How do you use the Disk method to set up the integral to find the volume of the solid generated by revolving about the y-axis the region bounded by the graphs of and the line #y = x#, and #y = x^3# between x = 0 and x = 1?

Answer 1

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Answer 2

To find the volume of the solid generated by revolving the region bounded by (y = x), (y = x^3), and the lines (x = 0) and (x = 1) around the y-axis using the Disk method, follow these steps:

  1. Graph the region: Plot (y = x) and (y = x^3) to identify the region being revolved. Here, since we are revolving around the y-axis and considering the region between (x = 0) and (x = 1), both functions and the specified x-range enclose the relevant area.

  2. Set up the integral using the Disk method:

    • The Disk method involves integrating the area of circular disks along the axis of revolution. Since the revolution is around the y-axis, we need to express the functions in terms of (y) (i.e., solve for (x) as a function of (y)). From (y = x), we get (x = y), and from (y = x^3), solving for (x), we get (x = y^{1/3}).
    • The radius of each disk is the horizontal distance between the two curves (x = y) and (x = y^{1/3}) at a given (y), which is (r(y) = y - y^{1/3}).
    • The area of a disk at height (y) is (π[r(y)]^2 = π(y - y^{1/3})^2).
    • Since we're looking at the region between (x = 0) and (x = 1), we consider the corresponding (y) values, which are also from (y = 0) to (y = 1), because both functions intersect and bound the same area in this range.
  3. Integrate: The volume (V) of the solid is obtained by integrating the area of these disks from the bottom to the top of the region in the (y)-direction: [ V = \int_{0}^{1} π(y - y^{1/3})^2 dy ] [ V = π\int_{0}^{1} (y^2 - 2y^{4/3} + y^{2/3}) dy ] [ V = π\left[\frac{y^3}{3} - \frac{3y^{7/3}}{7} + \frac{3y^{5/3}}{5}\right]_0^1 ] [ V = π\left(\frac{1}{3} - \frac{3}{7} + \frac{3}{5}\right) - 0 ] [ V = π\left(\frac{15}{45} - \frac{15}{35} + \frac{27}{45}\right) ] [ V = π\left(\frac{42}{105}\right) ] [ V = \frac{42π}{105} ]

This integral gives the volume of the solid of revolution.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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