Problem 8 - Thomas Calculus in SI Units

Question

Find the volumes of the solids. The base of a solid is the region bounded by the graphs of \(y=\sqrt{x}\) and \(y=x / 2 .\) The cross-sections perpendicular to the \(x\) -axis are a. isosceles triangles of height 6 b. semicircles with diameters running across the base of the solid.

Step-by-Step Solution

Answer

a. The volume is 4 cubic units. b. The volume is \(\frac{8\pi}{15}\) cubic units.

Step 1: Determine the Bounds of Integration

First, find the intersection points of the curves \(y=\sqrt{x}\) and \(y=\frac{x}{2}\). Set \(\sqrt{x} = \frac{x}{2}\). Solving for \(x\), square both sides to get \(x = \frac{x^2}{4}\). This leads to \(x(x-4) = 0\) so \(x = 0\) or \(x = 4\). Thus, the bounds of integration are from \(x=0\) to \(x=4\).

Step 2: Calculate the Length of the Base for Cross-Sections

For any \(x\) in [0,4], the length of the base of each cross-section is given by the difference of the functions: \(\sqrt{x} - \frac{x}{2}\).

Step 3: Volume of Isosceles Triangular Cross-Sections

The area of an isosceles triangle is \(\frac{1}{2}\) base \( \times\) height. Here, each triangle has a base of \(\sqrt{x} - \frac{x}{2}\) and a height of 6. Therefore, the area of a triangle is \(\frac{1}{2}(\sqrt{x} - \frac{x}{2}) \times 6 = 3(\sqrt{x} - \frac{x}{2})\). Integrate from 0 to 4: \(V = \int_{0}^{4} 3(\sqrt{x} - \frac{x}{2}) \, dx\).

Step 4: Evaluate the Integral for Triangular Cross-Sections

Evaluate \(\int_{0}^{4} 3(\sqrt{x} - \frac{x}{2}) \, dx\). This becomes \(\int_{0}^{4} (3\sqrt{x} - \frac{3x}{2}) \, dx\). Calculate each term separately: \( \int_0^4 3\sqrt{x} \, dx = 2x^{\frac{3}{2}}\Big|_0^4 = 16\) and \(\int_0^4 \frac{3x}{2} \, dx = \frac{3}{4}x^2\Big|_0^4 = 12\). Thus, the volume is \(16 - 12 = 4\) cubic units.

Step 5: Volume of Semicircular Cross-Sections

The area of a semicircle is given by \(\frac{1}{2} \pi r^2\). The diameter is \(\sqrt{x} - \frac{x}{2}\), so the radius \(r\) is \(\frac{\sqrt{x} - \frac{x}{2}}{2}\). Thus, the area is \(\frac{1}{2} \pi \left(\frac{\sqrt{x} - \frac{x}{2}}{2}\right)^2\). Integrate from 0 to 4: \(V = \int_{0}^{4} \frac{1}{2} \pi \left(\frac{\sqrt{x} - \frac{x}{2}}{2}\right)^2 \, dx\).

Step 6: Evaluate the Integral for Semicircular Cross-Sections

Evaluate the integral \(\int_{0}^{4} \frac{1}{8} \pi (\sqrt{x} - \frac{x}{2})^2 \, dx\). Expand \((\sqrt{x} - \frac{x}{2})^2\) to get \(x - x^{\frac{3}{2}} + \frac{x^2}{4}\). Thus, the integral becomes \(\frac{\pi}{8}\int_0^4 (x - x^{\frac{3}{2}} + \frac{x^2}{4}) \, dx\). Evaluate each term: \(\int_0^4 x \, dx = 8\), \(\int_0^4 x^{\frac{3}{2}} \, dx = \frac{16}{5}\), and \(\int_0^4 \frac{x^2}{4} \, dx = \frac{16}{3}\). The total is \(8 - \frac{16}{5} + \frac{16}{3} = \frac{64}{15}\). Thus, the volume is \(\frac{\pi}{8} \times \frac{64}{15} = \frac{8\pi}{15}\) cubic units.

Key Concepts

Cross-Sections

When we talk about cross-sections in geometry, we are discussing slices of a three-dimensional object that we get by cutting through it with a plane. Imagine slicing a loaf of bread; each slice is a cross-section.
For solids with diverse base structures, the shape and size of cross-sections can vary across the object. In the problem at hand, our cross-sections are perpendicular to the x-axis.
These cross-sections are more interesting as they can be triangles or semicircles depending on which part of the problem we're tackling. The key is to understand how these cross-sections change as they move along the base of our solid from one boundary to the other.

Definite Integrals

Definite integrals are a fundamental tool in calculus used to calculate the accumulation of quantities, which is perfect for finding volumes.
The definite integral essentially adds up tiny products of areas of cross-sections and small changes in x. Each of these contributes to the overall volume of the solid.
For any function and bounds, the definite integral provides the accumulated sum of the area "under the curve" in a precise interval. Here, we integrate specific functions from the intersection points, which give the bounds of 0 to 4.

Isosceles Triangles

Isosceles triangles are triangles with two equal sides and consequently, equal angle measures opposite those sides.
In our volume calculation problem, the cross-sections are vertical slices that form isosceles triangles. Each triangle has a base measured as the difference between two functions: \(y=\sqrt{x}\) and \(y=\frac{x}{2}\). The height is given as 6, meaning it is constant across the slices.
To find the volume, the area of each triangle is integrated across all slices from one boundary to the other. Mathematically, this means using a definite integral with the area formula for a triangle as the integrated function.

Semicircles

Semicircles are half-circles and are common in volume calculations for solids with such cross-sectional shapes.
Here, the semicircles have diameters that are stretched across the base of the solid. The radius thus becomes half this diameter, \( r = \frac{\sqrt{x} - \frac{x}{2}}{2}\).
The area of a semicircle \( \frac{1}{2}\pi r^2\) is calculated for each cross-section and accumulate through integration, providing us with a comprehensive volume of all semicircles stacked together from the beginning to the end of our domain.