# Finding limits of a volume triple integral in cylindrical coordinates

Find the volume between the cone $z=\sqrt{x^2+y^2}$ and the sphere $x^2+y^2+z^2=1$ that lies in the first octant (i.e., $x>0$, $y>0$, $z>0$) using cylindrical coordinates.

It is obvious that limits for $\theta$ are $[0,\pi/2]$. Can someone tell me a clear way to find out what the limits of $z$ and $r$ are?

Hint What are the equations of the cone and sphere in cylindrical coordinates?

• Perhaps the downvoter can explain? The hint seems both straightforward and helpful to me. – Travis Apr 9 '15 at 13:04
• z=r and z=sqrt(1-r^2) right? and dw ill up it – snowman Apr 9 '15 at 13:06
• Yes, that's right, and we can see that by substituting $r^2 = x^2 + y^2$ in the equations for the cone and cylinder: We get respectively $z^2 = r^2$ and $r^2 + z^2 = 1$; rearranging (and taking some care with signs) gives exactly your formulas. – Travis Apr 9 '15 at 13:19
• is this type of technique always the best approach to get the limits? Just sub it into the main equations and see what happens? I always try to sketch it and look what the limits can be but it is backfiring so much recently! – snowman Apr 9 '15 at 13:25
• It depends: In this case it's a natural thing to do, since the expressions $x^2 + y^2$, which we know can be written as $r^2$, actually occur in the defining equations of the boundary surfaces. This is not always nearly so straightfoward, though (sometimes this indicates that the coordinates might not be the best ones for evaluating the integral). In my experience it's valuable to have some facility for producing new limits both geometrically (i.e., with sketching) and algebraically. – Travis Apr 9 '15 at 13:32

Note that the region in the $xy$-plane is a circle and parametrized as $x=\frac{1}{\sqrt{2}}\cos \theta,\, y=\frac{1}{\sqrt{2}}\sin \theta$ and the volume is given by

$$V = \int_{0}^{\pi/2} \int_{0}^{1/ \sqrt{2}} \int_{r}^{\sqrt{1-r^2}}dz\,rdrd\theta$$

• why isn't one of the limits of z, 0? – snowman Apr 9 '15 at 13:07
• @snowman: Because $z$ is bounded by the cone from the bottom and from above by the sphere $\sqrt{x^2+y^2} \leq z \leq \sqrt{1-(x^2+y^2)}$. – science Apr 9 '15 at 13:08
• but the cone reaches the origin at z=0 though... (vertex) – snowman Apr 9 '15 at 13:09
• This is not how you look at it? You should look at the vertical strip and its bounds. – science Apr 9 '15 at 13:11
• wait volume we are trying to find is where exactly? I know its the first octant but it seems to me its like quarter of a cone with a circular top. – snowman Apr 9 '15 at 13:14