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2014-07-31 19:11:19 -0600 answered a question Final Review Number one
Ok, you take each point one at a time and get your values for a,b,c. Starting with $ax + by + cz = 4$ and the point $( ...
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2014-07-31 06:36:23 -0600 answered a question Final Review
If you bring the 4 to the other side you get $x^2 + y^2 = 4$, which is a circle of radius 2. For this reason, I set this ...
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2014-07-30 12:25:57 -0600 answered a question quiz 3 question 1
My answer was slightly different because when you u-sub you have to change your bounds of integration. I'll start here ...
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2014-07-29 20:01:54 -0600 answered a question Review for final Exam
This is how I solved this problem: I left $\vec{p}(t)$ in terms of $t$ and changed the parameter of $\vec{q}$ to $s$. I ...
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2014-07-29 09:38:44 -0600 answered a question Line Integrals
Anonymous, the first thing I noticed was your bounds of integration. I posted a question about this last night as I was ...
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2014-07-28 16:41:58 -0600 answered a question Number 6 on Exam 2
I actually missed this one on the exam, but since then have worked on what the contour diagram would look like. This sho ...
2014-07-28 12:06:32 -0600 asked a question How do I get the bounds of integration for line integrals?
I am currently working on the homework for 16.2. Number one, for example, states: "Compute $\int\limits_Cxy^2ds$ along t ...
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2014-07-25 16:06:35 -0600 answered a question Setting up spherical integrals
I don't really know if my answer will be easily explainable but I will try to explain it as I understand it. With spheri ...
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2014-07-24 18:58:35 -0600 answered a question Quiz question
Yes, I asked this question earlier because I was struggling with the set up for Cartesian land. It was one of the ones ...
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2014-07-24 14:52:49 -0600 asked a question Setting up exponential function in Cartesian coordinates
I am hoping for some help looking at the exponential function in Cartesian land. If we are given $e^{-(x^2+y^2)}$ and ...
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2014-07-23 11:45:41 -0600 answered a question In class Problem
The way our group set it up was different, and here is why. The domain is only the top half of the unit sphere, so $\d ...
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2014-07-19 06:19:32 -0600 answered a question Post-test analysis
Just wanted to chime in here, too. I also got (-1,0) and (1,0) for the maxima and I too noticed that this did not line u ...
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