# Volumes of solids of revolution with infinity

1. Mar 23, 2014

### cathy

1. The problem statement, all variables and given/known data
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.

2. Mar 23, 2014

### BvU

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) ?

3. Mar 23, 2014

### LCKurtz

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.

4. Mar 23, 2014

### BvU

Over to Lynn (bedtime for me...)

5. Mar 23, 2014

### cathy

for rotation about the x-axis, would it be
v=∫ (0 to infinity) of (2∏*x*e^-5x)dx?
since im 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 dont understand why the y axis one is correct.

Last edited: Mar 23, 2014
6. Mar 23, 2014

### LCKurtz

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?

7. Mar 23, 2014

### cathy

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

8. Mar 23, 2014

### cathy

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 dont 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: Mar 23, 2014
9. Mar 23, 2014

### LCKurtz

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 $(\pi r^2 \cdot \text{thickness})$.

10. Mar 23, 2014

### cathy

i think im 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?

11. Mar 24, 2014

### BvU

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