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Simple calculus volumes integration 
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#1
Apr312, 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. 


#2
Apr312, 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 2d object around the xaxis.
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? 


#3
Apr312, 08:57 AM

P: 106

yes the question gave me those two equations specifically. The picture of the answer shows two curves.



#4
Apr312, 11:25 AM

P: 1,195

Simple calculus volumes integration
3) (pi)((x^2)^2) ??????????????????????



#5
Apr612, 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. 


#6
Apr712, 12:23 AM

HW Helper
Thanks
P: 5,141

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. 


#7
Apr712, 09:53 AM

P: 106

question 7 


#8
Apr712, 10:29 AM

HW Helper
Thanks
P: 5,141

But you've cut off that part that was going to answer my question!



#9
Apr712, 12:23 PM

P: 106

lol sorry



#10
Apr712, 09:32 PM

HW Helper
Thanks
P: 5,141

Now that we all understand the question...are you right to finish it? 


#11
Apr812, 12:22 PM

P: 106

obviously not



#12
Apr912, 11:55 PM

HW Helper
Thanks
P: 5,141

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 circularbased 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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