Double Integrals

**Bracketology** · November 30th, 2011, 02:38 PM

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

I know the answer is 144pi, but how do I come about this answer?

Thanks for the help, I will definitely have more questions over the next week!

**HallsofIvy** · December 1st, 2011, 03:54 AM

Are you sure you have written that correctly? For (x,y) close to the boundary of that disk, $x^2+ y^2$ will be close to 36 which will make $32- x^2- y^2$ negative.

I am going to assume the function you want to integrate is $(36- x^2- y^2)^{1/2}$ . 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: $z= (36- x^2- y^2)^{1/2}$ gives $z^2= 36- x^2- y^2$ so $x^2+ y^2+ z^2= 36$ . Of course, on the xy-plane, z= 0 so we have $x^2+ y^2= 36$ , the given circle. The volume of a sphere is $\frac{4}{3}\pi r^3$ so the volume of the hemisphere is $\frac{2}{3}\pi r^3$ , which is, indeed, $\frac{2}{3}\pi 6^3= 144\pi$ .

It's very easy to do that in polar coordinates: $\int_{\theta= 0}^{2\pi}\int_{r= 0}^6 \sqrt{36- r^2} rdrd\theta$ . Let $u= 36- r^2$ and it becomes $\pi \int_{u= 0}^{36} u^{1/2}du$

Math Help Forum: Double Integrals

LinkBack

Thread Tools

Display