Calculus Volumes

[Signorino]

y=(1/4)x^2; y=5-x^2

Find the volume of the solid obtained by rotating the region bounded by the these curves about x=7.


Okay, I thought I needed to change the given equations to x= sqrt(4y) and x= sqrt(5-y), but I'm sure that's not right, because then there is no enclosed region. Now I have no idea how to do this one. Can anyone help me with this?
Thanks.

[EdEarl]

It might help you to graph these functions, then think about rotating the curve (around what) to form a solid.

[Signorino]

Yes, I graphed them and I know what the enclosed area looks like. I did the same problem revolving it around the x-axis, and around y=2. My problem is that I don't understand how to revolve it around x=7 because neither curve is closer or further away than the other, which is how you use the formula I was taught in class.

[qamaths01]

The final answer is 20pi. This can be shown in graph.

[EdEarl]

Yes, I graphed them and I know what the enclosed area looks like. I did the same problem revolving it around the x-axis, and around y=2. My problem is that I don't understand how to revolve it around x=7 because neither curve is closer or further away than the other, which is how you use the formula I was taught in class.

What formula were you given in class?

What are the limits for rotating?

I recommend you look at these lessons, which are very good and not very long.

https://www.khanacademy.org/math/calculus/solid_revolution_topic

Edited June 13, 2013 by EdEarl

[daniton]

Yes, I graphed them and I know what the enclosed area looks like. I did the same problem revolving it around the x-axis, and around y=2. My problem is that I don't understand how to revolve it around x=7 because neither curve is closer or further away than the other, which is how you use the formula I was taught in class.

let me help you with this

just rewrite the equations asif x=7 is x=0. i think you have learned about translation.

[Daedalus]

y=(1/4)x^2; y=5-x^2[/size]

 

Find the volume of the solid obtained by rotating the region bounded by the these curves about x=7.[/size]

 

 

Okay, I thought I needed to change the given equations to x= sqrt(4y) and x= sqrt(5-y), but I'm sure that's not right, because then there is no enclosed region. Now I have no idea how to do this one. Can anyone help me with this?

Thanks.[/size]

 

Actually there is an enclosed region. The problem states that you should rotate the region about / around [math]x=7[/math], which is a vertical line located at [math]x=7[/math]. It is this line that encloses the region. However, as daniton has suggested, it is easier to translate the function left along the [math]x[/math]-axis. The following graph should help you visualize the problem:

 

 

The next thing you have to do is determine the axis of integration. Since the region is rotated around a vertical line at [math]x=7[/math], or at [math]x=0[/math] if you translated the function to the left 7 units, the radius of the disks forming the solid will be values of [math]x[/math]. So we'll have to use the disk method and integrate along the [math]y[/math] axis. In other words, imagine adding up all of the volumes for the disks that have a radius of [math]x[/math] as you stack them along the [math]y[/math] axis.

 

Okay, I thought I needed to change the given equations to x= sqrt(4y) and x= sqrt(5-y), but I'm sure that's not right, because then there is no enclosed region.

 

You were correct in your assumption about using the inverse of the functions. However, these functions still describe the same region, but do so with respect to the [math]y[/math] axis. So you were not correct in your reasoning that there would be no enclosed region.

 

To solve this problem, you must figure out where these functions intersect the [math]y[/math] axis as well as each other. This will give you the limits of integration along the [math]y[/math] axis. Furthermore, you will end up with two integrals. One for the top and one for the bottom. The final solid will look like this:

 

 

Because this sounds like homework, I will not do the rest of the work for you. However, I will continue to guide you through solving the problem. The following is an example of how the solution should look like:

 

[math]V=\int_{y_0}^{y_1} f(y)\, dy+\int_{y_1}^{y_2} g(y)\, dy[/math]

 

Notice how the second integral's lower limit is the same as the first integral's upper limit. This is where the functions intersect each other.

Edited June 18, 2013 by Daedalus

[Daedalus]

Although your inactivity suggests that you already found the solution, I have another hint for you

 

The following is an example of how the solution should look like:

[math]V=\int_{y_0}^{y_1} f(y)\, dy+\int_{y_1}^{y_2} g(y)\, dy[/math]

Notice how the second integral's lower limit is the same as the first integral's upper limit. This is where the functions intersect each other.

