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seen Apr 18 '13 at 17:58

Apr
18
comment Center of Mass double integral
But it wouldn't be wrong to put 0 to 2sinθ as a bound?
Apr
18
comment Center of Mass double integral
Okay, I still don't get how you can just get rid of the 2 in r=2sinθ? It would result in a different answer so I don't see how it's unimportant.
Apr
18
awarded  Commentator
Apr
18
comment Triple Integral of $2y+x^2$
No problem, that was throwing me off but I get it now, thanks a lot.
Apr
18
comment Center of Mass double integral
I originally meant the equation for the mass because once you have that it's very easy to go from mass to center of mass...I don't understand why you went from (0 to pi) and(0 to sinθ)...if the lamina is restricted to the first quadrant, shouldn't it be (0 to pi/2)?
Apr
18
awarded  Student
Apr
18
comment Triple Integral of $2y+x^2$
But by that logic, x>√6 should make 2y+x^2>=0 because the minimum value of y is -3, and plugging that in gives -6+x^2>=0 which gives x>=√6
Apr
18
awarded  Editor
Apr
18
comment Center of Mass double integral
Sorry, forgot the (-2y), you have to complete the square so the radius becomes 1.
Apr
18
revised Center of Mass double integral
added 3 characters in body
Apr
18
asked Center of Mass double integral
Apr
18
comment 3-D Absolute Max/Min over closed&bounded region
Okay, so by factoring the cosines, I got the critical points of the circle occur when 2(cosθ-1)sinθ=0, so when cosθ=1 or when sinθ=0, which is pi and 0 for both, so the only points I test on the circle are (1,0) and (-1,0)? If so then I guess the min and max of the boundary are +/- 8 (not on the circle), which I sort of find strange because the question asked for the answer rounded to the nearest hundredths.
Apr
18
comment Triple Integral of $2y+x^2$
Sorry for needing to have this spoon-fed, but where does the 2root3 come from? I know that's x^2=12 but when would x^2 ever have to equal 12?
Apr
18
comment 3-D Absolute Max/Min over closed&bounded region
Ah, you plug in the 1...thanks for the help but I have one more question: I said the critical point of the function is (1,0) but I've been thinking about it and I don't think that's correct; the first partial wrt y is 2y, so y has to equal 0, and the first partial wrt x is 2, so does that make the critical point (2,0)?
Apr
18
comment Triple Integral of $2y+x^2$
Okay, so 2y+x^2 has to be greater than or equal to 0, so y has to be greater than or equal -x^2/2 so the y bounds should be changed to (-x^2/2,2), correct?
Apr
18
comment 3-D Absolute Max/Min over closed&bounded region
But your function doesn't have r's in it like it should. Shouldn't it be 2rcosθ+r^2sinθ-2?
Apr
17
comment Triple Integral of $2y+x^2$
So the y-bounds should be changed to [0,2]? Thanks for the responses guys, I completely forgot volume was just the triple integral of 1.
Apr
17
asked Triple Integral of $2y+x^2$
Apr
17
asked Triple Integral of $1+e^x\cos(y)$ bounded by planes
Apr
17
asked 3-D Absolute Max/Min over closed&bounded region