New Reply

Simple calculus volumes integration

 
Share Thread Thread Tools
Apr3-12, 08:46 AM   #1
 

Simple calculus volumes integration


1. The problem statement, all variables and given/known data
Find the volume of this equation, revolved around x axis


2. Relevant equations
y=x^2
y^2=x

3. The attempt at a solution
1) (pi)(r^2)
2) r = x^2
3) (pi)((x^2)^2)
4) (pi)(x^4)

now to integrate

5) (pi)(1/5(x^5))

since x = 1, and 1^5=1, 1/5=1/5

pi/5?

there are two equations here (y=x^2 & y^2=x), are these two somehow combined for the result

the answer is supposed to be 3pi/10

thanks.
PhysOrg.com
PhysOrg
science news on PhysOrg.com

>> Intel's Haswell to extend battery life, set for Taipei launch
>> Galaxies fed by funnels of fuel
>> The better to see you with: Scientists build record-setting metamaterial flat lens
Apr3-12, 08:52 AM   #2
 
You're not looking for the volume of the "equation", you're looking for the volume of the object that's formed when you revolve a 2-d object around the x-axis.

This may help:

http://www.wyzant.com/Help/Math/Calc...ng_Volume.aspx

Also, were those the actual equations given to you?

y = x^2 and y^2 = x?
Apr3-12, 08:57 AM   #3
 
yes the question gave me those two equations specifically. The picture of the answer shows two curves.
Apr3-12, 11:25 AM   #4
 

Simple calculus volumes integration


3) (pi)((x^2)^2) ??????????????????????
Apr6-12, 11:05 AM   #5
 
((x^2)^2) = x^4?

could someone just do this, I've wracked my brain on it hard enough already.
Apr7-12, 12:23 AM   #6
 
Recognitions:
Homework Helper Homework Help
Have you sketched the graphs? Are you finding the volume of material used in molding the walls of that 3D object?

Perhaps you should post the solution.
Apr7-12, 09:53 AM   #7
 


question 7
Apr7-12, 10:29 AM   #8
 
Recognitions:
Homework Helper Homework Help
But you've cut off that part that was going to answer my question!
Apr7-12, 12:23 PM   #9
 
lol sorry

Apr7-12, 09:32 PM   #10
 
Recognitions:
Homework Helper Homework Help
Quote by NascentOxygen View Post
Have you sketched the graphs? Are you finding the volume of material used in molding the walls of that 3D object?

Perhaps you should post the solution.
Okay, so from the solution we can see that the question does indeed require that you, for example, find the volume of clay needed to make the walls of that aforementioned jar.

Now that we all understand the question...are you right to finish it?
Apr8-12, 12:22 PM   #11
 
obviously not
Apr9-12, 11:55 PM   #12
 
Recognitions:
Homework Helper Homework Help
It might be clearer if we attack this in two steps:

① Find the volume of the generated solid enclosed within the outer curve (viz., x=y²) for 0≤ x ≥1,

② Find the volume of the generated solid enclosed within the inner curve (viz., y=x²) for that same domain.

Finally, subtract these volumes to determine the difference.

The volume of the disc shown shaded in your figure is a circular-based cylinder of thickness dx. For the moment, let's forget about the hole in the disc. At any distance x, the circular face of that disc is of radius = y. Using the area of a circle formula, and the thickness of the disc, what is the expression for the volume of just that thin disc shown shaded? (No calculus is involved in answering this.)
New Reply
Thread Tools


Similar Threads for: Simple calculus volumes integration
Thread Forum Replies
Simple calculus integration help Calculus & Beyond Homework 4
Volumes by Integration Calculus & Beyond Homework 2
Single Var Calculus -- Volumes of Revolution Calculus & Beyond Homework 3
Calculus Simple Integration by Parts question Calculus & Beyond Homework 3
Check plz, Volumes for Calculus 1 Calculus & Beyond Homework 5