# The volume of a solid bounded by two cylinders and two cones.

Is it possible to calculate the following with a triple integral, and if so, how?

The volume of a solid bounded by two cylinders: radius = 1 and radius = 2, AND by two cones: $\varphi$ = $\pi$/6 and $\varphi$ = $\pi$/3.

I don't believe it's possible using only spherical coordinates or only cylindrical coordinates, for the following reason: (it wouldn't allow me to post the image directly as this is my first post here).

2kreate.com/integral.png

When you go to attempt to pick a range for $\varphi$ you will always have some unaccounted for volume as is represented by the (poorly drawn) image above.

Could you complete the integral by making a change in variables from spherical to cylindrical halfway through the formula? Or would you be stuck computing two separate triple integrals (the second being to account for the volume represented in the image above)?

Any help or insight would be greatly appreciated.

You said:

When you go to attempt to pick a range for $\varphi$ you will always have some unaccounted for volume as is represented by the (poorly drawn) image above.

But we can try a trick. Since we have to find only the volume of the solid between cylinders and cones. We can find the volume in the first octant , then with the help of symmetry , we'll be able to find the total volume.

Now , see the exact region drawn below...

I think every thing is clear now.

So,

$$V=8\times \int_{\theta = 0}^{\pi /2}\int_{\varphi=\pi/6}^{\pi/3}\int_{\rho=\operatorname{cosec} \varphi}^{2 \operatorname{cosec} \varphi} \rho^2 \sin\varphi \ d\rho \, d\varphi\, d\theta$$

EDIT:

$r=1$ is the cylinder $x^2+y^2=1$ can be represented as $\rho=\operatorname{cosec}\varphi$ in spherical co-ordinate system.

• Could explain a little bit more why the $r-$coordinate in spherical coordinates goes from $1$ to $2$? Thanks ! – dem0nakos Apr 12 '18 at 7:55
• math.stackexchange.com/a/2729478/516174 , see here!! – освящение Apr 12 '18 at 7:56
• Ok i'll check it . I asked because i thought that $r$ must be depended from the angle $\phi$. – dem0nakos Apr 12 '18 at 7:58
• yes! you are right , there was a confusion between $r$ and $\rho$. Now it is clear. – освящение Apr 12 '18 at 8:12
• Nice ! now we are getting the same result ! – dem0nakos Apr 12 '18 at 8:19

I think is simpler to describe your region in cylindrical coordinates .

The cone with $\phi = \frac{\pi}{6}$ has equation $z = \sqrt{3(x^2+y^2)}$ and the other one $z = \dfrac{1}{\sqrt{3}}\sqrt{x^2+y^2}.$

So if we fix a point $(x,y)$ in the $xy$-plane and move straight up from that point we will 'hit' first the cone with $\phi = \dfrac{\pi}{3}$ and then the cone with $\phi = \dfrac{\pi}{6}.$ So our $z$ coordinate must be all the time between $$\dfrac{1}{\sqrt{3}}\sqrt{x^2+y^2} \leq z \leq \sqrt{3(x^2+y^2)}.$$

So we are done with $z-$coordinate.

Now the bounds for $(x,y)$ must be between the two circles of radius 1 and 2 , so $$1\leq \sqrt{x^2+y^2} \leq 2.$$

So our region in cartesian coordinates must be $$W = \biggl\{(x,y,z) : 1\leq \sqrt{x^2+y^2} \leq 2 \text{ and } \dfrac{1}{\sqrt{3}}\sqrt{x^2+y^2} \leq z \leq \sqrt{3(x^2+y^2)} \biggr\}.$$

Now converting in cylindrical coordinates our regions becomes

$$W = \biggl\{(r,\theta,z) : 1\leq r \leq 2 \ , 0\leq \theta \leq 2\pi \text{ and } \dfrac{1}{\sqrt{3}}r \leq z \leq \sqrt{3}r \biggr\}.$$

$$\mathcal{V}(W) = \int_{\theta = 0}^{2\pi}\int_{r=1}^{2}\int_{\frac{1}{\sqrt{3}}r}^{\sqrt{3}r} r \ dz\, dr\, d \theta$$
Ps.The result must be $2\times\mathcal{V}(W)$ since the region W is for $z>0$ but because the whole region is symmetrical to the $xy-$plane the case $z<0$ gives also volume equal to $\mathcal{V}(W).$
• Thanks ! i just noticed that i didnt involve the case $z<0$.So the result must be doubled ! – dem0nakos Apr 12 '18 at 7:56