Question

In Exercises 39-42, find the volume of the solid analytically. 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.

Short answer

The volume of the solid is \(4/3\) cubic units.

Step-by-step solution

Understand the solid

The given solid has its base as the disk defined by \(x^{2}+y^{2} \leq 1\). For any y-value between -1 and 1, the cross-section of the solid perpendicular to the y-axis is an isosceles right triangle, with its one leg along the x-axis.

Find the leg of the triangle

For any given y-value within the domain, the x-values extend from -sqrt(1-y^2) to sqrt(1-y^2). Hence, the length of one leg of the triangle, L, is double this value: \(L = 2 \times \sqrt {1-y^2}\).

Calculate the area of the triangle

Since the cross sections are isosceles right triangles, the area, A, can be calculated as \(A = 0.5 \times L^2 = 0.5 \times (2 \times \sqrt{1-y^2})^2 = 2 (1 - y^2) \).

Set up the integral

The volume of the solid, V, can be calculated by integrating the cross-sectional area, A, from -1 to 1, across the y-domain: \( V = \int_{-1}^{1} A \, dy = \int_{-1}^{1} 2(1 - y^2) \, dy \).

Solve the integral

You can compute the volume by evaluating this integral, with results in \(V = 2 \int_{-1}^{1} (1 - y^2) \, dy = 4/3\), after this, you can conclude the problem.