user73064
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 Apr18 comment Center of Mass double integral But it wouldn't be wrong to put 0 to 2sinθ as a bound? Apr18 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. Apr18 awarded Commentator Apr18 comment Triple Integral of $2y+x^2$ No problem, that was throwing me off but I get it now, thanks a lot. Apr18 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)? Apr18 awarded Student Apr18 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 Apr18 awarded Editor Apr18 comment Center of Mass double integral Sorry, forgot the (-2y), you have to complete the square so the radius becomes 1. Apr18 revised Center of Mass double integral added 3 characters in body Apr18 asked Center of Mass double integral Apr18 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. Apr18 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? Apr18 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)? Apr18 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? Apr18 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? Apr17 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. Apr17 asked Triple Integral of $2y+x^2$ Apr17 asked Triple Integral of $1+e^x\cos(y)$ bounded by planes Apr17 asked 3-D Absolute Max/Min over closed&bounded region