Volumes of solids of revolution with infinity

• cathy
In summary: You are correct that ##dV=\pi r^2##. Since the radius of the disk is ##r##, this would give you the volume of the disk.
cathy

Homework Statement

Consider the infinite region in the first quadrant between the curve y=e^-5x and the x-axis.
Find area= 1/5 (got this part)
Compute the volume of the solid generated by revolving the region about the x-axis:
Compute the volume of the solid generated by revolving the region about the y-axis:

2. The attempt at a solution
I do not understand how to do the last two parts. I tried to do
dV=base*height
dV=pi*x^2*dy

because area of circle is

base= pi*x^2
leaving me with
dv= pi*x^2*(-1/5e^(-5x)
and taking limit from 0 to inifity by taking limit from 0 to b, but i am not understanding how to get the answer. Please advise. I am so lost. Thank you in advance.

For the volume of the solid generated by revolving the region about the x-axis, you now integrate perpendicular to the axis of revolution (dy). It's not all that easy to express ##x^2## in y, but you do need it then.

What would you write down if you integrated parallel to the axis of revolution (i.e. integrate over dx) ?

cathy said:

Homework Statement

Consider the infinite region in the first quadrant between the curve y=e^-5x and the x-axis.
Find area= 1/5 (got this part)
Compute the volume of the solid generated by revolving the region about the x-axis:
Compute the volume of the solid generated by revolving the region about the y-axis:

2. The attempt at a solution
I do not understand how to do the last two parts. I tried to do
dV=base*height
dV=pi*x^2*dy

because area of circle is

base= pi*x^2
leaving me with
dv= pi*x^2*(-1/5e^(-5x)
and taking limit from 0 to inifity by taking limit from 0 to b, but i am not understanding how to get the answer. Please advise. I am so lost. Thank you in advance.

You have the choice of disks or shells and also the choice of dx or dy elements. That should be your first decision. Since ##y## is given in terms of ##x##, you will find using ##dx## elements to be most convenient. So, first question:

Revolving about the ##x## axis using ##dx## element, would you be using disks or shells? What is the corresponding dV?

Same questions for about the ##y## axis using a ##dx## element.

Show us those dV elements. The rest will be easy.

Over to Lynn (bedtime for me...)

for rotation about the x-axis, would it be
v=∫ (0 to infinity) of (2∏*x*e^-5x)dx?
since I am using washers, since the vertical stick is horizontal to axis of rotation?

im getting the answer as 2/25 pi which, for some reason is the answer to the rotation about the y axis, and not the x? Can you tell me why?

i still cannot get the x-axis one, and i don't understand why the y-axis one is correct.

Last edited:
cathy said:
for rotation about the x-axis, would it be
v=∫ (0 to infinity) of (2∏*x*e^-5x)dx?
since I am using washers, since the vertical stick is horizontal to axis of rotation?

im getting the answer as 2/25 pi which, for some reason is the answer to the rotation about the y axis, and not the x? Can you tell me why?

The ##x## factor in your integral is the distance of the vertical element from the ##y## axis. So ##2\pi x## is the distance around the ##y## axis and the ##e^{-5x}## is the height of the element. ##dx## is the thickness. So you have height*circumference*thickness of the shell revolving about the ##y## axis. You are getting the ##y## axis number because that is the integral you set up.

Now, going around the ##x## axis, using disks, what do you get?

1 person
is is simply v=integral(pi*e^-5x)dx to set up?

or would i have to square the inside of the integral making the inside e^-10x?

got answer as pi/10, but honestly I guessed. Did I set my integral up correctly? I still don't understand if I had to square it, or if I squared correctly. (If the integral I wrote on this post is correct, then I understand)

Last edited:
cathy said:
or would i have to square the inside of the integral making the inside e^-10x?

got answer as pi/10, but honestly I guessed. Did I set my integral up correctly? I still don't understand if I had to square it, or if I squared correctly. (If the integral I wrote on this post is correct, then I understand)

Remembering that the area of a disk of radius ##r## is ##\pi r^2##, does that give you a hint about your guess? So ##dV## would be ##(\pi r^2 \cdot \text{thickness})##.

i think I am getting confused with powers for some reason. could you clear something up? when you have something like e^x^2. Is that the same as e^2x? Well, I guess it's not. But is there some way you can rewrite the e^x^2 as an easier power?

Like ##(4^3)^2 = 4^3 \times 4^3 = 4^{3+3}## you write ## (e^y)^2 = e^y \times e^y = e^{y+y} = e^{2y}##

1. What is a solid of revolution with infinity?

A solid of revolution with infinity is a three-dimensional shape that is formed by rotating a two-dimensional shape around an axis that extends to infinity. This results in a shape that has no definite endpoints and continues infinitely in both directions.

2. How do you calculate the volume of a solid of revolution with infinity?

The volume of a solid of revolution with infinity can be calculated using the method of cylindrical shells. This involves dividing the shape into infinitely thin cylindrical shells and then integrating the surface area of each shell to find the total volume.

3. What is the significance of finding the volume of a solid of revolution with infinity?

Finding the volume of a solid of revolution with infinity is important in various fields of mathematics and science, such as calculus, physics, and engineering. It allows us to calculate the volume of complex shapes that cannot be easily measured in real life.

4. Can a solid of revolution with infinity have a finite volume?

No, a solid of revolution with infinity will always have an infinite volume. This is because the shape extends infinitely in both directions, so there is no limit to the amount of space it occupies.

5. What are some real-life examples of solids of revolution with infinity?

Some real-life examples of solids of revolution with infinity include a tornado, a spiral galaxy, and a water vortex. These shapes are formed by rotating a two-dimensional shape (air, stars, and water, respectively) around an axis that extends to infinity.

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