# How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region #y= sqrt(5x)#, #x=5# rotated about the y axis?

Please see the explanation section, below.

The region bounded by the functions is shown below in blue.

We are using the shell method, so we take a representative slice parallel to the axis of revolution. In this case, that is the

The dashed line shows the radius of revolution for the representative.

We will be integrating with respect to

From the given information, we conclude that

The volume of a cylindrical shell is the surface area of the cylinder times the thickness:

In the picture, the radius is shown by the dashed line, it is

The height of the slice is the upper

The volume of the representative shell is:

The volume of the resulting solid is:

# = 2pisqrt5 int_0^5 x^(3/2)dx#

# = 2pisqrt5[(2x^(5/2))/5]_0^5= 100pi#

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To use the shell method to set up and evaluate the integral for the volume of the solid generated by revolving the region about the y-axis, follow these steps:

- Determine the limits of integration by finding the intersection points of the curves. In this case, the curves are y = √(5x) and x = 5.
- Set up the integral using the shell method formula: (V = 2\pi \int_a^b x \cdot f(x) , dx), where (a) and (b) are the limits of integration and (f(x)) is the height of the shell.
- Substitute (x = \frac{y^2}{5}) into the formula to express (x) in terms of (y).
- Determine the limits of integration by finding where the curves intersect and integrate with respect to (y).
- Evaluate the integral.

The integral setup for the shell method in this case is:

(V = 2\pi \int_0^{\sqrt{25}} y \cdot \left(\frac{y^2}{5}\right) , dy)

After integrating and evaluating this expression, you will find the volume of the solid generated by revolving the given region about the y-axis.

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