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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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hmm you can convert this in polar coordinates right?
x=rcostheta y = r sin theata if i am not wrong.. limit of r would be 0-4 and theta limit 0-2pi
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: \[\int\limits_0^{2 \pi} \int\limits_0^4 (16-r^2)r dr d \phi\] As: \[r^2=x^2+y^2\]

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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
ok let z = ur integrand = sqrt(16-x^2-y^2) what's the geometric relationship between x,y,and z?
k one more hint: i said z = sqrt(16-x^2-y^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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