Problem 62 - Calculus: Early Transcendentals

Question

The base of \( S \) is a circular disk with radius \( r \). Parallel cross- sections perpendicular to the base are isosceles triangles with height \( h \) and unequal side in the base. (a) Set up an integral for the volume of \( S \). (b) By interpreting the integral as an area, find the volume of \( S \).

Step-by-Step Solution

Answer

The volume of the solid is \( \frac{h \pi r^2}{2} \).

Step 1: Understand the Solid's Cross-Section

The solid given is structured by taking parallel slices perpendicular to its circular base. Each of these cross-sections is an isosceles triangle with a constant height, \( h \), and a base equal to the diameter of the circle, \( 2r \). This means every cross-section at a particular height is an isosceles triangle with a width that changes depending on the position within the circle but has a maximum of \( 2r \).

Step 2: Express Triangle Base as a Function of Position

Consider a slice at a position \( x \) from the center of the circle (base). At a distance \( x \) from the center, the width \( x' \) of the base of the triangle slice can be described by the equation \( x' = 2 \sqrt{r^2 - x^2} \) using the circle equation \( x^2 + y^2 = r^2 \). This ensures that the slice's base ranges from \(-r\) to \(r\) within the circle.

Step 3: Calculate Area of Cross-Section

The area \( A(x) \) of each triangular slice at position \( x \) is calculated using the formula for the area of a triangle: \( A(x) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2\sqrt{r^2 - x^2} \times h \). Simplifying, this becomes \( A(x) = h \sqrt{r^2 - x^2} \).

Step 4: Set Up the Integral for Volume

The volume \( V \) of the solid is calculated by integrating the area of the cross-sections along the base of the circular disk. This results in: \[ V = \int_{-r}^{r} A(x) \, dx = \int_{-r}^{r} h \sqrt{r^2 - x^2} \, dx \]. This integral represents the sum of all triangular slices that comprise the solid.

Step 5: Solve the Integral for the Volume

Recognizing the integral \( \int \sqrt{r^2 - x^2} \, dx \) as the formula for the area of a semicircle, modify the limits and evaluate: The integral from \(-r\) to \(r\) of \( \sqrt{r^2 - x^2} \) is: \( \frac{\pi r^2}{2} \), which corresponds to the area of a semicircle with radius \( r \). Therefore, \[ V = h \cdot \frac{\pi r^2}{2} = \frac{h \pi r^2}{2} \]. This is the volume of the solid \( S \).

Key Concepts

Integral Calculus

Integral calculus is a crucial part of understanding the volume of complex shapes. It involves summing infinitesimally small quantities to find wholes, such as areas under a curve or the volume of solids. In this exercise, we use integral calculus to find the volume of the solid structure described.
When dealing with volumes, you often need to compute an integral over a given domain. For this particular solid with circular base and isosceles triangle-shaped cross-sections, integral calculus helps us to sum up the areas of all these triangles through the thickness of the solid.
In the step-by-step solution, we set up the integral equation: \[V = \int_{-r}^{r} A(x) \, dx = \int_{-r}^{r} h \sqrt{r^2 - x^2} \, dx \]which sums the areas of each isosceles triangle slice across the diameter of the circular disk. This expression essentially combines all areas to give the total volume.

Cross-Sections

Understanding cross-sections is essential for determining the volume of a 3D object. A cross-section refers to the intersection of a solid and a plane, creating a 2D shape such as a triangle, circle, or rectangle. In this problem, each cross-section of the solid is an isosceles triangle.
Cross-sections vary depending on the shape of the solid and the angle of the intersecting plane. Here, these cuts are perpendicular to the circular base and consistently form isosceles triangles with a fixed height. The base of each triangular slice changes, determined by how far the slice is from the center of the circle.
To find the base's length at position \( x \) from the center, we use the formula: \[x' = 2 \sqrt{r^2 - x^2} \]This makes it clear that the triangles get narrower as they move towards the edges of the circle.

Isosceles Triangles

An isosceles triangle is a type of triangle that has at least two sides of equal length. The problem involves these types of triangles as cross-sections of the solid.
The height \( h \) of these triangles remains constant, whereas the base width changes. At the very center of the circular base, the isosceles triangle's base is at its maximum width (the diameter of the circle, \(2r\)).
The area of each cross-section is calculated as:\[A(x) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2\sqrt{r^2 - x^2} \times h = h \sqrt{r^2 - x^2}\]The isosceles triangles form a crucial part of determining the solid's volume, as they are the fundamental building blocks stacked to form the entire structure. Understanding how the base changes according to its position ensures accurate calculation of the volume.