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Chapter 6: Problem 37

Let \(R\) be the region bounded by the following curves. Use the disk or washer method to find the volume of the solid generated when \(R\) is revolved about the \(y\) -axis. $$y=x^{3}, y=0, x=2$$

Short Answer

Expert verified
The volume of the solid generated when the region R is revolved around the y-axis is V = 96蟺/5.

Step by step solution

01

Sketch the region

First, sketch the region bounded by the curves. The curve \(y=x^3\) is a monotonically increasing function that passes through the origin. The curve \(y=0\) is the x-axis, and \(x=2\) is a vertical line intersecting the x-axis at x = 2. The enclosed region is a finite area between these three curves.
02

Determine the method

Since the region is revolved around the \(y\)-axis, we can use the washer method. The washer method involves finding the outer radius and inner radius of each washer and integrate the difference in their areas over the interval of integration.
03

Find the radii

In this case, the inner radius is formed by revolving the curve \(y=0\) around the \(y\)-axis, which results in a constant inner radius of 0. The outer radius is formed by revolving the curve \(y=x^3\) around the \(y\)-axis. To find the outer radius as a function of y, we must first solve the equation \(y=x^3\) for x: $$x = y^{\frac{1}{3}}$$ Now we have the function for the outer radius, \(R(y) = y^{\frac{1}{3}}\).
04

Find the limits of integration

We need to determine the limits of integration. Since we are integrating along the \(y\)-axis, our limits will be the minimum and maximum values of \(y\) in the region. We find these values by evaluating the endpoints of the region: $$y = x^3; x = 0 \Rightarrow y = 0$$ $$y = x^3; x = 2 \Rightarrow y = 2^3 = 8$$ So, the limits of integration are \(y = 0\) and \(y = 8\).
05

Set up the integral

Now, set up the integral using the washer method formula: Volume = \(\pi \int_{a}^{b} [R(y)]^2 \, dy = \pi \int_{0}^{8} (y^{\frac{1}{3}})^2 \, dy\)
06

Evaluate the integral

Next, evaluate the integral: $$\pi \int_{0}^{8} y^{\frac{2}{3}} \, dy = \pi \left[ \frac{3}{5}y^{\frac{5}{3}}\right]_{0}^{8}$$ Now, just substitute the limits of integration and calculate the volume: $$\pi \left( \frac{3}{5}(8)^{\frac{5}{3}} - \frac{3}{5}(0)^{\frac{5}{3}} \right) = \pi \left(\frac{3}{5}\cdot 32 \right) = \pi\left( \frac{96}{5}\right)$$
07

Write down the final answer

The volume of the solid generated when the region \(R\) is revolved around the \(y\)-axis is: $$V = \frac{96\pi}{5}$$

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