Problem 10 - Thomas Calculus in SI Units

Question

Find the volumes of the solids. 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: Understand the Base and Cross-Section

The base of the solid is a circle given by the equation \(x^2 + y^2 \leq 1\). The cross-sections perpendicular to the \(y\)-axis are isosceles right triangles. The legs of these triangles extend from \(x = -\sqrt{1-y^2}\) to \(x = \sqrt{1-y^2}\). The length of the leg in the \(x\)-direction is therefore \(2\sqrt{1-y^2}\).

Step 2: Determine the Area of the Cross-Section

An isosceles right triangle with legs of length \(l\) has an area \(A = \frac{1}{2} l^2\). The length \(l\) in this case is \(2\sqrt{1-y^2}\). Thus, the area of the cross-section is \(A(y) = \frac{1}{2} (2\sqrt{1-y^2})^2 = 2(1-y^2)\).

Step 3: Set Up the Integral for Volume

The volume of the solid can be found by integrating the area of the cross-sections from \(y = -1\) to \(y = 1\). This gives the integral for the volume: \[ V = \int_{-1}^{1} A(y) \ dy = \int_{-1}^{1} 2(1-y^2) \ dy. \]

Step 4: Evaluate the Integral

Compute the integral: \[ V = 2 \int_{-1}^{1} (1 - y^2) \ dy. \] First, integrate \(1\) with respect to \(y\), giving \(y\), and \(-y^2\), giving \(-\frac{y^3}{3}\). Evaluate from \(-1\) to \(1\): \[ V = 2 \left[ y - \frac{y^3}{3} \right]_{-1}^{1} = 2 \left( \left[ 1 - \frac{1^3}{3} \right] - \left[ -1 + \frac{(-1)^3}{3} \right] \right). \]

Step 5: Simplify to Find the Final Volume

Calculating the above expression, \[ V = 2 \left( \left[ 1 - \frac{1}{3} \right] - \left[ -1 + \left(-\frac{1}{3}\right) \right] \right) = 2 \left( \frac{2}{3} + \frac{2}{3} \right). \] So, \[ V = 2 \times \frac{4}{3} = \frac{8}{3}. \] The volume of the solid is \(\frac{8}{3}\).

Key Concepts

Isosceles Right Triangles

An isosceles right triangle is a type of right triangle that has two sides of equal length and one right angle (90 degrees). In our exercise, these triangles are used in the cross-sections of the solid. These triangles have their legs along the x-direction. The hypotenuse is facing outwards in a way that makes it ideal for forming a solid through rotation or integration.

Key features of an isosceles right triangle include:

• Two sides (legs) are the same length.
• The hypotenuse is different from the legs and is the longest side.
• Angles are 45°, 45°, and 90°.

In this problem, we find the length of the legs within the base disk by calculating from \(-\sqrt{1 - y^2}\) to \(\sqrt{1 - y^2}\), making their combined length \(2\sqrt{1-y^2}\). This structure is foundational in setting up the geometry needed to solve for the volume of the solid of revolution.

Integral Calculus

Integral calculus is a branch of calculus, crucial for calculating areas under curves, among other applications. It's central to finding volumes of solids of revolution, like in the current exercise. Here, we use the integral to accumulate the cross-sectional areas along an axis.

The process involves using the formula:

• Find the expression for the area of the cross-section, which is the isosceles right triangle in this case: \(A(y) = 2(1-y^2)\).
• Set up the definite integral over the interval of interest (from \(y = -1\) to \(y = 1\)).

In our problem:\[ V = \int_{-1}^{1} 2(1-y^2) \, dy \]This integral computes the sum of the areas along the y-axis, giving the total volume. Evaluating this integral uses the antiderivative calculations: \[ V = \left[ y - \frac{y^3}{3} \right]_{-1}^{1} \], which after substitution and simplification yields the final volume. Understanding each step in the integration process provides a deep insight into how volume is calculated using calculus.

Cross-Sectional Area

The cross-sectional area is a key factor when calculating the volume of a solid. It essentially slices the solid into manageable parts that can be easily added together through integration. In this task, determining the correct cross-sectional area of the isosceles right triangles is essential.

To understand and calculate the cross-sectional area in this exercise:

• Recognize that the base of each triangle leg spans from \(-\sqrt{1-y^2}\) to \(\sqrt{1-y^2}\), giving it a length of \(2\sqrt{1-y^2}\).
• Calculate the area using the triangle area formula \(\frac{1}{2} \times \text{base} \times \text{height}\), which simplifies to \(2(1-y^2)\).

Once this area expression is integrated along the y-axis, the entire volume of the solid is obtained. Visualizing each triangle as a slice of the volume helps in understanding how sequential slices add up to the full volume, showcasing the elegance of using cross-sectional areas in calculus.