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Simple calculus volumes integration

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togo
#1
Apr3-12, 08:46 AM
P: 106
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.
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NewtonianAlch
#2
Apr3-12, 08:52 AM
P: 440
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?
togo
#3
Apr3-12, 08:57 AM
P: 106
yes the question gave me those two equations specifically. The picture of the answer shows two curves.

LawrenceC
#4
Apr3-12, 11:25 AM
P: 1,195
Simple calculus volumes integration

3) (pi)((x^2)^2) ??????????????????????
togo
#5
Apr6-12, 11:05 AM
P: 106
((x^2)^2) = x^4?

could someone just do this, I've wracked my brain on it hard enough already.
NascentOxygen
#6
Apr7-12, 12:23 AM
HW Helper
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P: 5,243
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.
togo
#7
Apr7-12, 09:53 AM
P: 106


question 7
NascentOxygen
#8
Apr7-12, 10:29 AM
HW Helper
Thanks
P: 5,243
But you've cut off that part that was going to answer my question!
togo
#9
Apr7-12, 12:23 PM
P: 106
lol sorry

NascentOxygen
#10
Apr7-12, 09:32 PM
HW Helper
Thanks
P: 5,243
Quote 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?
togo
#11
Apr8-12, 12:22 PM
P: 106
obviously not
NascentOxygen
#12
Apr9-12, 11:55 PM
HW Helper
Thanks
P: 5,243
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=y) for 0≤ x ≥1,

② Find the volume of the generated solid enclosed within the inner curve (viz., y=x) 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.)


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