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Simple calculus volumes integration |
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| Apr3-12, 08:46 AM | #1 |
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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. |
| Apr3-12, 08:52 AM | #2 |
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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 |
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yes the question gave me those two equations specifically. The picture of the answer shows two curves.
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| Apr3-12, 11:25 AM | #4 |
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Simple calculus volumes integration
3) (pi)((x^2)^2) ??????????????????????
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| Apr6-12, 11:05 AM | #5 |
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((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 |
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Recognitions:
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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 |
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![]() question 7 |
| Apr7-12, 10:29 AM | #8 |
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But you've cut off that part that was going to answer my question!
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| Apr7-12, 12:23 PM | #9 |
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lol sorry
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| Apr7-12, 09:32 PM | #10 |
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Recognitions:
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![]() Now that we all understand the question...are you right to finish it? |
| Apr8-12, 12:22 PM | #11 |
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obviously not
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| Apr9-12, 11:55 PM | #12 |
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Recognitions:
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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.) |
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