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anonymous
 4 years ago
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.
anonymous
 4 years ago
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.

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Nubeer
 4 years ago
Best ResponseYou've already chosen the best response.0hmm you can convert this in polar coordinates right?

Nubeer
 4 years ago
Best ResponseYou've already chosen the best response.0x=rcostheta y = r sin theata if i am not wrong.. limit of r would be 04 and theta limit 02pi

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah, 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: \[r^2=x^2+y^2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what do i do with the square root? also the problem says to use geometry. but i'm not sure what the height would be

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok let z = ur integrand = sqrt(16x^2y^2) what's the geometric relationship between x,y,and z?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0k 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?
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