How do you use the Disk method to set up the integral to find the volume of the solid generated by revolving about the yaxis the region bounded by the graphs of and the line #y = x#, and #y = x^3# between x = 0 and x = 1?
See the answer below:
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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 yaxis using the Disk method, follow these steps:

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

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 yaxis, 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.

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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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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