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## zhiyuan3yu5 2 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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1. nubeer

hmm you can convert this in polar coordinates right?

2. nubeer

x=rcostheta y = r sin theata if i am not wrong.. limit of r would be 0-4 and theta limit 0-2pi

3. malevolence19

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: $r^2=x^2+y^2$

4. zhiyuan3yu5

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

5. cnknd

ok let z = ur integrand = sqrt(16-x^2-y^2) what's the geometric relationship between x,y,and z?

6. cnknd

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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