Find the Centroid
- Thread starter Jason76
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What step did they show you in the book for finding the center of mass for a specified region? How far have you gotten in applying the processes they taught you? You found the area (standing in for the mass), found the moments, and... then what?
Please be complete. Thank you!
Find the centroid of the region bounded by the curve x=4-y^2 and the y-axis
You might find this informative:
http://www.youtube.com/watch?v=NYUyHj3c1Xg
And for a more complete/rigorous explanation:
http://www.youtube.com/watch?v=ntvJbZEd2eU
D
Deleted member 4993
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That is not necessary.
The area is symmetric about x-axis →YG = 0
To find XG:
\(\displaystyle \displaystyle{X_G \ = \ \dfrac{\displaystyle \int_0^2x^2dy}{2\displaystyle \int_0^2xdy} }\)
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Then why did you raise the issue of negative areas?
Then why did you ask for the "first step"? Where are you now stuck?
As far as negative areas, at first I thought it had something to do with the problem, but now I see otherwise.
Now, the first step would be putting the equation in a \(\displaystyle y = \) form.
\(\displaystyle x=4-y^{2}\)
\(\displaystyle x - 4 = -y^{2}\)
\(\displaystyle -y^{2} = x - 4\)
\(\displaystyle y^{2} = -x + 4\)
\(\displaystyle y = \pm \sqrt{-x + 4}\)
But also is the problem of not knowing the interval. I believe to find that, you have to set two equations equal to each other, and solve for x.
Now, the first step would be putting the equation in a \(\displaystyle y = \) form.
\(\displaystyle x=4-y^{2}\)
\(\displaystyle x - 4 = -y^{2}\)
\(\displaystyle -y^{2} = x - 4\)
\(\displaystyle y^{2} = -x + 4\)
\(\displaystyle y = \pm \sqrt{-x + 4}\)
But also is the problem of not knowing the interval. I believe to find that, you have to set two equations equal to each other, and solve for x.
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D
Deleted member 4993
Guest
As far as negative areas, at first I thought it had something to do with the problem, but now I see otherwise.
Now, the first step would be putting the equation in a \(\displaystyle y = \) form.
\(\displaystyle x=4-y^{2}\)
\(\displaystyle x - 4 = -y^{2}\)
\(\displaystyle -y^{2} = x - 4\)
\(\displaystyle y^{2} = -x + 4\)
\(\displaystyle y = \pm \sqrt{-x + 4}\)
But also is the problem of not knowing the interval. I believe to find that, you have to set two equations equal to each other, and solve for x.
In these problems, first sketch the function/s (here I speak of functions loosely).
That is what the "Graphic Calculators" were manufactured for - or use wolframalfa.
Then ask questions......
The region is bounded by the y-axis. This means that x = 0 is one endpoint of the interval.
What's the x-intercept, for the two equations shown in the quote above?
That x-intercept is the other endpoint. :cool:
\(\displaystyle x=4-y^{2}\)
\(\displaystyle x - 4 = -y^{2}\)
\(\displaystyle -y^{2} = x - 4\)
\(\displaystyle y^{2} = -x + 4\)
\(\displaystyle y = \pm \sqrt{-x + 4}\)
\(\displaystyle y = -(x + 2)\)
\(\displaystyle 0 = -x - 2\)
\(\displaystyle -x = 2\)
\(\displaystyle x = -2\)
\(\displaystyle y = x + 2\)
\(\displaystyle 0 = x + 2\)
\(\displaystyle x = -2\)
\(\displaystyle x - 4 = -y^{2}\)
\(\displaystyle -y^{2} = x - 4\)
\(\displaystyle y^{2} = -x + 4\)
\(\displaystyle y = \pm \sqrt{-x + 4}\)
\(\displaystyle y = -(x + 2)\)
\(\displaystyle 0 = -x - 2\)
\(\displaystyle -x = 2\)
\(\displaystyle x = -2\)
\(\displaystyle y = x + 2\)
\(\displaystyle 0 = x + 2\)
\(\displaystyle x = -2\)
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D
Deleted member 4993
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\(\displaystyle x=4-y^{2}\)
\(\displaystyle x - 4 = -y^{2}\)
\(\displaystyle -y^{2} = x - 4\)
\(\displaystyle y^{2} = -x + 4\)
\(\displaystyle y = \pm \sqrt{-x + 4}\)
\(\displaystyle y = -(x + 2)\) ..............How did you get that??
\(\displaystyle 0 = -x - 2\)
\(\displaystyle -x = 2\)
\(\displaystyle x = -2\)
\(\displaystyle y = x + 2\)
\(\displaystyle 0 = x + 2\)
\(\displaystyle x = -2\)
Can you please tell us - what you are trying to do???
Find the x intercept
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y = sqrt(4 - x)
x-intercepts occur where y = 0
0 = sqrt(4 - x)
There exists only one quantity whose square root is zero. That quantity is zero, itself.
Therefore, we know that the blue equation is true only when the radicand is zero.
4 - x = 0
The x-intercept must be 4.
x-intercepts occur where y = 0
0 = sqrt(4 - x)
There exists only one quantity whose square root is zero. That quantity is zero, itself.
Therefore, we know that the blue equation is true only when the radicand is zero.
4 - x = 0
The x-intercept must be 4.
D
Deleted member 4993
Guest
Which could have been (should have been) found from the original equation x = 4 - y2 ..... that would be too simple for Jason.
For sure.
Also, it could have been easily seen by graphing, which is the very first suggestion that you posted!
Close to 1,100 posts since October 2012, and the only thing I'm certain of is that Jason insists on "my way or the highway" approach to studying math.
When are you going to learn algebra, Jason?