Ch. 6 - Applications of Definite Integrals

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 6 - Applications of Definite Integrals

Problem 6.PE.5

Chapter 6, Problem 6.PE.5

Volumes
Find the volumes of the solids in Exercises 1–18.
The solid lies between planes perpendicular to the x-axis at x = 0 and x = 4. The cross-sections of the solid perpendicular to the x-axis between these planes are circular disks whose diameters run from the curve x² = 4y to the curve y² = 4x.

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First, identify the region bounded by the curves given: \(x^{2} = 4y\) and \(y^{2} = 4x\). These curves define the limits of the diameter of the circular cross-sections perpendicular to the x-axis between \(x=0\) and \(x=4\).

Rewrite each curve to express \(y\) in terms of \(x\) to find the upper and lower boundaries of the diameter at each \(x\). From \(x^{2} = 4y\), solve for \(y\) to get \(y = \frac{x^{2}}{4}\). From \(y^{2} = 4x\), solve for \(y\) to get \(y = \sqrt{4x} = 2\sqrt{x}\) (considering the positive root since diameters are positive lengths).

At each \(x\) between 0 and 4, the diameter of the circular cross-section is the vertical distance between the two curves: \(D(x) = 2\sqrt{x} - \frac{x^{2}}{4}\).

The radius of the circular cross-section is half the diameter: \(r(x) = \frac{D(x)}{2} = \frac{1}{2} \left( 2\sqrt{x} - \frac{x^{2}}{4} \right)\).

The volume of the solid is found by integrating the area of the circular cross-sections along the \(x\)-axis from 0 to 4. The area of each cross-section is \(A(x) = \pi [r(x)]^{2}\). Therefore, the volume is \(V = \int_{0}^{4} \pi \left( \frac{2\sqrt{x} - \frac{x^{2}}{4}}{2} \right)^{2} dx\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Volume of Solids with Known Cross-Sections

This concept involves finding the volume of a solid by integrating the area of its cross-sections perpendicular to an axis. When cross-sections have a known shape, such as circles, the volume is the integral of the area function along the axis of interest.

Finding the Diameter from Curves

To determine the diameter of each circular cross-section, you must find the vertical distance between the two curves at a given x. This requires solving each curve for y in terms of x and subtracting to get the length of the segment forming the diameter.

Setting up and Evaluating Definite Integrals

Once the area of each cross-section is expressed as a function of x, the volume is found by integrating this area from the lower to upper bounds (x=0 to x=4). Proper setup of the integral and accurate evaluation are essential to find the exact volume.

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Find the volumes of the solids in Exercises 1–18.

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