# Double Integral over general region

1. Nov 18, 2012

### Math11

1. The problem statement, all variables and given/known data

Find the integral using a geometric argument.

∫∫D√(16 - x2 - y2)dA
over the region D where D = {(x,y) : x2 + y2 ≤ 16}

By the way, the subscript D next to the integral refers to the region over which the function is integrated.

2. Relevant equations
∫∫f(x,y)dxdy = ∫∫f(x,y)dydx

Don't think it's a polar coordinate question but here is the equation for a double integral in polar coordinates anyway:
∫∫Df(x,y)r dA

3. The attempt at a solution
Attempted to solve the equation using a double integral in polar coordinates with:
0 ≤ r ≤ 4
0 ≤ θ ≤ pi
got 64π/3 but I'm not certain whether it is correct. I'm not sure whether the graph approaches the x-y plane asymptotically or not. The equation for f(x,y) = √(16 - x2 - y2) describes what seems like the positive z half of a sphere

Edit: Made a second attempt. Assumed the shape to be the top half of a sphere so the equation became:
(0.5)∫∫D(16 - x2 - y2)dA
Where x ranges from 0 to 4 and y ranges from -√(16-x2) to √(16-x2)

Any help would be much appreciated.
Thank you

Last edited: Nov 18, 2012
2. Nov 18, 2012

### Zondrina

Draw your region, its a circle with radius 4 with a center at (0,0).

There are four curves you need to consider when setting up your integral. You can find these curves by solving x2 + y2 = 16.