Volume of Revolution Problem: Calculating the Total Volume of a Haystack

[tweety1234]

Homework Statement

The height of an (axially symmetric) pipe z as a function of the distance from the axis of symmetry is [tex] z = 2-2x^{2} [/tex] , where both z and x are measured in metres, and where [tex] 0 \leq z \leq 2 [/tex] and [tex] 0 \leq  x  \leq 1 [/tex] What is the total volume of hay in cubic metres
in the haystack?

The correect answer is [tex] \pi [/tex]

I am told this is a volumes of revolutions problem, how would I go about solving it?

Not sure I know the equation for volumes of revolution is [tex] V = \int^{b}_{a} \pi y^{2} dx [/tex]

Any help appreciated.

 

[cragar]

solve for x in the z=2-2x^2 , and then you will integrate over dz and z will go from 0 to 2 , and when you do the pi(r^2) the square root will go away and it will be straight forward to integrate and you should get pi.

 

[tweety1234]

solve for x in the z=2-2x^2 , and then you will integrate over dz and z will go from 0 to 2 , and when you do the pi(r^2) the square root will go away and it will be straight forward to integrate and you should get pi.

Thank you,

Just got a few questions, why do I have to rearrange for 'x'? Cause at first I thought of simply integrating z = 2 - 2x^{2}, I know z is the height and r is the radius, right?

[tex] \int^{2}_{0} -\sqrt{\frac{z}{2}} + 1 dz [/tex] Is this the expression I am meant to ingrate?

Also not sure what you meant by pir^{2} ? I know the volume of the pipe is going to be that of a cylinder sp, [tex] V = \pi r^{2} h [/tex], Am I meant to use this?

 

[dmriser]

Close.. Try solving for x again.

[tex]x = \sqrt{- \frac{z}{2} + 1}[/tex]

Then, following the formula you provided, try integrating. Remember that this is the radius function, so you'll need to square it. The outcome should be directly integrable.

 

[tweety1234]

Close.. Try solving for x again.

[tex]x = \sqrt{- \frac{z}{2} + 1}[/tex]

Then, following the formula you provided, try integrating. Remember that this is the radius function, so you'll need to square it. The outcome should be directly integrable.

hmm okay, so I have [tex] V = \pi (\sqrt{- \frac{z}{2} + 1})^{2} z [/tex] so is this what I am suppose to integrate? from 2, to 0?

[tex] \int^{2}_{0} \pi (\sqrt{- \frac{z}{2} + 1})^{2} z dz [/tex]

Thakn you

 

[dmriser]

Let's back up from the equations for a moment.

The concept of solids of revolution lies in the fact that the area of a flat disc is

[tex] \pi r^2 [/tex]

When we take a function and use it as a boundary point, we are saying that the radius of out flat disc is equal to some function.

Integrating over

[tex] \pi r^2 [/tex]

using our function as the radius gives us a volume.

So.. pi may be factored out as it's a constant. Leaving the formula.

[tex] \pi \int_{0}^{2} r^2 dr [/tex]

So, the squaring should be applied to the entire function that x is set equal to. Effectively removing the square root and leaving you with something simple to integrate from 0 to 2 that should evaluate to 1.. Which can then be multiplied by pi for your final answer.

Hope this helps

 

[tweety1234]

Let's back up from the equations for a moment.

The concept of solids of revolution lies in the fact that the area of a flat disc is

[tex] \pi r^2 [/tex]

When we take a function and use it as a boundary point, we are saying that the radius of out flat disc is equal to some function.

Integrating over

[tex] \pi r^2 [/tex]

using our function as the radius gives us a volume.

So.. pi may be factored out as it's a constant. Leaving the formula.

[tex] \pi \int_{0}^{2} r^2 dr [/tex]

So, the squaring should be applied to the entire function that x is set equal to. Effectively removing the square root and leaving you with something simple to integrate from 0 to 2 that should evaluate to 1.. Which can then be multiplied by pi for your final answer.

Hope this helps

I think have got it, but don't get the right answer, what am I doing wrong?

[tex] \pi \int^{2}_{0} \frac{z}{2} + 1 \\\\\\\\\\\\\\\\dz [/tex]

that gives me = [tex] \frac{z^{2}}{4} + z [/tex]

and evaluating that from 2-0 I get 3?

 

[cragar]

should be -z^2/4 you are missing A NEGATIVE sign in front of the first term.

 

[tweety1234]

should be -z^2/4 you are missing A NEGATIVE sign in front of the first term.

oh right, I thought since your squaring a negative expression, it will become positive?

Thank you very much!