http://calc3.askbot.com/questions/Open source question and answer forum written in Python and DjangoenCopyright Askbot, 2010-2011.Fri, 25 Jul 2014 13:40:06 -0500http://calc3.askbot.com/question/179/question-2-on-quiz/I don't remember the exact wording of question #2 on the quiz but I think it asked what the area above the curve $z=-\sqrt{x^2+y^2}$ and below $z=\sqrt{x^2+y^2}$ inside the cylinder $x^2+y^2=1$; evaluate $\int\int\int (z) dV$. My answer came out to be zero but I don't think that is correct. If anybody understands this problem or has any ideas, please post them.Fri, 25 Jul 2014 09:05:21 -0500http://calc3.askbot.com/question/179/question-2-on-quiz/http://calc3.askbot.com/question/179/question-2-on-quiz/?answer=182#post-id-182I agree with Gear Junky. I believe that zero is the correct answer. I would like to offer a geometric perspective on this problem that does not require calculus (but can be easily verified using calculus by following the steps that Gear Junky took). Sometimes you can save a lot of work—and panic—by thinking visually and this is a good example of such a problem. I have provided an image of the graph of the equations in question for reference. We want to find the area within the blue cylinder that is not taken up by the cones.  The side view of this graph looks like this (this is the $xz$ plane):  You will notice that $z$ ranges from $-r$ to $r$, as Gear Junky said. Both this graph and the integrand ($z$) has symmetry about the $xy$ plane, and for every value above the $xy$ plane (positive $z$), there is an equal but opposite value below the plane. This means that the integral cancels itself and gives us our scary yet beautiful answer of $0$. Here is a view that illustrates the symmetry: Fri, 25 Jul 2014 13:40:06 -0500http://calc3.askbot.com/question/179/question-2-on-quiz/?answer=182#post-id-182http://calc3.askbot.com/question/179/question-2-on-quiz/?answer=181#post-id-181I ended up with $\pi$ as my answer. Although since talking to people, i'm not sure if that is correct. What I did was... $$\int _0 ^{2\pi} \int _0 ^1 \int _{-r^2} ^{r^2}\quad z \quad r\delta z \delta r \delta \theta $$ $$= \int _0 ^{2\pi} \int _0 ^1 \quad \frac 12 z^2 r \delta r \delta\theta\bigg| _{-r}^r $$ $$= \int_0^{2\pi} \int _0 ^1 \frac 12 r^2 - \frac 12 r^2 r\delta r \delta \theta$$ $$= \int _0 ^{2\pi} \int _0 ^1 r \delta r \delta\theta$$ $$ = \int _0 ^{2\pi} \frac 12 r^2 \delta \theta\bigg |_0 ^1$$ $$ = \int _0 ^{2\pi} \frac 12 \delta \theta $$ $$ = \frac 12 * 2\pi$$ $$=\pi$$ I know in class when we were talking we were saying that the volume function = 1, and that we could solve out for it and find that out, but I don't know if this is right or not. Or if this will even help at all, but this is how I went about it!Fri, 25 Jul 2014 10:18:52 -0500http://calc3.askbot.com/question/179/question-2-on-quiz/?answer=181#post-id-181http://calc3.askbot.com/question/179/question-2-on-quiz/?answer=180#post-id-180Actually that's the answer that I got as well. I thought about it for a while and it sorta makes sense to me when I drew out the graphs on my paper. When I followed through the problem I got:$$\int_0^{2\pi}\int_0^1\int_{(-r)}^{(r)}z\delta zr\delta r\delta \theta$$ Which is equal to saying:$$2\pi\int_0^1\int_{(-r)}^{(r)}z\delta z r \delta r$$ Following through the integration you get:$$2\pi\int_0^1\frac{z^2}{2}\biggr|_{(-r)}^{(r)}r\delta r$$ Which is equal to:$$\pi\int_0^1z^2\biggr|_{(-r)}^{(r)}r\delta r$$ Ending up with:$$\pi\int_0^1((r)^2-(-r)^2)r\delta r$$ Which is also:$$\pi\int_0^1(0)r\delta r$$ Which is also: $$\Huge-$$$$\Huge=$$$$\Huge(0)$$ Which are the graphs of $(r^2)$ and $(-r)^2$ and from examination it can be noticed that you are subtracting the function from itself, ending up with $\biggr.{(0)}$ and then following through with the rest of the integral which is multiplied by $\biggr.{(0)}$, ending up with $\biggr.{(0)}$.Fri, 25 Jul 2014 09:34:30 -0500http://calc3.askbot.com/question/179/question-2-on-quiz/?answer=180#post-id-180