Question

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.

Answer 1

hmm you can convert this in polar coordinates right?

Answer 2

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

Answer 3

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:
\[r^2=x^2+y^2\]

Answer 4

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

Answer 5

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

Answer 6

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?