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Chapter 6: Problem 4

Find the volumes of the solids. The solid lies between planes perpendicular to the \(x\) -axis at \(x=-1\) and \(x=1 .\) The cross-sections perpendicular to the \(x\) -axis between these planes are squares whose diagonals run from the semicircle \(y=-\sqrt{1-x^{2}}\) to the semicircle \(y=\sqrt{1-x^{2}}\).

Short Answer

Expert verified
The volume of the solid is \( \frac{8}{3} \).

Step by step solution

01

Identify Cross-Section Area

The diagonal of the square is equal to the diameter of the semicircle. Since the semicircle's equation is \( y = \pm \sqrt{1-x^2} \), the length of the diagonal of the square is \( 2\sqrt{1-x^2} \). The area of a square with diagonal \( d \) is \( \frac{d^2}{2} \). Hence the area, \( A(x) \), of the cross-section is \( \frac{(2\sqrt{1-x^2})^2}{2} = 2(1-x^2) \).
02

Set Up the Volume Integral

To find the volume of the solid, we integrate the area of the cross-sections from \( x = -1 \) to \( x = 1 \). This is given by the integral: \[ V = \int_{-1}^{1} A(x) \ dx = \int_{-1}^{1} 2(1-x^2) \ dx \].
03

Evaluate the Integral

Solve the integral \[ V = \int_{-1}^{1} 2(1 - x^2) \, dx \]. Start by integrating the function: \[ \int 2 - 2x^2 \, dx = 2x - \frac{2}{3}x^3 \]. Evaluate this from \( x = -1 \) to \( x = 1 \): \[ V = \left[2x - \frac{2}{3}x^3\right]_{-1}^{1} \].
04

Calculate Definite Integral

Substitute the limits into the integrated function: \[ V = \left(2(1) - \frac{2}{3}(1)^3\right) - \left(2(-1) - \frac{2}{3}(-1)^3\right) \]. Simplify each part: \(2 - \frac{2}{3} = \frac{6}{3} - \frac{2}{3} = \frac{4}{3}\), and \(-2 + \frac{2}{3} = -\frac{6}{3} + \frac{2}{3} = -\frac{4}{3}\). Total Volume is: \[ V = \frac{4}{3} - (-\frac{4}{3}) = \frac{4}{3} + \frac{4}{3} = \frac{8}{3} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integral Calculus
Integral calculus is all about accumulation and finding totals. It helps us calculate quantities like areas, volumes, and more by summing infinitesimally small parts. Picture it as adding up tiny slices to get a full pie.
This branch of mathematics is contrasted with differentiation, which focuses on rates of change. Simply put, if you think of differentiation as slicing a cake to see each layer, integration is about putting all those slices back together to see the whole cake.
  • When we talk about finding volumes using integrals, we accumulate the cross-sectional areas along a specific axis.
  • This is useful for finding volumes of irregular shapes that can't be calculated using simple formulas.
In our exercise, integral calculus is critical for summing the areas of squared-cross sections from one side of the solid to the other. Thereby giving us the full volume of the solid in question.
Cross-Sectional Area
Cross-sectional area refers to the shape you see when you take a cut through a solid object. Think of it as a slice of bread from a loaf. It's crucial for understanding volumes because these sections help us figure out how to "stack" them to get the whole solid.
In this exercise, our cross-sections are squares. Each square is oriented such that its diagonal runs along the semicircle from one edge to the other. To find the area of a square given its diagonal:
  • First, calculate the diagonal using the semicircle's geometry. Here, it's the full width of the semicircle: \(2\sqrt{1-x^2}\).
  • To determine the area, use the formula for the area of a square based on the diagonal: \(A = \frac{d^2}{2}\), leading us to the cross-sectional area as \(2(1-x^2)\).
This area function, \(A(x) = 2(1-x^2)\), is what we use in the integral to ultimately find the volume.
Definite Integral
The definite integral provides a way to find the accumulation of quantities across an interval. While indefinite integrals provide a general form of accumulation, definite integrals specify limits, allowing us to measure exact quantities or areas between two points.
To compute the volume of the solid, we employ the definite integral\[ V = \int_{-1}^{1} 2(1-x^2) \, dx \] that calculates the total of all cross-sectional areas between \(x = -1\) and \(x = 1\).
  • One critical aspect is substituting these limits to evaluate the integral.
  • This translates the continuous area function into a summation of volumes across that interval.
The integration delivers the solid's volume effectively, solving through evaluation and subtraction, as seen in our exercise.
Semicircle Geometry
Understanding semicircle geometry is key in this problem, as the semicircle defines the size and shape of the cross-sections. A semicircle is simply half of a circle, which can be expressed in mathematical form using the equation \( y^2 = r^2 - x^2 \), where \( r \) is the radius.
In the given problem, the semicircle serves as a boundary for the square's diagonal, so:
  • It helps outline the square's dimensions that change smoothly along the \(x\)-axis.
  • The semicircle's diameter \(2\sqrt{1-x^2}\) dictates how large (or small) each square slice becomes.
This equation reflects the constraints posed by the semicircle. By linking semicircle geometry to square dimensions, the problem connects geometric understanding with algebraic expression, crucial for wrapping up our volume determination.

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Most popular questions from this chapter

Find the lateral (side) surface area of the cone generated by revolving the line segment \(y=x / 2,0 \leq x \leq 4,\) about the \(x\) -axis. Check your answer with the geometry formula Lateral surface area \(=\frac{1}{2} \times\) base circumference \(\times\) slant height.

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