Problem 7 - The Calculus with Analytic Geometry

Question

The base of a solid is a circle with a radius of 9 in., and each plane section perpendicular to a fixed diameter of the base is a square having a chord of the circle as a diagonal. Find the volume of the solid.

Step-by-Step Solution

Answer

The volume of the solid is 1944 cubic inches.

Step 1: Understand the Geometry

The base of the solid is a circle with a radius of 9 inches. A key detail is that each cross-section perpendicular to a fixed diameter of the circle is a square whose diagonal matches a chord of the circle.

Step 2: Express a Chord as a Function of Distance

For an arbitrary point at distance x from the center along the diameter, a chord perpendicular to the diameter and centered at this point has a length determined by the circle's equation: \[ \sqrt{9^2 - x^2} \]. The chord length is 2\sqrt{9^2 - x^2}.

Step 3: Relate the Chord to the Square's Diagonal

Since the chord is the diagonal of the square, we can relate the side length s of the square to the chord length: \[ s\sqrt{2} = 2\sqrt{9^2 - x^2} \]. So, the side length s of the square is \[ s = \sqrt{2} \cdot \sqrt{9^2 - x^2} = \sqrt{2(81 - x^2)} \].

Step 4: Calculate the Area of the Square

The area A of the square as a function of x is: \[ A(x) = s^2 = 2(81 - x^2) = 162 - 2x^2. \]

Step 5: Set Up the Volume Integral

The volume V of the solid is found by integrating the area of the square from -9 to 9 along the diameter: \[ V = \int_{-9}^{9} A(x)\, dx = \int_{-9}^{9} (162 - 2x^2)\, dx. \]

Step 6: Evaluate the Integral

Evaluate the integral to find the volume: \[ V = \int_{-9}^{9} 162\, dx - \int_{-9}^{9} 2x^2\, dx. \] These can be solved separately as: \[ V = 162x \Big|_{-9}^{9} - 2\int_{-9}^{9} x^2\, dx. \] \[ = 162\cdot[9 - (-9)] - 2\cdot\frac{x^3}{3}\Big|_{-9}^{9} \] \[ = 162\cdot 18 - 2\cdot \frac{(9)^3 - (-9)^3}{3} \] \[ = 2916 - \left(2 \cdot \frac{729 - (-729)}{3}\right) \] \[ = 2916 - 2\cdot \frac{1458}{3} = 2916 - 2\cdot 486 = 2916 - 972 = 1944. \]

Key Concepts

Integrals

Integrals are a fundamental concept in calculus that help us understand and calculate areas, volumes, and accumulations. In this exercise, we use integrals to determine the volume of a solid. When setting up an integral to find volume, we need a function that represents the cross-sectional area of the solid at different points along an axis. In general, if you have a function that describes the area A(x) of a cross-section at position x along the axis, the volume V of the solid can be found using the integral:




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