http://calc3.askbot.com/questions/Open source question and answer forum written in Python and DjangoenCopyright Askbot, 2010-2011.Thu, 24 Jul 2014 17:43:32 -0500http://calc3.askbot.com/question/165/graphical-explanation/Can someone give a graphical explanation (pictures please) as to how you can find the answer to question 1b of todays worksheet? $$\int_0^1\int_0^{\sqrt{1-x^2}}\int_0^{\sqrt{1-x^2-y^2}}1dzdydx$$ No one in my group could really grasp what McClure was saying in class.Thu, 24 Jul 2014 10:40:31 -0500http://calc3.askbot.com/question/165/graphical-explanation/http://calc3.askbot.com/question/165/graphical-explanation/?answer=171#post-id-171 Hopefully this image helps with Dylan's great explanation. This should be the final product.Thu, 24 Jul 2014 17:43:32 -0500http://calc3.askbot.com/question/165/graphical-explanation/?answer=171#post-id-171http://calc3.askbot.com/question/165/graphical-explanation/?answer=167#post-id-167I'll try my best! Because we're just integrating over $\mathrm{d}z\mathrm{d}y\mathrm{d}x$, this means that we're only finding the volume of whatever our domain is, so theoretically if we know what the domain is and already know a formula for the volume of that shape, then we can just write down the answer without integrating anything. The first thing we can see is that $z$ is varying from $z=0$ to $z=\sqrt{1-x^2-y^2}$. Manipulating the second equation by squaring both sides and moving $x$ and $y$ terms over, we get $x^2+y^2+z^2 = 1$, which is just the equation for the unit sphere. We know $z$ can't get smaller than $0$, so we are limited to the half of the sphere where $z$ is positive; we're slicing away the half below the $z=0$ plane. Next we look at the $y$ bounds: $y$ ranges from 0 to $\sqrt{1-x^2}$. Similarly to before, if square both sides and put the $x$ and $y$ terms on one side, we get an equation for a unit *circle*; $x^2+y^2 = 1$. This is still tracing out a domain in our sphere, but now $y$ starts at $0$ and can only be positive, so we're slicing our half-circle into a quarter circle. If anyone with better Mathematica skills than I could make some pretty pictures, that would show what I'm saying better than words can. Finally, we need to look at the $x$ bounds, which are from $0$ to $1$: This is still within our little quarter sphere, but now $x$ is being limited to positive values, so we're slicing the quarter into an eighth, giving us just the part of the unit sphere in the positive octant. We know the volume of a sphere is just $\frac{4}{3} \pi r^3$, so we can just divide this by $8$ and plug in $1$ for the radius, giving a volume of $\frac{\pi}{6}$. Hopefully that might help. Thu, 24 Jul 2014 11:22:09 -0500http://calc3.askbot.com/question/165/graphical-explanation/?answer=167#post-id-167