield the total volume, demonstrating the power of calculus in solving real-world geometric problems.

Triangular base

The triangular base serves as the foundational aspect guiding the shape and size of the solid in this problem. This triangle has vertices at (0,0), (1,0), and (0,1), establishing a right triangle with specific boundary lines. Here's what makes the triangular base essential:

• It determines the location and length of the semicircular diameters.
• The coordinates of the triangle provide the limits of integration in the volume calculation.
• Based on the rise from (0,0) to the point (0,1), the line calculated as y = 1 - x dictates the length of each semicircle along the x-axis.

The triangle thus plays a critical role in determining not just the base area for each slice, but also inspires the whole approach to using calculus in solving this volume problem.