meta user
network profile
| bio | website | gulshansingh.com |
|---|---|---|
| location | ||
| age | ||
| visits | member for | 1 year, 8 months |
| seen | May 13 at 23:25 | |
| stats | profile views | 27 |
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May 13 |
awarded | Caucus |
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Feb 21 |
awarded | Popular Question |
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Dec 3 |
awarded | Teacher |
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Aug 31 |
awarded | Yearling |
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May 1 |
awarded | Nice Question |
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Apr 12 |
accepted | Surface Integral of $xy$ over the surface $y^2+z^2=36$ |
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Apr 12 |
comment |
Evaluating the integral $\int x\,dV$ where $V$ is the region bounded the surface $x^2+y^2+z^2=1$ and the planes $x = 0$, $y = x$, $z=0$ @ZevChonoles I think you mean phi goes from 0 to pi/2. Theta should be between 0 and pi/4 correct? |
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Apr 10 |
comment |
Surface Integral of $xy$ over the surface $y^2+z^2=36$ So first of all, your upper bound for x is always negative, so it should really be a lower bound. But x should never go below zero because it's in the first quadrant... |
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Apr 10 |
awarded | Critic |
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Apr 10 |
comment |
Surface Integral of $xy$ over the surface $y^2+z^2=36$ Oh, I think I see. The maximum value of y is when x=0. Then $0+y^2=25$, so $6\cos(\theta)=5$. Thanks for the answer, I'll get back to you after I complete the problem. |
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Apr 10 |
awarded | Commentator |
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Apr 10 |
comment |
Surface Integral of $xy$ over the surface $y^2+z^2=36$ I might be missing something obvious here, but how did you figure out that $\cos(\theta)\le \frac{5}{6}$? |
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Apr 10 |
revised |
Surface Integral of $xy$ over the surface $y^2+z^2=36$ added 4 characters in body |
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Apr 10 |
asked | Surface Integral of $xy$ over the surface $y^2+z^2=36$ |
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Mar 20 |
comment |
Proof that $\pi$ is rational Haha, it's meant to be a joke, don't take it seriously :) |
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Mar 15 |
comment |
Proof that $\pi$ is rational @PeterT.off The link is in the post. |
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Mar 15 |
accepted | Proof that $\pi$ is rational |
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Mar 15 |
comment |
Proof that $\pi$ is rational Ahh, I completely forgot that induction only applied for the natural numbers! Thanks! |
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Mar 15 |
asked | Proof that $\pi$ is rational |
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Mar 9 |
accepted | Prove $\binom{n}{2k+1}=\sum_{i=1}^n{\binom{i-1}{k}\binom{n-i}{k}}$ |
