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Chapter 6: Problem 6

A solid has a circular base and cross sections perpendicular to the base are squares. What method should be used to find the volume of the solid?

Short Answer

Expert verified
Answer: The volume of the given solid is \(16r^3\).

Step by step solution

01

Identify variables

Let the radius of circular base be r. The volume (V) we want to find will be obtained by adding the areas of all the square cross-sections.
02

Set up the integral

Since volume (V) is a summation of areas of cross-sectional squares, we will use definite integral to represent it. Let x be the distance from the center of the circular base to any point on the circumference. \[ V = \int_{-r}^{r}A(x) dx \] Where the A(x) is the area of the square cross section at a distance x.
03

Find a formula for A(x)

Using the Pythagorean theorem, we can find the side (s) of the square cross-section at a distance x from the center. The side of the square is twice the height at that point (x) which we shall call H; this is the line segment from the circumference of the circle to the center (from the point P to the center C): \[ s(x) = 2H = 2(r^2 - x^2)^{1/2} \] Area of the square (A) is the square of the side length (s) so: \[ A(x) = s^2(x) = 4(r^2 - x^2) \]
04

Substitute A(x) in integral

Now, we substitute the A(x) formula into the volume integral: \[ V = \int_{-r}^{r} 4(r^2 - x^2) dx \]
05

Evaluate the integral

Integrate the function with respect to x and apply the limits -r to r: \[ V = 4 \int_{-r}^{r} (r^2 - x^2) dx = 4 \left [ r^2x - \frac{1}{3}x^3 \right ]_{-r}^{r} \] After evaluating, we get: \[ V = 8r^3 - (-8r^3) = 16r^3 \] The volume of the given solid is \(16r^3\).

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