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zhiyuan3yu5
Group Title
Using geometry (the interpretation of a double integral as a volume), evaluate the double integral
∫∫D sqrt(16−x^2−y^2) dA over the circular disk D: x^2+y^2≤16.
 one year ago
 one year ago
zhiyuan3yu5 Group Title
Using geometry (the interpretation of a double integral as a volume), evaluate the double integral ∫∫D sqrt(16−x^2−y^2) dA over the circular disk D: x^2+y^2≤16.
 one year ago
 one year ago

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nubeer Group TitleBest ResponseYou've already chosen the best response.0
hmm you can convert this in polar coordinates right?
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.0
x=rcostheta y = r sin theata if i am not wrong.. limit of r would be 04 and theta limit 02pi
 one year ago

malevolence19 Group TitleBest ResponseYou've already chosen the best response.1
Yeah, so remember: \[dA= \mathcal{J}(r,\phi)dr d \phi=r dr d \phi\] \[x=r \cos \phi; y=r \sin \phi\] \[D: \left\{ r^2 \le 16 \right\} \implies D: \left\{ r \le 4 \right\}\] Full circle implies \[0 \le \phi \le 2 \pi\] So your integral would be: \[\int\limits_0^{2 \pi} \int\limits_0^4 (16r^2)r dr d \phi\] As: \[r^2=x^2+y^2\]
 one year ago

zhiyuan3yu5 Group TitleBest ResponseYou've already chosen the best response.0
what do i do with the square root? also the problem says to use geometry. but i'm not sure what the height would be
 one year ago

cnknd Group TitleBest ResponseYou've already chosen the best response.0
ok let z = ur integrand = sqrt(16x^2y^2) what's the geometric relationship between x,y,and z?
 one year ago

cnknd Group TitleBest ResponseYou've already chosen the best response.0
k one more hint: i said z = sqrt(16x^2y^2). let me rewrite this as: x^2+y^2+z^2 = 16 (with the understanding that z can't be negative). does this equation ring a bell?
 one year ago
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