Recent content by c.dube
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Finding the Mass of a Solid in 3d
Nevermind, I got an answer. Can anyone confirm 6\pi?- c.dube
- Post #2
- Forum: Calculus and Beyond Homework Help
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Finding the Mass of a Solid in 3d
Homework Statement Find the mass of the solid bounded by the cylinder x^2+y^2=2x and the cone z^2=x^2+y^2 if the density is \delta = \sqrt{x^2+y^2} [b2. The attempt at a solution[/b] I had some trouble looking at how to set up the limits on this integral. What I came up with was: 2...- c.dube
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- 3d Mass Solid
- Replies: 1
- Forum: Calculus and Beyond Homework Help
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Convergence of Improper Integrals: Exploring the Power of p
Ah sweet! Thanks so much for your help it was much appreciated.- c.dube
- Post #22
- Forum: Calculus and Beyond Homework Help
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Convergence of Improper Integrals: Exploring the Power of p
That it diverges, yes? And so the final answer would be 0\leq p<1?- c.dube
- Post #20
- Forum: Calculus and Beyond Homework Help
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Convergence of Improper Integrals: Exploring the Power of p
So, on 0\leq p<1, is \frac{e^{-x}}{x^p} integrable and therefore converges? As for p=1, we have the indefinite integral of lnx. But that's undefined at x=0, so am I missing your hint?- c.dube
- Post #18
- Forum: Calculus and Beyond Homework Help
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Convergence of Improper Integrals: Exploring the Power of p
Ohhhhh ok so we have: 1-p<0 Now, that will go to infinity. The other case: 1-p>0 Will always be zero. So, the function doesn't blow up when p<1. Would that indicate that the answer is 0\leq p<1? And what about the case where p=1? Thanks again for your patience it is very much appreciated.- c.dube
- Post #15
- Forum: Calculus and Beyond Homework Help
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Convergence of Improper Integrals: Exploring the Power of p
So the indefinite integral is: \frac{x^{-p+1}}{-p+1} + c When x=0, doesn't the integral become 0 for every p except p=1, where you get the indeterminate form 0^0? Gah I really don't know why I'm not getting this.- c.dube
- Post #13
- Forum: Calculus and Beyond Homework Help
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Convergence of Improper Integrals: Exploring the Power of p
Oh ok so if p=0 then the x^p just becomes one, leaving only e^{-x}, which converges, so the answer would be p=0?- c.dube
- Post #12
- Forum: Calculus and Beyond Homework Help
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Convergence of Improper Integrals: Exploring the Power of p
From 0 to 1, then, don't we have the exact opposite logic, e.g. \frac{1}{x^p} will always be greater than or equal to one? And therefore it diverges for all p, leaving me with an answer p = null set? Haha thanks for bearing with me I don't know why my brain isn't getting this one.- c.dube
- Post #9
- Forum: Calculus and Beyond Homework Help
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Convergence of Improper Integrals: Exploring the Power of p
OK that makes sense to me, but where does it leave me? How do I bridge the gap to finding what, if any, p works? Thanks- c.dube
- Post #7
- Forum: Calculus and Beyond Homework Help
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Convergence of Improper Integrals: Exploring the Power of p
OK, so \int_1^\infty e^{-x} converges because the limit as x goes to infinity is finite. Now, because \frac{1}{x^{p}} is decreasing and the largest case would be one, it is always less than or equal to one (is there a more rigorous way of saying this?). Anyways, moving on to case between 0 and...- c.dube
- Post #4
- Forum: Calculus and Beyond Homework Help
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Convergence of Improper Integrals: Exploring the Power of p
I'm trying to refresh my knowledge of Calc II, and I'm going through improper integrals right now. The problem I am trying to solve is: For which numbers p\geq0 does \int_0^\infty \frac{e^{-x}}{x^{p}} converge? Justify your answer. So far, I've split up the integral into two halves (0 to 1...- c.dube
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- Convergence Integral
- Replies: 22
- Forum: Calculus and Beyond Homework Help
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Solving integral with substitution
Look at the part e^{-\ln2} and how you could re-express that. Then run with it. EDIT: Haha as Bohrok said.- c.dube
- Post #16
- Forum: Calculus and Beyond Homework Help
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Solving integral with substitution
Into two fractions - you have addition on top - and then attack them both separately.- c.dube
- Post #6
- Forum: Calculus and Beyond Homework Help
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Solving integral with substitution
Hahah I don't know how your book wants you to do it but try splitting it into two integrals.- c.dube
- Post #4
- Forum: Calculus and Beyond Homework Help