1. Find the centroid of the region in the first quadrant bounded by the x-axis, the parabola y^2=x, and the line x+y=6. I would like th confirm my answer since i can't check. the answer i get is (26/11,8/11)
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anyone?
Originally Posted by Belowzero78 anyone? Definitely check your x. Mine is higher and looks much more likely. Will check the y, but yours looks reasonable. Show your work?
Ok. So, for Mx=0,when i integrate with respect to dx first, my integral comes to ( 36-12y+y^2)/2 -(y^4/2)dy with a lower limit of 0 and upper limit of 2. After i simplify i get 18-6y+(y^2/2)-(y^4/2)dy (same limits.)
The mass i found was 22/3 and My=0 = 16/3
Are you using something like and ? Then, and ... no? http://www.mathhelpforum.com/math-he...-centroid.html Centroid - Wikipedia, the free encyclopedia
why are your limits of integration different?
So, what answers did you get? If i can compare i will know if im right or wrong. Thanks! Also, im using exactly what you said.
Originally Posted by Belowzero78 why are your limits of integration different?
Originally Posted by Belowzero78 So, what answers did you get? If i can compare i will know if im right or wrong. Thanks! Also, im using exactly what you said. OK, and your x value now is ...?
yeah so then for y its from 0 to 2 and x from y^2 to 6-y.
i get like 332/15, which makes no sense. Ive checked all my work 10 times.... so i dont know lol
Originally Posted by Belowzero78 yeah so then for y its from 0 to 2 and x from y^2 to 6-y. 6 + y
Originally Posted by Belowzero78 i get like 332/15, which makes no sense. Ive checked all my work 10 times.... so i dont know lol Fine - now divide by the area! (as per the formula)
ok for x i get 166/55
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