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Chapter 6: Problem 11

Let \(R\) be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when \(R\) is revolved about the \(y\) -axis. $$y=x^{3}-x^{8}+1, y=1$$

Short Answer

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Based on the given step by step solution, write a short answer question. Question: Find the volume of the solid generated when the region \(R\), bounded by the curves \(y = x^3 - x^8 + 1\) and \(y = 1\), is revolved about the y-axis using the shell method. Answer: The volume of the solid is \(\pi\) cubic units.

Step by step solution

01

Find the Intersection Points of the Curves

To find the points where the curves meet, we need to set the equations equal to each other and solve for x. $$x^3 - x^8 + 1 = 1$$ We can subtract 1 from both sides: $$x^3 - x^8 = 0$$ Factor x^3 out, we get $$x^3(1-x^5)=0$$ This gives us the intersection points at \(x=0\) and \(x=1\) (do not worry about other possible solutions because these are boundary points which we are interested).
02

Set up the Shell Method Formula

The volume of the solid generated by revolving the region R around the y-axis can be found using the shell method. The formula for this is: $$V = 2 \pi \int_a^b x * h(x) dx$$ Here, \(V\) represents the volume, \(h(x)\) represents the height of the shell at location x, and a and b represent the interval [a, b] along the x-axis that contains the region R.
03

Find the Height Function h(x)

In this case, the height h(x) of the shell at any x-value within the region R is the distance between the two curves. Since the second curve is just a horizontal line at \(y=1\), the height of the shell is simply the vertical distance from the first curve down to the horizontal line: $$h(x) = (x^3 - x^8 + 1) - 1 = x^3 - x^8$$
04

Set up the Integral

Now we can set up the integral to find the volume of the solid. From Step 1, we know the intersection points on x-axis are at \(x=0\) and \(x=1\). Then the integral is: $$V = 2 \pi \int_0^1 x(x^3 - x^8) dx$$
05

Simplify and Evaluate the Integral

First, let's distribute the x inside the parentheses: $$V = 2 \pi \int_0^1 (x^4 - x^9) dx$$ Now we need to evaluate this integral. Using the power rule for integration, we get: $$V = 2 \pi\left[\frac{1}{5}x^5 - \frac{1}{10}x^{10}\right]_0^1$$ Now, we'll plug in our limits of integration and subtract: $$V = 2 \pi\left[\left(\frac{1}{5} - \frac{1}{10}\right) - (0)\right] = 2 \pi \left(\frac{1}{2}\right)$$ Finally, we find the volume of the solid: $$V = \pi$$ The volume of the solid generated when the region R is revolved about the y-axis is \(\pi\) cubic units.

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