Shell Method: Find Vol. of Solid Rotated Around x-Axis

That is what you integrate to get the volume of that portion of the solid.On the lower portion of the graph, the width of a horizontal strip is 1/y (just the radius) and the thickness is dy. That is what you integrate to get the volume of that portion of the solid.In summary, when using the shell method to find the volume of a solid generated by revolving a plane region about the x-axis, you will need to evaluate two different integrals with different limits of integration. In both integrals, the typical volume element will be 2π⋅radius⋅width⋅dy, where the radius is the y-coordinate on the horizontal strip being revolved. However, the width of the
  • #1
Jovy
17
2

Homework Statement



Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the x-axis.
$$y=\frac 1 x$$
se07c01017.png

Homework Equations



$$Volume=2\pi\int_a^b p(y)h(y)dy$$

The Attempt at a Solution


[/B]
I see that there are two shells, therefore, I would do the integral twice. Meaning, ##Volume=2\pi\int_a^b p(y)h(y)dy+2\pi\int_a^b p(y)h(y)dy##

I'm having trouble identifying what h(x) are for both integrals. I know that you can change ##y=\frac 1 x## to be ##x=\frac 1 y## and I know that for both p(y)=y. I think the dashed lines indicates that ##y=\frac 1 2## is the axis in which it is being rotated.
How do you determine h(x)? Once I understand how to determine h(x) for both integrals, I know how to solve the rest of the problem.

this website has an image of the graph, incase the one I uploaded doesn't show:
http://www.calcchat.com/book/Calculus-ETF-6e/7/3/17/
 
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  • #2
Jovy said:

Homework Statement



Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the x-axis.
$$y=\frac 1 x$$

Homework Equations



$$Volume=2\pi\int_a^b p(y)h(y)dy$$

The Attempt at a Solution


[/B]
I see that there are two shells, therefore, I would do the integral twice. Meaning, ##Volume=2\pi\int_a^b p(y)h(y)dy+2\pi\int_a^b p(y)h(y)dy##
You won't do the integral twice -- you need to evaluate two different integrals with different limits of integration. Also, it would help you to have more meaningful expressions than p(y) and h(y). For each integral the typical volume element will be ##2\pi \cdot \text{radius} \cdot \text{width} \cdot dy##
In both integrals the radius will be y (i.e., the y-coordinate on the horizontal strip being revolved), but the width of the element that is being revolved is different in the two integrals.

In the lower region, the width is very simple. In the upper region, the width varies, but it will always be the x-value on the curve (expressed in terms of y) minus the x-value on the vertical line. Since you know the equation of the curve, you can write the x-value on the curve in terms of the y-coordinate there. You're on the right track below.

Jovy said:
I'm having trouble identifying what h(x) are for both integrals. I know that you can change ##y=\frac 1 x## to be ##x=\frac 1 y## and I know that for both p(y)=y. I think the dashed lines indicates that ##y=\frac 1 2## is the axis in which it is being rotated.
How do you determine h(x)? Once I understand how to determine h(x) for both integrals, I know how to solve the rest of the problem.

this website has an image of the graph, incase the one I uploaded doesn't show:
http://www.calcchat.com/book/Calculus-ETF-6e/7/3/17/
 
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  • #3
Mark44 said:
You won't do the integral twice -- you need to evaluate two different integrals with different limits of integration. Also, it would help you to have more meaningful expressions than p(y) and h(y). For each integral the typical volume element will be ##2\pi \cdot \text{radius} \cdot \text{width} \cdot dy##
In both integrals the radius will be y (i.e., the y-coordinate on the horizontal strip being revolved), but the width of the element that is being revolved is different in the two integrals.

In the lower region, the width is very simple. In the upper region, the width varies, but it will always be the x-value on the curve (expressed in terms of y) minus the x-value on the vertical line. Since you know the equation of the curve, you can write the x-value on the curve in terms of the y-coordinate there. You're on the right track below.

the x-value on the curve is ##\frac 1 y## but what is the x-value on the vertical line? Do you mean what x is in ##y=\frac 1 x##?
 
  • #4
Jovy said:
the x-value on the curve is ##\frac 1 y## but what is the x-value on the vertical line? Do you mean what x is in ##y=\frac 1 x##?
The vertical line is x = 1. On the curve, x = 1/y. So on that upper portion of the graph in your image, a horizontal strip has a width of (1/y - 1) and a thickness of dy.
 
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Related to Shell Method: Find Vol. of Solid Rotated Around x-Axis

1. What is the Shell Method and how is it used to find the volume of a solid rotated around the x-axis?

The Shell Method is a mathematical technique used to find the volume of a solid formed by rotating a 2-dimensional shape around an axis. To use the Shell Method, we divide the solid into thin, cylindrical shells and integrate their volumes with respect to the axis of rotation (in this case, the x-axis).

2. How do you set up the integral for the Shell Method?

To set up the integral for the Shell Method, we use the formula V = 2π∫(radius)(height)dx, where the radius is the distance from the axis of rotation to the shell, and the height is the height of the shell. We integrate this formula with respect to x, the axis of rotation.

3. What are the limits of integration for the Shell Method?

The limits of integration for the Shell Method are the bounds of the region being rotated around the x-axis. These can be found by setting up and solving equations for the intersection points of the curves that define the region.

4. Can the Shell Method be used to find the volume of a solid rotated around a different axis?

Yes, the Shell Method can be used to find the volume of a solid rotated around any axis. The only change that needs to be made is to integrate with respect to the appropriate axis, and adjust the formula to use the appropriate radius and height values.

5. Are there any limitations to using the Shell Method to find volume?

The Shell Method can only be used to find the volume of solids with rotational symmetry around an axis. It also requires that the solid can be divided into thin, cylindrical shells, which may not be possible for all shapes. Additionally, the region being rotated must have a bounded area, otherwise the integral will not converge.

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