Problem 10 - Thomas Calculus

Question

Find the volumes of the solids in Exercises \(3-10\) . The base of the solid is the disk \(x^{2}+y^{2} \leq 1 .\) The cross-sections by planes perpendicular to the \(y\) -axis between \(y=-1\) and \(y=1\) are isosceles right triangles with one leg in the disk.

Step-by-Step Solution

Answer

The volume of the solid is \( \frac{8}{3} \).

Step 1: Understanding the Problem

We have a solid whose base is a disk defined by the inequality \( x^2 + y^2 \leq 1 \). The cross-sections of this solid, perpendicular to the \( y \)-axis, are isosceles right triangles. Our goal is to find the volume of this solid by integrating across the relevant domain.

Step 2: Determine the Area of Cross-Section

For any cross-section at a particular \( y \)-value, the base of the isosceles right triangle is given by the diameter of the disk's cross-section at that \( y \). The length of this base is \( b = 2\sqrt{1-y^2} \), since that's the length from \(-\sqrt{1-y^2}\) to \(\sqrt{1-y^2}\). The area \( A \) of an isosceles right triangle with leg length \( b \) is \( \frac{1}{2}b^2 \). Inserting for \( b \), we get: \[ A(y) = \frac{1}{2} (2\sqrt{1-y^2})^2 = 2(1-y^2) \]

Step 3: Set Up the Integral for the Volume

The volume of the solid can be found by integrating the area of the cross-sections with respect to \( y \) over the interval from \(-1\) to \(1\). Thus, we set up the integral for the volume \( V \) as follows: \[ V = \int_{-1}^{1} A(y) \, dy = \int_{-1}^{1} 2(1-y^2) \, dy \]

Step 4: Evaluate the Integral

Evaluate the integral: \[ V = 2 \int_{-1}^{1} (1-y^2) \, dy \] This can be split into two separate integrals:\[ V = 2 \left( \int_{-1}^{1} 1 \, dy - \int_{-1}^{1} y^2 \, dy \right) \]Calculate each separately:- \( \int_{-1}^{1} 1 \, dy = [y]_{-1}^{1} = 1 - (-1) = 2 \)- \( \int_{-1}^{1} y^2 \, dy = \left[ \frac{y^3}{3} \right]_{-1}^{1} = \frac{1}{3} - \frac{-1}{3} = \frac{2}{3} \)Finally, substitute back:\[ V = 2 \left( 2 - \frac{2}{3} \right) = 2 \times \frac{4}{3} = \frac{8}{3} \]

Step 5: Conclusion

The volume of the solid is \( \frac{8}{3} \). This result is obtained by integrating the area of the cross-sections over the specified range.

Key Concepts

Integration

Integration is a powerful mathematical technique used to find the accumulation of quantities, such as area or volume. In the context of finding volumes of solids, integration helps us sum up infinitesimally small slices or cross-sections of the solid to determine the entire volume. Each slice is evaluated as we move along a particular axis. In this exercise, we calculate the area of each isosceles right triangle, which forms the cross-section, and then integrate these areas over the given range of the variable along the y-axis. Thus, integration transforms these small areas into a collective whole—essentially rebuilding the shape in terms of volume.

Isosceles Right Triangle

An isosceles right triangle is a special type of triangle where two sides (the legs) are equal in length, and the angle between them is 90 degrees. In the problem, each cross-section of the solid is an isosceles right triangle with one of its legs lying within the disk's boundary. The area formula for this triangle is provided as:

• Area: \( A = \frac{1}{2} b^2 \)

where \( b \) is the length of the leg. By understanding this geometry, we can systematically determine the cross-sectional area as we move across the disk, which is crucial in calculating the total volume of the solid.

Cross-sections

Cross-sections are essentially the slices or sections of a solid that are perpendicular to an axis. In this problem, the cross-sections are taken perpendicular to the y-axis. As we move from \( y = -1 \) to \( y = 1 \), each cross-section reveals the shape of an isosceles right triangle. The complexity of this exercise lies in determining how these individual cross-sections change as a function of \( y \). The length of the leg of the triangle depends on the value of \( y \) and is calculated from the equation of the circle, giving rise to the unique triangular cross-section at each point.

Disk Geometry

Disk geometry plays a pivotal role in understanding the foundational shape of this problem. The base of the solid is defined by the inequality \( x^2 + y^2 \leq 1 \), which describes a disk of radius 1 centered at the origin. The disk's geometry helps us ascertain the bounds for the base of the isosceles right triangles. As each cross-section is taken across the y-axis, the diameter of the circle at each slice position \( y \) is calculated. This diameter determines the length of the sliced triangle's base. Specifically, it extends from \( -\sqrt{1-y^2} \) to \( \sqrt{1-y^2} \), and thus the diameter of the circle at any height \( y \) is \( 2\sqrt{1-y^2} \). Understanding this ensures accurate integration of cross-sectional areas to derive the solid's volume.