The functions [math]f(y)[/math] and [math]g(y)[/math] do not reflect the actual functions specified in the problem. However, If we let:

[math]f(y)=2\sqrt{y}-7[/math]

[math]g(y)=\sqrt{5-y}-7[/math]

Then our integrals would be:

[math]V=\int_{y_0}^{y_1} \pi\,f(y)^2\, dy+\int_{y_1}^{y_2} \pi\,g(y)^2\, dy[/math]

Edited June 20, 2013 by Daedalus

[alexwang32]

Since this thread has been here a long time, I'd like to give it a try. My answer is not [latex]20\pi[/latex]... here's how I did it:

 

First, translate the functions so that x=7 becomes the 'y-axis'( the parabola is placed at x = 7).

 

[latex]y=\frac{1}{4}(x-7)^2[/latex]

[latex]y=5-(x-7)^2[/latex]

 

The upper and lower bond are 9 and 5.

 

As illustrated here:

 

Then by using the volume of revolution about the y-axis formula:

We get [latex]2\pi\int_{5}^{9} x(5-\frac{5}{4}(x-7)^2)dx=\frac{560\pi}{3}[/latex].

 

Am I right?

 

[latex]20\pi[/latex] is not correct because you can simply estimate the volume by calculating [latex]2\pi\int_{5}^{9}5x~dx[/latex], which equals [latex]280\pi[/latex]

 

 

Edited July 26, 2013 by alexwang32

[Signorino]

Hello,
Sorry for the extended inactivity. I was unable to get 20pi as an answer, and eventually gave up. This was a bonus problem, so I wasn't that worried about it, as I figured it would be a better idea to study for my upcoming final than spend too much time on this problem. Then the class was over and I started worrying about other classes, and eventually completely forgot about this thread. Anyway, I'm still unable to get 20pi as an answer, but I did understand alexwang32's explanation, and I thought that seemed right. Anyway, thanks for the help on this problem everyone.

 

On a side note, I thought it was worth mentioning that I think Daedalus, judging from his picture, looks a lot like Ryan Haywood from Roosterteeth.
http://images3.wikia.nocookie.net/__cb20121229060260/roosterteeth/images/1/1b/Ryan50ba8afb9fedf.jpg

[Daedalus]

On a side note, I thought it was worth mentioning that I think Daedalus, judging from his picture, looks a lot like Ryan Haywood from Roosterteeth.

http://images3.wikia.nocookie.net/__cb20121229060260/roosterteeth/images/1/1b/Ryan50ba8afb9fedf.jpg

How funny! All my life, people tell me I look like some famous actor or musician. I used to have really long hair, and when I wore my Lennon sunglasses, people said I looked like Ozzy. Without them, I used to get Dave Mustaine from Megadeath but with dark hair. Now that I've had short hair for over 15 years, people say I look like Charlie Sheen. Personally, I think they are all crazy... Perhaps, I should've tried to be an actor or a stunt double.

[alexwang32]

Thanks for agreeing with me. As far as I can see, my solution seems to be correct. As a matter of fact I did study pretty hard on 'Volume of revolution' in my Calculus course

 

Another way of solving this... though is may seem rather too complicated, is to roatate the parabolas 90 degrees, so that one can calculate it's volume of revolution about the x-axis. It's a good practice for students who want to master translation and volume of revolution,

[Signorino]

How funny! All my life, people tell me I look like some famous actor or musician. I used to have really long hair, and when I wore my Lennon sunglasses, people said I looked like Ozzy. Without them, I used to get Dave Mustaine from Megadeath but with dark hair. Now that I've had short hair for over 15 years, people say I look like Charlie Sheen. Personally, I think they are all crazy... Perhaps, I should've tried to be an actor or a stunt double.

Charlie Sheen? I don't see that one at all. Nope, the only one I can see is Ryan Haywood. Though I doubt being his stunt double would be very interesting. I'm not sure how many stunts he needs to do playing video games all day. I guess you could press space for him when he goes to jump off something high in Minecraft.