Problem 8 - Thomas Calculus

Question

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 \begin{equation} \begin{array}{l}{\text { a. isosceles triangles of height } 6 \text { . }} \\\ {\text { b. semicircles with diameters running across the base of the solid. }}\end{array} \end{equation}

Step-by-Step Solution

Answer

Volumes are 4 for part a and \( \frac{28\pi}{15} \) for part b.

Step 1: Find Intersection Points

To determine the region's boundaries, set the equations equal to each other to find their points of intersection: \( \sqrt{x} = \frac{x}{2} \). Squaring both sides results in \( x = \frac{x^2}{4} \). Rearranging gives \( x^2 - 4x = 0 \), i.e., \( x(x - 4) = 0 \). Thus \( x = 0 \) and \( x = 4 \) are the intersection points.

Step 2: Calculate the Area of a Triangular Cross-Section (Part a)

For part a, the cross-section is an isosceles triangle with its base along the distance between the curves \( y=\sqrt{x} \) and \( y=\frac{x}{2} \), and height 6. The base length at any \( x \) is \( \sqrt{x} - \frac{x}{2} \). The area of the triangle is \( \frac{1}{2} \times \text{(base)} \times \text{(height)} = \frac{1}{2} \times (\sqrt{x} - \frac{x}{2}) \times 6 \). Simplifying, the area is \( 3(\sqrt{x} - \frac{x}{2}) \).

Step 3: Integrate to Find Volume of Solid with Triangular Cross-Sections

The volume is the integral of the cross-sectional area along the x-axis from 0 to 4. Thus, \( V = \int_{0}^{4} 3(\sqrt{x} - \frac{x}{2}) \, dx \). Solve the integral: \( \int_{0}^{4} 3\sqrt{x} \, dx = 3 \times \frac{2}{3}x^{3/2} \bigg|_{0}^{4} \) and \( \int_{0}^{4} -\frac{3x}{2} \, dx = -\frac{3}{2} \times \frac{x^2}{2} \bigg|_{0}^{4} \). Calculating gives \( V = 2 \times 8 - \frac{3}{4} \times 16 = 16 - 12 = 4 \).

Step 4: Calculate the Radius of Semicircles (Part b)

For part b, the diameter of the semicircle at any \( x \) is \( \sqrt{x} - \frac{x}{2} \). The radius \( r \) is half of the diameter: \( r = \frac{1}{2}(\sqrt{x} - \frac{x}{2}) \).

Step 5: Find Area of Semicircular Cross-Section (Part b)

The area of a semicircle is \( \frac{1}{2} \pi r^2 \). Thus, the area of a semicircular cross-section at \( x \) is \( \frac{1}{2} \pi \left(\frac{1}{2}(\sqrt{x} - \frac{x}{2})\right)^2 \). Simplifying, we get \( \frac{\pi}{8} (\sqrt{x} - \frac{x}{2})^2 \).

Step 6: Integrate to Find Volume of Solid with Semicircular Cross-Sections

The volume is the integral of the cross-sectional area along the x-axis from 0 to 4. Thus, \( V = \int_{0}^{4} \frac{\pi}{8} (\sqrt{x} - \frac{x}{2})^2 \, dx \). Simplify and solve this integral: first expand \( (\sqrt{x} - \frac{x}{2})^2 = x - x\sqrt{x} + \frac{x^2}{4} \). Substitute back into the integral and compute each term separately to find the volume.

Step 7: Compute the Integral for Volume with Semicircular Cross-Sections

The integral splits as follows: \( V = \frac{\pi}{8} [\int_{0}^{4} x \, dx - 2 \int_{0}^{4} x\sqrt{x} \, dx + \frac{1}{4} \int_{0}^{4} x^2 \, dx] \). Compute each term: \( \frac{1}{2}x^2 \bigg|_{0}^{4} = 8 \), \( \frac{2}{5}x^{5/2} \bigg|_{0}^{4} = \frac{64}{5} \), and \( \frac{1}{12}x^3 \bigg|_{0}^{4} = \frac{64}{3} \). Thus, the volume \( V = \frac{\pi}{8} (8 - \frac{128}{5} + \frac{64}{3}) \). Simplify to find the final volume.

Key Concepts

Intersection Points

Finding the intersection points of two curves is like discovering where their paths cross. It tells us the boundary of the region we are dealing with. Consider the functions given in the exercise:

• The first function is the square root function, represented by, \( y = \sqrt{x} \).
• The second function is a straight line, given by \( y = \frac{x}{2} \).

To find where these two meet, we set them equal to each other, resulting in the equation \( \sqrt{x} = \frac{x}{2} \). Squaring both sides helps eliminate the square root, shaping the equation to \( x = \frac{x^2}{4} \). Rearranging gives us two solutions: \( x = 0 \) and \( x = 4 \). These values are our intersection points, marking the boundaries of the solid's base.

Cross-Sectional Area

Cross-sectional areas are the surfaces that are perpendicular to a particular direction of growth in a solid. In this problem, we find the cross-sections by looking in the direction of the x-axis.These cross-sections are different shapes:

• Part a: Isosceles triangles
• Part b: Semicircles

For isosceles triangles, the base lies between \( y=\sqrt{x} \) and \( y=\frac{x}{2} \). The length of this base at any point, \( x \), is \( \sqrt{x} - \frac{x}{2} \). For semicircles, the diameter is the same as the base length. These measurements will be used to determine each shape's area, a crucial step for finding the volume.

Isosceles Triangles

An isosceles triangle is symmetrical around its height. Its shape is easy to divide into two equal right triangles, which helps in calculating its area. Consider that each cross-section is an isosceles triangle:

• The height of these triangles is constant at 6 units.
• The base changes according to the position along the x-axis and is \( \sqrt{x} - \frac{x}{2} \).

The area of each triangle, therefore, is \( \frac{1}{2} \times (\text{base}) \times (\text{height}) = 3(\sqrt{x} - \frac{x}{2}) \). Calculating this for each cross-section across the x-axis from \( x=0 \) to \( x=4 \), will help us find the solid's total volume through integration.

Semicircles

Semicircles are half-circles, and their geometrical properties derive from their full-circle cousins. At any slice through the solid, the semicircle's diameter extends from \( y=\sqrt{x} \) to \( y=\frac{x}{2} \).The radius \( r \) of the semicircle is half the length of this diameter, given by \( r = \frac{1}{2}(\sqrt{x} - \frac{x}{2}) \).The formula for the area of a semicircle is rooted in the area formula for a full circle, \( \frac{\pi r^2}{2} \). Thus, for a cross-sectional semicircle: \( \text{Area} = \frac{\pi}{8} (\sqrt{x} - \frac{x}{2})^2 \). Integration of these areas across the solid's base provides the total volume.

Integration

Integration is the magical tool that sums up all the little pieces to find a whole. In this case, we use integration to compute the volume of the solid, by adding up all the infinitely thin cross-sectional areas along the x-axis.For the isosceles triangles, the volume is given by the integration of their area from \( x=0 \) to \( x=4 \).\[ V = \int_{0}^{4} 3(\sqrt{x} - \frac{x}{2}) \, dx \]This helps us understand how each slice contributes to the overall volume.Similarly, for the semicircles, the integration also happens over the same interval:\[ V = \int_{0}^{4} \frac{\pi}{8} (\sqrt{x} - \frac{x}{2})^2 \, dx \]By solving these integrals separately for each shape, we calculate the total volume of the entire solid, obtaining the fusion of mathematics and geometry that accurately reveals its structure.