1. ## Double Integrals

Can't find an example of this, this was in the previous final exams, and with finals only a week away, I need an explaination of a few problems from this last section of the book.

Using geometry, evaluate the double integral over D of [(32-x^2-y^2)^.5]dA over the circular disk D: x^2+y^2<=36

Are you sure you have written that correctly? For (x,y) close to the boundary of that disk, $\displaystyle x^2+ y^2$ will be close to 36 which will make $\displaystyle 32- x^2- y^2$ negative.
I am going to assume the function you want to integrate is $\displaystyle (36- x^2- y^2)^{1/2}$. With that, since the problem says "using geometry", we can interpret this integral as the volume of a the upper half of a sphere of radius 6: $\displaystyle z= (36- x^2- y^2)^{1/2}$ gives $\displaystyle z^2= 36- x^2- y^2$ so $\displaystyle x^2+ y^2+ z^2= 36$. Of course, on the xy-plane, z= 0 so we have $\displaystyle x^2+ y^2= 36$, the given circle. The volume of a sphere is $\displaystyle \frac{4}{3}\pi r^3$ so the volume of the hemisphere is $\displaystyle \frac{2}{3}\pi r^3$, which is, indeed, $\displaystyle \frac{2}{3}\pi 6^3= 144\pi$.
It's very easy to do that in polar coordinates: $\displaystyle \int_{\theta= 0}^{2\pi}\int_{r= 0}^6 \sqrt{36- r^2} rdrd\theta$. Let $\displaystyle u= 36- r^2$ and it becomes $\displaystyle \pi \int_{u= 0}^{36} u^{1/2}du$